RD Sharma XII Vol 2 2019 Solutions for Class 12 Science Math Chapter 5 Scalar Or Dot Product are provided here with simple step-by-step explanations. These solutions for Scalar Or Dot Product are extremely popular among class 12 Science students for Math Scalar Or Dot Product Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 2 2019 Book of class 12 Science Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 2 2019 Solutions. All RD Sharma XII Vol 2 2019 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 24.29:

Question 1:

Find a ·b when
(i) a =i^-2j^+k^ and b =4i^-4j^+7k^

(ii) a =j^+2k^ and b =2i^+k^

(iii) a =j^-k^ and b =2i^+3j^-2k^

Answer:

i We havea=i-2j+k and b=4i-j+7kab=i-2j+k.4i-j+7k      =14+-2-4+17      =4+8+7      =19

ii We havea=j^+2k^=0 i+j+2k and b=2i+k=2i+0j+kab=0 i+j+2k 2i+0j+k      =02+10+21      =0+0+2      =2

iii We havea=j-k=0 i+j-k and b=2i+3j-2kab=0 i+j-k 2i+3j-2k      =02+13+-1-2      =3+2      =5



Page No 24.30:

Question 2:

For what value of λ are the vectors a and b perpendicular to each other if
(i) a =λi^+2j^+k^ and b =4i^-9j^+2k^

(ii) a =λi^+2j^+k^ and b =5i^-9j^+2k^

(iii) a =2i^+3j^+4k^ and b =3i^+2j^-λk^

(iv) a =λi^+3j^+2k^ and b =i^-j^+3k^

Answer:

i If the vectors a and b are perpendicular to each other, thena. b=0λi+2j+k . 4i-9j+2k=04λ-18+2=04λ-16=04λ=16λ=4

ii If the vectors a and b are perpendicular to each other, thena. b=0λi+2j+k . 5i-9j+2k=05λ-18+2=05λ-16=05λ=16λ=165

iii If the vectors a and b are perpendicular to each other, thena. b=02i+3j+4k . 3i+2j-λk=06+6-4λ=012-4λ=04λ=12λ=3

iv If the vectors a and b are perpendicular to each other, thena. b=0λi+3j+2k . i-1j+3k=0λ-3+6=0λ+3=0λ=-3

Page No 24.30:

Question 3:

If a and b are two vectors such that  a =4,  b =3 and a ·b =6, find the angle between a and b .

Answer:

Let θ be the angle between a and b.Given thata . b=6a b cos θ=643 cos θ=612 cos θ=6cos θ=612=12θ=cos-1 12=π3

Page No 24.30:

Question 4:

 If a =i^-j^ and b =-j^+2k^, find a -2b ·a +b .

Answer:

We havea=i-j and b=-j+2ka-2b=i-j -2 -j+2k=i-j+2j-4k=i+j-4ka+b=i-j-j+2k=i-2j+2ka-2b . a+b=i+j-4k . i-2j+2k=1-2-8=-9

Page No 24.30:

Question 5:

Find the angle between the vectors a and b , where
(i) a =i^-j^ and b =j^+k^

(ii) a =3i^-2j^-6k^ and b =4i^-j^+8k^

(iii) a =2i^-j^+2k^ and b =4i^+4j^-2k^

(iv) a =2i^-3j^+k^ and b =i^+j^-2k^

(v) a =i^+2j^-k^, b =i^-j^+k^

Answer:

i Let θ be the angle between a and b.a=12+-12=2b=12+12=2a . b=0-1+0=-1cos θ=a . ba b=-122=-12θ=cos-1 -12=2π3

ii Let θ be the angle between a and b.a=32+-22+-62=49=7b=42+-12+82=81=9a . b=12+2-48=-34cos θ=a . ba b=-3479=-3463θ=cos-1 -3463

iii Let θ be the angle between a and b.a=22+-12+ 22=9=3b=42+42+-22=36=6a . b=8-4-4=0cos θ=a . ba b=036=0θ=cos-1 0=π2

iv Let θ be the angle between a and b.a=22+-32+ 12=14b=12+12+-22=6a . b=2-3-2=-3cos θ=a . ba b=-3146=-384θ=cos-1 -384

v Let θ be the angle between a and b.a=12+22+ -12=6b=12+-12+12=3a . b=1-2-1=-2cos θ=a . ba b=-263=-218=-2×22×9=-23θ=cos-1 -23

Page No 24.30:

Question 6:

Find the angles which the vector a =i^-j^+2k^ makes with the coordinate axes.

Answer:

Let θ1 be the angle between a and  x-axis.a=12+-12+22=4=2b=i(Because i is the unit vector along x-axis)b=12=1=1a . b=1+0+0=1cos θ1=a . ba b=121=12θ1=cos-1 12=π3Let θ2 be the angle between a and  y-axis.a=12+-12+22=4=2b=j(Because j is the unit vector along y-axis)b=12=1=1a . b=0-1+0=-1cos θ2=a . ba b=-121=-12θ2=cos-1 -12=2π3Let θ3 be the angle between a and  z-axis.a=12+-12+22=4=2b=k(Because k is the unit vector along z-axis)b=12=1=1a . b=0+0+2=2cos θ=a . ba b=221=12θ=cos-1 12=π4

Page No 24.30:

Question 7:

(i) Dot product of a vector with i^+j^-3k^, i^+3j^-2k^ and 2i^+j^+4k^ are 0, 5 and 8 respectively. Find the vector.
(ii) Dot products of a vector with vectors i^-j^+k^, 2i^+j^-3k^ and i^+j^+k^ are respectively 4, 0 and 2. Find the vector.

Answer:

i Let ai+bj+ck be the required vector.Given thatai+bj+ck.i+j-3k=0a+b-3c=0 ... 1ai+bj+ck.i+3j-2k=5a+3b-2c=5 ... 2ai+bj+ck.2i+j+4k=52a+b+4c=8 ... 3Solving (1), (2) and (3), we geta=1, b=2, c=1So, ai+bj+ck=i+2j+k

ii Let ai+bj+ck be the required vector.Given thatai+bj+ck.i-j+k=4a-b+c=4 ... 1ai+bj+ck.2i+j-3k=02a+b-3c=0 ... 2ai+bj+ck.i+j+k=2a+b+c=2 ... 3Solving (1), (2) and (3), we geta=2; b=-1; c=1So, ai+bj+ck=2i-j+k

Page No 24.30:

Question 8:

If a ^ and b ^are unit vectors inclined at an angle θ, prove that
(i) cosθ2=12 a ^+ b ^

(ii) tanθ2= a ^- b ^  a ^+b ^

Answer:

Given that a and b are unit vectors.So, a^=1, b^=1We havea^+b^2=a^2+b^2+2 a^.b^         =1+1+2 a^ b^ cos θ         =2+2cos θcosθ=a^+b^2-22  ... 1a^-b2=a^2+b^2-2 a^.b^         =1+1-2 a^ b^ cos θ         =2-2cos θcosθ=2-a^-b^22  ... 2i Now,cos θ2=1+cos θ2=1+a^+b^2-22 2 From 1=2+a^+b^2-24=a^+b^24=12a^+b^ii sin θ2=1-cos θ2=1-2-a^-b^22 2[From (2)]=2+a^-b^2-24=a^-b^24=12a^-b^Now,tan θ2=sin θ2cos θ2=12a^-b^12a^+b^ =a^-b^a^+b^

Page No 24.30:

Question 9:

If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is 3.

Answer:

Given that ab and a^+b^ are unit vectors.So, a^=1, b^=1 and a^+b^=1We have   a^+b^2+a^-b^2=2a^2+b^2 1+a^-b^2=21+1 1+a^-b^2=4  a ^-b^2=3 a^-b^=3         

Page No 24.30:

Question 10:

If a, b, c are three mutually perpendicular unit vectors, then prove that  a +b +c =3.

Answer:

Given that ab and c are unit vectors.So, a=1, b=1 and c=1Since they are mutually perpendicular,a.b=b.c=c.a=0Now,a+b+c2 =a2+b2+c2+2 a. b+2 b.c+2 c.a                  =1+1+1+0+0+0                  =3 a+b+c=3

Page No 24.30:

Question 11:

If  a +b =60,  a -b =40 and  b =46, find  a 

Answer:

We know that a+b2+a-b2=2a2+b2 602+402=2a2+462     (Given) 3600+1600=2a2+4232 968=2a2 a2=484 a=22

Page No 24.30:

Question 12:

Show that the vector i^+j^+k^ is equally inclined to the coordinate axes.

Answer:

Let θ1 be the angle between a and  x-axis.a=12+12+12=3b=i(Because i is the unit vector along x-axis)b=12=1=1a . b=1+0+0=1cos θ1=a . ba b=131=13θ1=cos-1 13...1Let θ2 be the angle between a and  y-axis.a=12+12+12=3b=j(Because j is the unit vector along y-axis)b=12=1=1a . b=0+1+0=1cos θ2=a . ba b=131=13θ2=cos-1 13...2Let θ3 be the angle between a and  z-axis.a=12+12+12=3b=k(Because k is the unit vector along z-axis)b=12=1=1a . b=0+0+1=1cos θ=a . ba b=131=13θ=cos-1 13...3From (1), (2) and (3), the given vector is equally inclined to the coordinate axes.

Page No 24.30:

Question 13:

Show that the vectors a =172i^+3j^+6k^, b =173i^-6j^+2k^, c =176i^+2j^-3k^ are mutually perpendicular unit vectors.

Answer:

We havea=1722+32+62=1749=77=1b=1732+-62+22=1749=77=1c=1762+22+-32=1749=77=1Anda. b=172i^+3j^+6k^. 173i^-6j^+2k^=1496-18+12=0b. c=173i^-6j^+2k^. 176i^+2j^-3k^=14918-12-6=0c. a=176i^+2j^-3k^.  172i^+3j^+6k^=14912+6-18=0So, a=b=c=1 and a. b=b. c=c. a=0So, the given vectors are mutually perpendicular unit vectors.

Page No 24.30:

Question 14:

For any two vectors a and b , show that  a +b · a -b =0 a = b .

Answer:

We havea+b .a-b=0a2-b2=0a2=b2a=b

Page No 24.30:

Question 15:

If a=2i^-j^+k^b=i^+j^-2k^ and c=i^+3j^-k^, find λ such that a is perpendicular to λb+c.                 [NCERT EXEMPLAR]

Answer:


The given vectors are a=2i^-j^+k^b=i^+j^-2k^ and c=i^+3j^-k^.

Now,

λb+c=λi^+j^-2k^+i^+3j^-k^=λ+1i^+λ+3j^-2λ+1k^

It is given that

aλb+ca.λb+c=02i^-j^+k^.λ+1i^+λ+3j^-2λ+1k^=02λ+1-λ+3-2λ+1=02λ+2-λ-3-2λ-1=0λ=-2

Thus, the value of λ is −2.

Page No 24.30:

Question 16:

If p =5i^+λj^-3k^ and q =i^+3j^-5k^, then find the value of λ, so that p +q  and p -q are perpendicular vectors.

Answer:

Given thatp=5i^+λj^-3k^and q=i^+3j^-5k^p+q=5i^+λj^-3k^+i^+3j^-5k^=6i^+λ+3j^-8k^p-q=5i^+λj^-3k^-i^+3j^-5k^=4i^+λ-3j^+2k^Given that p+q is orthogonal to p-q.p+q. p-q=06i^+λ+3j^-8k^.4i^+λ-3j^+2k^=024+λ2-9-16=0λ2=1 λ=±1

Page No 24.30:

Question 17:

If α =3i^+4j^+5k^ and β =2i^+j^-4k^, then express β in the form of β =β1+β2, where β1 is parallel to α and β2 is perpendicular to α .

Answer:

Given that α=3i^+4j^+5k^ and β=2i^+j^-4k^ Also,β=β1+β2β2=β-β1 ... 1Since β1 is parallel to α,β1=t αβ1=t 3i^+4j^+5k^ =3t i^+4t j^+5t k^   ...(2)Substituting the values of β1 andα in (1), we getβ2=2i^+j^-4k^ -3t i^+4t j^+5t k^=2-3t i^+1-4tj^+-4-5t k^ ... 3Since β2 is perpendicular to α,β2. α=02-3t i^+1-4tj^+-4-5t k^. 3i^+4j^+5k^=03 2-3t +4 1-4t+5 -4-5t=06-9t+4-16t-20-25t=0-50t=10t=-15From (2) and (3), we getβ1=-15 3i^+4j^+5k^ β2=135i^+95j^-3k^=1513i^+9j^-15k^



Page No 24.31:

Question 18:

If either a =0  or b =0 , then a ·b =0. But the converse need not be true. Justify your answer with an example.

Answer:

Let us assume that eithera=0 or  b=0Then, a. b=a b cos θ=0 (θ is the angle between a and b)Now, let us assume that a. b=0a b cos θ=0But here we cannot say that eithera=0 or  b=0. (Because even cos θ can be zero)For example, leta=2i^+j^+3k^ and b=-3i^+2k^Here, a=4+1+9=14≠0b=9+4=13≠0But a. b=2i^+j^+3k^ . -3i^+2k^=-6+0+6=0

Page No 24.31:

Question 19:

Show that the vectors a =3i^-2j^+k^, b =i^-3j^+5k^, c =2i^+j^-4k^ form a right-angled triangle.

Answer:

Let ABC be the given triangle and AC=b=i^-3j^+5k^CB=a=3i^-2j^+k^AB=c=2i^+j^-4k^a. b=3+6+5=14b. c=2-3-20=-21c. a=6-2-4=0So, AB is perpendicular to CB.Thus, ∆ABC is a right-angled triangle.

Page No 24.31:

Question 20:

If a =2i^+2j^+3k^, b =-i^+2j^+k^ and c =3i^+j^ are such that a +λb is perpendicular to c , then find the value of λ.

Answer:

We havea=2i^+2j^+3k^b=-i^+2j^+k^andc=3i^+j^ a+λb=2i^+2j^+3k^+λ -i^+2j^+k^=2-λ i^+2+2λ j^+3+λ k^Given that a+λb is perpendicular to c.a+λb. c=02-λ i^+2+2λ j^+3+λ k^ .  3i^+j^+0k^=03 2-λ+1 2+2λ+0=06-3λ+2+2λ=0 λ=8

Page No 24.31:

Question 21:

Find the angles of a triangle whose vertices are A (0, −1, −2), B (3, 1, 4) and C (5, 7, 1).

Answer:

Given thatOA=0i^-1j^-2k^; OB=3i^+1j^+4k^; OC=5i^+7j^+1k^AB=OB-OA=3i^+2j^+6k^ AB=9+4+36=7BA=OA-OB=-3i^-2j^-6k^ BA=9+4+36=7BC=OC-OB=2i^+6j^-3k^BC=4+36+9=7CB=OB-OC=-2i^-6j^+3k^CB=4+36+9=7CA=OA-OC=-5i^-8j^-3k^CA=25+64+9=98=72AC=OC-OA=5i^+8j^+3k^AC=25+64+9=98=72cos A=AB. ACABAC=15+16+18772=49492=12A=cos-112=π4cos B=BA. BCBABC=-6-12+1877=049=0B=cos-10=π2cos C=CB. CACBCA=10+48-9772=49492=12C=cos-112=π4

Page No 24.31:

Question 22:

Find the magnitude of two vectors a and b that are of the same magnitude, are inclined at 60° and whose scalar product is 1/2.

Answer:

Given that the angle between a and b is 300.Also,a= b; a. b=12We know that a. b=a b cos θ12=aa cos 6012=a212 a2=1 a=1 a= b=1

Page No 24.31:

Question 23:

Show that the points whose position vectors are a =4i^-3j^+k^, b =2i^-4j^+5k^, c =i^-j^form a right triangle.

Answer:

Given thata=OA=4i^-3j^+k^; b=OB=2i^-4j^+5k^; c=OC=i^-j^+0k^AB=OB-OA=-2i^-j^+4k^ BC=OC-OB=-i^+3j^-5k^CA=OA-OC=3i^-2j^+k^AB. BC=2-3-20=-210BC. CA=-3-6-5=-140AB. CA=-6+2+4=0So, AB is perpendicular to  CA.So, ∆ABC is a right-angled triangle.

Page No 24.31:

Question 24:

If the vertices A, B and C of ∆ABC have position vectors (1, 2, 3), (−1, 0, 0) and (0, 1, 2), respectively, what is the magnitude of ∠ABC?

Answer:

Given thatOA=i^+2j^+3k^; OB=-1i^+0j^+0k^; OC=0i^+1j^+2k^AB=OB-OA=-2i^-2j^-3k^ AB=4+4+9=17BC=OC-OB=i^+j^+2k^BC=1+1+4=6CA=OA-OC=i^+j^+k^CA=1+1+1=3cos ABC=AB. BCABBC=-2-2-6176=10102ABC=cos-110102

Page No 24.31:

Question 25:

If A, B and C have position vectors (0, 1, 1), (3, 1, 5) and (0, 3, 3) respectively, show that ∆ ABC is right-angled at C.

Answer:

Given thatOA=0i^+j^+k^; OB=3i^+j^+5k^; OC=0i^+3j^+3k^BC=OC-OB=-3i^+2j^-2k^CA=OA-OC=0i^-2j^-2k^Now,BC. CA=0-4+4=0So, BC is perpendicular to CA.So, ∆ABC is right-angled at C.

Page No 24.31:

Question 26:

Find the projection of b +c  on a , where a =2i^-2j^+k^, b =i^+2j^-2k^ and c =2i^-j^+4k^.

Answer:

Given thata=2i^-2j^+k^b=i^+2j^-2k^and c=2i^-j^+4k^ b+c=i^+2j^-2k^+2i^-j^+4k^=3i^+j^+2k^Projection of b+c on a isb+c.aa=3i^+j^+2k^. 2i^-2j^+k^4+4+1=6-2+23=2

Page No 24.31:

Question 27:

If a =5i^-j^-3k^ and b =i^+3j^-5k^, then show that the vectors a +b and a -b are orthogonal.

Answer:

Given thata=5i^-j^-3k^; b=i^+3j^-5k^ a+b=5i^-j^-3k^+i^+3j^-5k^=6i^+2j^-8k^And a-b=5i^-j^-3k^-i^+3j^-5k^=4i^-4j^+2k^Now,a+b. a-b=6i^+2j^-8k^.4i^-4j^+2k^=24-8-16=0 So, a+b is orthogonal to a-b.

Page No 24.31:

Question 28:

A unit vector a makes angles π4andπ3 with i^ and j^ respectively and an acute angle θ with k^. Find the angle θ and components of a .

Answer:

Let a=a1i^+a2j^+a3k^, where a1, a2 and a3 are components of a.a12+a22+a32=1 (Because  a is a unit vector)... 1Now,a. i^=a1ai^ cos π4=a1 (Because the angle between a and i^ is π4)1 1 12=a1 (Because a and i^ are unit vectors)a1=12Again,a. j^=a2ai^ cos π3=a2 (Because the angle between a and i^ is π3)1 1 12=a2(Because a and i^ are unit vectors)a2=12Now from (1),122+122+a32=112+14+a32=134+a32=1a32=14a3=12Now,a. k^=a3ak^ cos θ=12 (Because the angle between a and k^ is θ)1 1 cos θ=12(Because a and i^ are unit vectors)θ=cos-112=π3And a=12i^+12j^+12k^

Page No 24.31:

Question 29:

If two vectors a and b are such that  a =2, b =1 and a ·b =1, then find the value of 3a -5b ·2a +7b .

Answer:

Given thata=2, b=1 and  a. b=1 ... 1Now,3a-5b.2a+7b=6a2+21a. b-10 b. a-35b2=6a2+21a. b-10 a. b-35b2   (We know that a. b=b. a)=6a2+11a. b-35b2=622+11 1-3512  From (1)=24+11-35=0

Page No 24.31:

Question 30:

If a is a unit vector, then find  x  in each of the following.
(i)  x -a · x +a =8

(ii)  x -a · x +a =12
  

Answer:

Given that a is a unit vector.a=1 ... 1i x-a.x+a=8x2-a2=8x2-12=8       From (1)x2=9x=3ii x-a.x+a=12x2-a2=12x2-12=12     From (1)x2=13x=13

Page No 24.31:

Question 31:

Find  a  and  b  if
(i)  a +b · a -b =12 and  a =2 b 

(ii)  a +b · a -b =8 and  a =8 b 

(iii)  a +b · a -b =3 and  a =2 b 

Answer:

i Given thata=2 b ... 1And a+b.a-b=12a2-b2=122 b2-b2=12    From (1)4 b2-b2=123 b2=12b2=4b=2a=2 b=22=4a=4 and b=2

ii Given thatAnd a=8 b ... 1a+b.a-b=8a2-b2=88 b2-b2=8 From (1)64 b2-b2=863 b2=8b2=863b=863a=8 b=8 863=8863a=8863 and b= 863

iii Given thata=2 b ... 1And a+b.a-b=3a2-b2=32 b2-b2=3  From (1)4 b2-b2=33 b2=3b2=1b=1a=2 b=2 1=2a=2 and b=1

Page No 24.31:

Question 32:

Find  a -b  if
(i)  a =2, b =5 and a ·b =8

(ii)  a =3, b =4 and a ·b =1

(iii)  a =2, b =3 and a ·b =4

Answer:

i Given thata=2, b=5 and a. b=8 ...1We know thata-b2=a2+b2-2 a. b          =22+52-2 8   Using 1          =4+25-16          =13 a-b=13

ii Given thata=3, b=4 and a. b=1 ...1We know thata-b2=a2+b2-2 a. b          =32+42-2 1    Using 1          =9+16-2          =23 a-b=23

iii Given thata=2, b=3 and a. b=4  ...1We know thata-b2=a2+b2-2 a. b          =22+32-2 4  Using 1          =4+9-8          =5 a-b=5

Page No 24.31:

Question 33:

Find the angle between two vectors a and b if
(i)  a =3, b =2 and a ·b =6

(ii)  a =3, b =3 and a ·b =1

Answer:

i Let θ be the angle between a and b.Given thata=3, b=2 and a. b=6 ...1We know that a. b=a b cos θ6=32 cos θ   Using 1cos  θ=623=12 θ=cos-112=π4

ii Let θ be the angle between a and b.Given thata=3, b=3 and a. b=1 ...1We know that a. b=a b cos θ1=33 cos θ  Using 1cos  θ=133=19 θ=cos-119



Page No 24.32:

Question 34:

Express the vector a =5i^-2j^+5k^ as the sum of two vectors such that one is parallel to the vector b =3i^+k^ and other is perpendicular to b .

Answer:

Given that a=5i^-2j^+5k^ and b=3i^+k^Let x and y be such thata=x+yy=a-x ... 1Since x is parallel to b,x=t b    t is constantx=t 3i^+k^=3t i^+t k^Substituting the values of x and a in (1), we gety=5i^-2j^+5k^-3t i^+t k^=5-3t i^-2j^+5-t k^ ... 2Since y is perpendicular to b,y. b=05-3t i^-2j^+5-t k^. 3i^+k^=03 5-3t+0+5-t=015-9t+5-t=020-10t=0t=2From (1) and (2), we getx=6 i^+2 k^y=- i^-2j^+3 k^

Page No 24.32:

Question 35:

If a and b are two vectors of the same magnitude inclined at an angle of 30°, such that a ·b =3, find  a , b .

Answer:

Given that the angle between a and b is 300.Also,a= b and a. b=3We know that a. b=a b cos θ3=aa cos 303=a232 a2=63=23 a=23= b a=b=23

Page No 24.32:

Question 36:

Express 2i^-j^+3k^ as the sum of a vector parallel and a vector perpendicular to 2i^+4j^-2k^.

Answer:

Let a=2i^-j^+3k^ and b=2i^+4j^-2k^ and x and y be such thata=x+yy=a-x  ...(1)Since x is parallel to b,x=t bx=t 2i^+4j^-2k^ =2t i^+4t j^-2t k^ ...(2)Substituting the values of x and a in (1),y=2i^-j^+3k^-2t i^+4t j^-2t k^=2-2t i^+-1-4tj^+3+2t k^ ... 3Since y is perpendicular to b,y. b=02-2t i^+-1-4tj^+3+2t k^. 2i^+4j^-2k^ =02 2-2t+4 -1-4t-2 3+2t=04-4t-4-16t-6-4t=0-24t=6t=-14From (2) and (3),x=2-14 i^+4-14 j^-2-14 k^=-12i^-j^+12k^y=2-2-14 i^+-1-4-14j^+3+2-14 k^ =52i^+52k^=52i^+k^So,a=x+y=-12i^-j^+12k^+52i^+k^

Page No 24.32:

Question 37:

Decompose the vector 6i^-3j^-6k^ into vectors which are parallel and perpendicular to the vector i^+j^+k^.

Answer:

Let a=6i^-3j^-6k^ and b=i^+j^+k^ and x and y be such thata=x+yy=a-x ... 1Since x is parallel to b,x=t bx=t i^+j^+k^=t i^+t j^+t k^  ...(2)Substituting the values of x and a in (1), we gety=6i^-3j^-6k^-t i^+t j^+t k^  =6-t i^+-3-tj^+-6-t k^ ... 3Since y is perpendicular to b,y. b=06-t i^+-3-tj^+-6-t k^. i^+j^+k^=01 6-t+1-3-t + 1 -6-t =0-3-3t=0t=-1From (2) and (3), we getx=-i^-j^-k^y=7 i^-2j^-5 k^So,a=x+y=-i^-j^-k^+7 i^-2j^-5 k^

Page No 24.32:

Question 38:

Let a =5i^-j^+7k^ and b =i^-j^+λk^. Find λ such that a +b is orthogonal to a -b .

Answer:

Given thata=5i^-j^+7k^; b=i^-j^+λk^ a+b=5i^-j^+7k^+i^-j^+λk^=6i^-2j^+7+λk^and a-b=5i^-j^+7k^-i^-j^+λk^=4i^+0j^+7-λk^Given that a+b is orthogonal to a-b.a+b. a-b=06i^-2j^+7+λk^.4i^+0j^+7-λk^=024+0+49-λ2=0λ2=73λ=73

Page No 24.32:

Question 39:

If a ·a =0 and a ·b =0, what can you conclude about the vector b ?

Answer:

Given that a. a=0a2=0a=0   ...1Also given thata. b=0a b cos θ=0   (Where θ is the angle between a and b)0 b cos θ=0    [From (1)]0=0So, it means that for any vector b, the given equation a. b=0 is satisfied.

Page No 24.32:

Question 40:

If c is perpendicular to both a and b , then prove that it is perpendicular to both a +b and a -b .

Answer:

Given that c is perpendicular to both  a and  b. c. a=0 and  c.  b=0 ... 1Now, c.  a+ b= c.  a+ c.  b=0+0=0 From 1So,  c is perpendicular to  a+ b.Again, c.  a -b= c.  a- c.  b=0-0=0 From 1So,  c is perpendicular to  a- b.

Page No 24.32:

Question 41:

If  a =a and  b =b, prove that a a2-b b22=a -b ab2.

Answer:

aa2-bb22=aa22+bb22-2a.ba2b2=a2a4+b2b4- 2a.ba2b2=a2a4+b2b4- 2a.ba2b2(From the given information)=1a2+1b2- 2a.ba2b2=b2+a2-2a.ba2b2=a2+b2-2a.ba2b2=a2+a2-2a.ba2b2(From the given information)=a-b2a2b2=a-bab2

Page No 24.32:

Question 42:

If a, b, c are three non-coplanar vectors, such that d ·a =d ·b =d ·c =0, then show that d  is the null vector.

Answer:

Given that: d·a=0
so, either d = 0 or da

similarly, d·b=0
so, d = 0 or db

Also, d·c=0
so, d = 0 or dc

But d cannot be perpendicular to a, b, c as  a, b, c are non-coplanar.
 
so, d=0. d is a null vector. 

Page No 24.32:

Question 43:

If a vector a is perpendicular to two non-collinear vectors b and c , then show that a is perpendicular to every vector in the plane of b and c .

Answer:

Given that a is perpendicular to b and c.a. b=0 and ac=0 ... (1)Now, let r be any vector in the plane of b and c.Then, r is the linear combination of b and c.r=xb+yc, for some x and y.Now,a. r=a. xb+yc=x a. b+y a. c=x0+y0 [From (1)]=0Thus, a is perpendicular to r.That is, a is perpendicular to every vector in the plane of b and c .

Page No 24.32:

Question 44:

If a +b +c =0 , show that the angle θ between the vectors b and c is given by

cos θ =  a  2- b  2- c 22 b   c .

Answer:

Given,a+b+c=0b+c=-ab+c2=-a2b2+c2+2b. c=a22b. c=a2-b2-c22 b c cos θ=a2-b2-c2 cos θ=a2-b2-c22 b c 

Page No 24.32:

Question 45:

Let u, v and w be vectors such u +v +w =0 . If  u =3, v =4 and  w =5, then find u  ·v  +v  ·w +w ·u .

Answer:

Given thatu+v+w=0u+v+w=0u+v+w2=0u2+v2+w2+2u. v+v. w+w. u=032+42+52+2u. v+v. w+w. u=0    (Given:u=3, v=4 and w=5)9+16+25+2u. v+v. w+w. u=050+2u. v+v. w+w. u=02u. v+v. w+w. u=-50 u. v+v. w+w. u=-502=-25

Page No 24.32:

Question 46:

Let a =x2i^+2j^-2k^, b =i^-j^+k^ and c =x2i^+5j^-4k^ be three vectors. Find the values of x for which the angle between a and b is acute and the angle between b and c is obtuse.

Answer:

We havea= =x2i+2j^-2k^, b=i^-j^+k^ and c=x2i^+5j^-4k^Let θ1 be the angle between a and b  and θ2 be the angle between b and c.Given that θ1 is acute and θ2 is obtuse.cos θ1>0 and cos θ2<0a.ba.b>o and b.cb.c<0x2-4x4+4+41+1+1>0 and x2-91+1+1x4+25+16<0x2-4>0 and x2-9<0x-, -22,  and x-3, 3x-3, -2  2, 3

Page No 24.32:

Question 47:

Find the values of x and y if the vectors a =3i^+xj^-k^ and b =2i^+j^+yk^ are mutually perpendicular vectors of equal magnitude.

Answer:

We havea=3i^+xj^-k^ and b=2i^+j^+yk^It is given that the vectors are perpendicular.a. b=06+x-y=0x-y=-6 ... 1Also, it is given thata=b9+x2+1=4+1+y210+x2=5+y2 10+x2=5+y2 x2-y2=-5x+yx-y=-5-6 x+y=-5         Using 1x+y=56 ... 2(1)+(2) gives2x=-316x=-3112From (1),-3112-y=-6y=-3112+6=4112x=-3112 and y=4112

Page No 24.32:

Question 48:

If a and b are two non-collinear unit vectors such that  a +b =3, find  2a -5b · 3a +b .

Answer:

We havea+b=3Squaring both sides, we geta+b2=3a2+b2+2 a. b=31+1+2 a. b=3 (Because a and b are unit vectors)2+ 2 a. b=32 a. b=12 a. b=1 a. b=12... 1Now,2a-5b.3a+b=6a2+2a. b-15 b. a-5b2=6a2+2a. b-15 a. b-5b2         (a. b=b. a)=6a2-13a. b-5b2=61-13 12-51   From (1)=1-132=-112



Page No 24.33:

Question 49:

If a , b  are two vectors such that  a +b = b , then prove that a +2b is perpendicular to a .

Answer:

Given thata+b=bSquaring both sides, we geta+b2=b2a2+b2+2 a. b=b2a2+2 a. b=0 ... 1Now,a+2b.a=a.a+2 b. a=a2+2 a. b=0       Using 1So, a+2b is perpendicular to a.



Page No 24.46:

Question 1:

In a triangle OAB, AOB = 90º. If P and Q are points of trisection of AB, prove that OP2+OQ2=59AB2.

Answer:




In triangle OAB, AOB = 90º. P and Q are points of trisection of AB.

Taking O as the origin, let the position vectors of A and B be a and b, respectively.

Since P and Q are the points of trisection of AB, so AP : PB = 1 : 2 and AQ : QB = 2 : 1.

Position vector of P, OP=2a+b3                (Using section formula)
Position vector of Q, OQ=a+2b3

Also, OAOB.

a.b=0                .....(1)

Now,

OP2+OQ2=OP2+OQ2=2a+b3.2a+b3+a+2b3.a+2b3=4a2+4a.b+b2+a2+4a.b+4b29
=5a2+5b29                        Using 1=59a2+b2=59AB2                                   Using Pythagoras Theorem=59AB2

Page No 24.46:

Question 2:

Prove that: If the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:




Let OACB be a quadrilateral such that diagonals OC and AB bisect each other at 90º.

Taking O as the origin, let the poisition vectors of A and B be a and b, respectively. Then,

OA=a and OB=b
Position vector of mid-point of AB, OE=a+b2

∴ Position vector of C, OC=a+b

By the triangle law of vector addition, we have

OA+AB=OBAB=OB-OA=b-a

Since ABOC,

AB.OC=0b-a.a+b=0b2-a2=0a2=b2a=bOA=OB

In a quadrilateral if diagonals bisects each other at right angle and adjacent sides are equal, then it is a rhombus.

Page No 24.46:

Question 3:

(Pythagoras's Theorem) Prove by vector method that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Answer:


Let ABC be a right triangle with BAC = 90º. Taking A as the origin, let the position vectors of B and C be b and c, respectively. Then,

AB=b and AC=c

Since ABACb.c=0          .....(1)

Now,

AB2+AC2=b2+c2              .....(2)

Also,

BC2=c-b2          =c-b.c-b          =c2-2b.c+b2          =c2+b2                  .....3                  Using 1

From (2) and (3), we have

AB2+AC2=BC2

Page No 24.46:

Question 4:

Prove by vector method that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer:




Let ABCD be a parallelogram such that AC and BD are its two diagonals. Taking A as the origin, let the position vectors of B and D be b and d, respectively. Then,

AB=b and AD=d

Using triangle law of vector addition, we have

AD+DB=ABDB=b-d

In ∆ABC,

AC=AB+BC=AB+AD=b+d

Now,

AB2+BC2+CD2+DA2=AB2+AD2+-AB2+-AD2=2AB2+2AD2=2 b2+2d2                                 .....1

Also,

DB2+AC2=b-d2+b+d2=b-d.b-d+b+d.b+d=b2-2b.d+d2+b2+2b.d+d2=2b2+2d2                                   .....2

From (1) and (2), we have

AB2+BC2+CD2+DA2=DB2+AC2

Page No 24.46:

Question 5:

Prove using vectors: The quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.

Answer:



ABCD is a rectangle. Let P, Q, R and S be the mid-points of the sides AB, BC, CD and DA, respectively.

Now,

PQ=PB+BQ=12AB+12BC=12AB+BC=12AC              .....(1)

SR=SD+DR=12AD+12DC=12AD+DC=12AC            .....(2)

From (1) and (2), we have

PQ=SR

So, the sides PQ and SR are equal and parallel. Thus, PQRS is a parallelogram.

Now,

PQ2=PQ.PQPQ2=PB+BQ.PB+BQPQ2=PB2+2PB.BQ+BQ2PQ2=PB2+0+BQ2                    PBBQPQ2=PB2+BQ2                    .....3

Also,

PS2=PS.PSPS2=PA+AS.PA+ASPS2=PA2+2PA.AS+AS2PS2=PB2+0+BQ2                    PAASPS2=PB2+BQ2                    .....4

From (3) and (4), we have

PQ2=PS2PQ=PS

So, the adjacent sides of the parallelogram are equal. Hence, PQRS is a rhombus.

Page No 24.46:

Question 6:

Prove that the diagonals of a rhombus are perpendicular bisectors of each other.

Answer:



Let OABC be a rhombus, whose diagonals OB and AC intersect at D. Suppose O is the origin.

Let the position vector of A and C be a and c, respectively. Then,

OA=a and OC=c

In ∆OAB,

OB=OA+AB=OA+OC=a+c                  AB=OC

Position vector of mid-point of OB=12a+c

Position vector of mid-point of AC=12a+c                (Mid-point formula)

So, the mid-points of OB and AC coincide. Thus, the diagonals OB and AC bisect each other.

Now,

OB.AC=a+c.c-a            =c+a.c-a            =c2-a2            =OC2-OA2            =0                                    OC=OAOBAC

Hence, the diagonals OB and AC are perpendicular to each other.

Thus, the diagonals of a rhombus are perpendicular bisectors of each other.

Page No 24.46:

Question 7:

Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.

Answer:



Let ABCD be a rectangle. Take A as the origin.

Suppose the position vectors of points B and D be a and b, respectively.

Now,

AC=AB+BC=AB+AD=a+b

Also, BD=a-b

Since ABCD is rectangle, so ABAD.

a.b=0               .....(1)

Now, diagonals AC and BD are perpendicular

iff AC.BD=0

iff a+b.a-b=0iff a2-b2=0iff a=biff AB=AD

iff ABCD is a square

Thus, the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.

Page No 24.46:

Question 8:

If AD is the median of ∆ABC, using vectors, prove that AB2+AC2=2AD2+CD2.

Answer:



Taking A as the origin, let the position vectors of B and C be b and c, respectively.

It is given that AD is the median of ∆ABC.

∴ Position vector of mid-point of BC = AD=b+c2             (Mid-point formula)

Now,

AB2+AC2=AB2+AC2=b2+c2         .....(1)

Also,

2AD2+CD2=2AD2+CD2=2b+c2.b+c2+b+c2-c.b+c2-c=2b+c2.b+c2+b-c2.b-c2=b2+2b.c+c22+b2-2b.c+c22=2b2+2c22=b2+c2                              .....2

From (1) and (2), we have

AB2+AC2=2AD2+CD2

Page No 24.46:

Question 9:

If the median to the base of a triangle is perpendicular to the base, then triangle is isosceles.

Answer:



Let ∆ABC be a triangle such that AD is the median. Taking A as the origin, let the position vectors of B and C be b and c, respectively. Then,

Position vector of D = b+c2               (Mid-point formula)

Now,

AD = Position vector of D − Position vector of A = b+c2

BC = Position vector of C − Position vector of B = c-b

Since ADBC,

AD.BC=012b+c.c-b=0c+b.c-b=0c2-b2=0c=bAC=AB

Hence, the ∆ABC is an isosceles triangle.

Page No 24.46:

Question 10:

In a quadrilateral ABCD, prove that AB2+BC2+CD2+DA2=AC2+BD2+4PQ2, where P and Q are middle points of diagonals AC and BD.

Answer:



Let ABCD be the quadrilateral. Taking A as the origin, let the position vectors of B, C and D be b,c and d, respectively. Then,

Position vector of P = c2                       (Mid-point formula)

Position vector of Q = b+d2                (Mid-point formula)

Now,

AB2+BC2+CD2+DA2=AB2+BC2+CD2+DA2=b2+c-b2+d-c2+d2=b2+c2-2c.b+b2+d2-2d.c+c2+d=2b2+2c2+2d2-2b.c-2c.d            .....1

Also,

AC2+BD2+4PQ2=AC2+BD2+4PQ2=c2+d-b2+4b+d2-c22=c2+d-b2+b+d2-2b+d.c+c2=2c2+2d2+2b2-2b.c-2d.c=2b2+2c2+2d2-2b.c-2c.d         .....2

From (1) and (2), we have

AB2+BC2+CD2+DA2=AC2+BD2+4PQ2

Page No 24.46:

Question 1:

What is the angle between vectors a and b with magnitudes 2 and 3 respectively? Given a .b =3.

Answer:

 Let θ be the angle between a and b.Given thata=2, b=3 and a. b=3We know that a. b=a b cos θ3=23 cos θcos  θ=323cos  θ=12 θ=cos-112=π3

Page No 24.46:

Question 2:

a and b are two vectors such that a .b =6, a =3 and  b =4. Write the projection of a on b .

Answer:

We havea. b=6 and b=4The projection of a on b isa. bb=64=32

Page No 24.46:

Question 3:

Find the cosine of the angle between the vectors 4i^-3j^+3k^ and 2i^-j^-k^.

Answer:

Let, a=4i^-3j^+3k^and b=2i^-j^-k^Let θ be the angle between a and b.a=42+-32+32=34b=22+-12+-12=6 a . b=8+3-3=8cos θ=a . ba b=8346=8251=451

Page No 24.46:

Question 4:

If the vectors 3i^+mj^+k^ and 2i^-j^-8k^ are orthogonal, find m.

Answer:

It is given that the vectors are othgonal. So, their dot product is zero.3i^+mj^+k^ . 2i^-j^-8k^=06-m-8=0-m-2=0m=-2

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Question 5:

If the vectors 3i^-2j^-4k^ and 18i^-12j^-mk^ are parallel, find the value of m.

Answer:

The given vectors are parallel.  3i^-2j^-4k^=t 18i^-12j^-mk^3i^-2j^-4k^=18t i^-12t j^-tm k^Comparing both sides, we get  18t=3, -12t=-2, -4=-tmt=16 Substituting the value of m in -4=-tm, we get-4=-m16 m=24

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Question 6:

If a and b are vectors of equal magnitude, write the value of  a +b . a -b .

Answer:

We havea=b     ...iNow,a+b . a-b=a2-b2=a2-a2     Using i=0



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Question 7:

If a and b are two vectors such that  a +b . a -b =0, find the relation between the magnitudes of a and b .

Answer:

Given thata+b . a-b=0a2-b2=0a2=b2a=b

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Question 8:

For any two vectors a and b , write when  a +b = a + b  holds.

Answer:

Given thata+b=a+bSquaring both sides,we geta+b2=a+b2a2+b2+2 a. b=a2+b2+2a ba. b=a ba b cos θ=a b (where θ is the angle between a and b)cos θ=1θ=0oa and b are parallel.

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Question 9:

For any two vectors a and b , write when  a +b = a -b  holds.

Answer:

Given thata+b=a-bSquaring both sides, we geta+b2=a-b2a2+b2+2 a. b=a2+b2-2 a. b4a. b=0a. b=0a and b are perpendicular.

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Question 10:

If a and b are two vectors of the same magnitude inclined at an angle of 60° such that a .b =8, write the value of their magnitude.

Answer:

Given thata=b and a and  b are inclined at an angle of 60°Also, given thata. b=8a b cos 60°=8aa 12=8a2=16a=4

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Question 11:

If a .a =0 and a .b =0, what can you conclude about the vector b ?

Answer:

Given that a. a=0a2=0a=0 ... 1Also, given thata. b=0a b cos θ=0 (where θ is the angle between a and b)0 b cos θ=0 [From (1)]0=0So, it means that for any vector b, the given equation a. b=0 is satisfied.

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Question 12:

If  b is a unit vector such that  a +b . a -b =8, find  a .

Answer:

 Given that b is a unit vector. b=1Anda+b.a-b=8     Givena2-b2=8a2-12=8a2=9  a=3

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Question 13:

If a ^, b ^ are unit vectors such that a ^+b ^is a unit vector, write the value of  a ^-b ^.

Answer:

Given that a^ and b^ are unit vectors such that a^+b^ is a unit vector.a^=b^=a^+b^=1 ... 1Now,a^+b^=1Squaring both sides, we geta^2+b^2+2a^. b^ =11+1+2a^. b^ =1      From (1)a^. b^ =-12 ... 2Now,a^-b^2=a^2+b^2-2a^. b^         =1+1-2-12=3      From (1) and (2) a^-b^=3

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Question 14:

If  a =2, b =5 and a .b =2, find  a -b .

Answer:

We havea=2, b=5 and a. b=2Now,a-b2 =a2+b2-2 a. b            =22+52-2 2            =4+25-4            =25 a-b=25=5

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Question 15:

If a =i^-j^ and b =-j^+k^, find the projection of a on b .

Answer:

We havea=i^-j^ and b=-j^+k^The projection of a on b isa. bb=i^-j^.-j^+k^-j^+k^=0+1+01+1=12

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Question 16:

For any two non-zero vectors, write the value of  a +b 2+ a -b 2 a 2+ b 2.

Answer:

We havea+b2+a-b2a2+b2=a2+b2+2 a. b+a2+b2-2 a. ba2+b2=2a2+b2a2+b2=2

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Question 17:

Write the projections of r =3i^-4j^+12k^ on the coordinate axes.

Answer:

We haver=3i^-4j^+12k^Projection of r on x-axis = r. i^i^=31=3Projection of r on y-axis = r. j^j^=-41=-4Projection of r on z-axis = r. k^k^=121=12

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Question 18:

Write the component of  b along a .

Answer:

Component of b on a is a . baa^=a. ba2a=a. baa2

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Question 19:

Write the value of a .i^ i^+a .j^ j^+a .k^ k^, where a is any vector.

Answer:

Let a=a1i+a2j+a3kNow,a. i i+a. j j+a. k k=a1i+a2j+a3k=a

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Question 20:

Find the value of θ ∈(0, π/2) for which vectors a =sin θ i^+cos θ j^ and b =i^-3j^+2k^ are perpendicular.

Answer:

We havea=sin θi^+cos θj^andb=i^-3 j^+2k^It is given that the vectors are perpendicular.a. b=0sin θ-3 cos θ=0sin θ=3 cos θtan θ=3 θ=π3

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Question 21:

Write the projection of i^+j^+k^ along the vector j^.

Answer:

Projection of a on b =a. bbProjection of i+j+k along j=i+j+k. jj=11=1

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Question 22:

Write a vector satisfying a .i^=a .i^+j^=a .i^+j^+k^=1.

Answer:

Let a=a1 i^+a2 j^+a3 k^a. i^=a1a.  i^+ j^=a1+a2a.  i^+ j^+k^=a1+a2+a3Given thata. i^=a.  i^+ j^=a.  i^+ j^+k^=1a1=a1+a2=a1+a2+a3=1a1=1;a1+a2=1; a1+a2+a3=1a1=1; 1+a2=1; 1+a2+a3=1a1=1; a2=0; 1+0+a3=1a1=1; a2=0; a3=0So, a=a1 i^+a2 j^+a3 k^=1 i^+0 j^+0 k^=i^

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Question 23:

If a and b are unit vectors, find the angle between a +b and a -b .

Answer:

We havea=1 and b=1  ...iNow,a+b . a-b=a2-b2                          =12-12             Using i                          =0a+b and a-b are perpendicular. Angle between a+b and a-b is 900.

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Question 24:

If a and b are mutually perpendicular unit vectors, write the value of  a +b .

Answer:

a and b are unit vectors and they are perpendicular.a=b=1; a. b=0           ...iNow,a+b2=a2+b2+2a. b           =1+1+2 0    Using i           =2  a+b=2

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Question 25:

If a , b and c are mutually perpendicular unit vectors, write the value of a +b +c .

Answer:

Given that ab and c are unit vectors.So, a=1, b=1 and c=1  ...iSince they are mutually perpendicular, a.b=b.c=c.a=0  ...iiNow,a+b+c2 =a2+b2+c2+2 a. b+2 b.c+2 c.a                  =1+1+1+0+0+0      Using iand ii                   =3 a+b+c=3

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Question 26:

Find the angle between the vectors a =i^-j^+k^ and b =i^+j^-k^.

Answer:

We havea=i^-j^+k^ and b=i+j-k Let θ be the angle between a and b.a=12+-12+ 12=3b=12+12+-12=3anda . b=1-1-1=-1Now,cos θ=a . ba b=-133=-13 θ=cos-1 -13

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Question 27:

For what value of λ are the vectors a =2i^+λj^+k^ and b =i^-2j^+3k^ perpendicular to each other?

Answer:

We havea=2i^+λj^+k^ and b=i^-2j^+3k^Given that a and b are perpendicular.a. b=02i^+λj^+k^.i^-2j^+3k^=02-2λ+3=05-2λ=0 λ=52

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Question 28:

Find the projection of a on b if a ·b =8 and b =2i^+6j^+3k^.

Answer:

We havea. b=8 and b=2i^+6j^+3k^The projection of a on b isa. bb=84+36+9=87

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Question 29:

Write the value of p for which a =3i^+2j^+9k^ and b =i^+pj^+3k^ are parallel vectors.

Answer:

We havea=3i^+2j^+9k^ and b= i^+pj^+3k^Given that a and b are parallel.a=tb for some t.3i^+2j^+9k^=t i^+pj^+3k^3i^+2j^+9k^=t i^+pt j^+3t k^Comparing both sides, we get3=t, 2=pt and 9=3tt=3 and pt=23t=2t=23

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Question 30:

Find the value of λ if the vectors 2i^+λj^+3k^ and 3i^+2j^-4k^ are perpendicular to each other.

Answer:

Given: 2i^+λj^+3k^ and 3i+2j-4k are perpendicular to each other.So, their dot product is zero.2i^+λj^+3k^.3i+2j-4k6+2λ-12=02λ-6=0λ=3



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Question 31:

If  a =2, b =3 and a ·b =3, find the projection of b on a .

Answer:

We havea=2 and a. b=3So,the projection of b on a isa. ba=32

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Question 32:

Write the angle between two vectors a and b with magnitudes 3 and 2 respectively if a ·b =6.

Answer:

Let θ be the angle between a and b.Given,a=3; b=2; a. b=6We know that a. b=a b cos θ6=32 cos θcos  θ=623=12 θ=cos-112=π4

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Question 33:

Write the projection of the vector i^+3j^+7k^ on the vector 2i^-3j^+6k^.

Answer:


We know that projection of a on b = a.bb.

Let a=i^+3j^+7k^ and b=2i^-3j^+6k^.

∴ Projection of a on b

=i^+3j^+7k^.2i^-3j^+6k^2i^-3j^+6k^=1×2+3×-3+7×622+-32+62=2-9+4249=357=5

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Question 34:

Find λ when the projection of a =λi^+j^+4k^ on b =2i^+6j^+3k^ is 4 units.

Answer:

We havea=λ i^+j^+4k^ and b=2i^+6j^+3k^The projection of a on b is a. bbGiven thata. bb=4λ i^+j^+4k^.2i^+6j^+3k^2i^+6j^+3k^2λ+6+124+36+9=42λ+187=42λ+18=282λ=10λ=5

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Question 35:

For what value of λ are the vectors a =2i^+λj^+k^ and b =i^-2j^+3k^ perpendicular to each other?

Answer:

We havea=2i^+λ j^+k^ and b=i^-2j^+3k^Given, a and b are perpendicular. So, their dot product is zero.2i^+λ j^+k^.i^-2j^+3k^=02-2λ+3=0-2λ+5=0λ=52

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Question 36:

Write the projection of the vector 7i^+j^-4k^ on the vector 2i^+6j^+3k^.

Answer:

Let a=7i^+j^-4k^; b=2i^+6j^+3k^The projection of a on b isa. bb=7i^+j^-4k^.2i^+6j^+3k^2i^+6j^+3k^=14+6-124+36+9=87

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Question 37:

Write the value of λ so that the vectors a =2i^+λj^+k^ and b =i^-2j^+3k^ are perpendicular to each other.

Answer:

We havea=2i^+λj^+k^ and b=i^-2j^+3k^The given vectors are perpendicular. So, their dot product is zero.2i^+λj^+k^ . i^-2j^+3k^2-2λ+3=05-2λ=0-2λ=-5λ=52

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Question 38:

Write the projection of b +c on a  when a =2i^-2j^+k^, b =i^+2j^-2k^ and c =2i^-j^+4k^.

Answer:

Given thata=2i^-2j^+k^b=i^+2j^-2k^c=2i^-j^+4k^b+c=i^+2j^-2k^+2i^-j^+4k^=3i^+j^+2k^Projection of b+c on a isb+c.aa=3i^+j^+2k^. 2i^-2j^+k^2i^-2j^+k^=6-2+24+4+1=63=2

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Question 39:

If a and b are perpendicular vectors, a+b=3 and a=5, find the value of b.                           [CBSE 2014]

Answer:


Disclaimer: a+b=13 has been taken in order to solve the question.

It is given that a and b are perpendicular vectors.

a.b=0           .....(1)

a+b=13a+b2=169a2+2a.b+b2=16925+2×0+b2=169               Using1
b2=169-25=144b=12

Thus, the value of b is 12.

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Question 40:

If the vectors a and b are such that a=3,b=23 and a×b is a unit vector, then write the angle between a and b.          [CBSE 2014]

Answer:


Let the angle between a and b be θ.

It is given that a×b is a unit vector.

a×b=1absinθ=13×23×sinθ=1sinθ=12θ=π6

Thus, the angle between a and b is π6.

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Question 41:

If a and b are two unit vectors such that a+b is also a unit vector, then find the angle between a and b.              [CBSE 2014]

Answer:


Let the angle between a and b be θ.

It is given that a=b=a+b=1.

a+b=1a+b2=1a2+2abcosθ+b2=11+2×1×1×cosθ+1=12cosθ=-1cosθ=-12=cos2π3θ=2π3

Thus, the angle between a and b is 2π3.

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Question 42:

If a and b are unit vectors, then find the angle between a and b, given that 3a-b is a unit vector.                [CBSE 2014]

Answer:


Let the angle between a and b be θ.

It is given that a=b=3a-b=1.

3a-b=13a-b2=13a2-23a.b+b2=13a2-23abcosθ+b2=1
3×1-23×1×1×cosθ+1=123cosθ=3cosθ=32=cosπ6θ=π6

Thus, the angle between a and b is π6.

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Question 43:

Find the magnitude of each of the two vectors a and b, having the same magnitude such that the angle between them is 60° and their scalar product is 92.

Answer:

Let the two vectors be a and b
Since the vectors have same magnitude so, 
a=b
Scalar product of two vectors = a·b=abcosθ
92=abcos60°92=ab×12ab=9aa=9                        From ia2=9a=b=3



Page No 24.49:

Question 1:

The vectors a and b satisfy the equations 2a +b =p and a +2b =q ,where p =i^+j^ and q =i^-j^. If θ is the angle between a and b , then
(a) cos θ=45

(b) sin θ=12

(c) cos θ=-45

(d) cos θ=-35

Answer:

(c) cos θ=-45

Given that2a+b=p ... 1a+2b=q ... 2Solving these two we geta=2p-q3, b=2q-p3And we havep=i^+j^ and q=i^-j^Substituting the values of p and q,  we geta=2p-q3=2i^+j^-i^-j^3 =i^+3j^3a=131+9=103b=2q-p3=2i^-j^-i^+j^3=i^-3j^3b=131+9=103a. b=19 1-9=-89We know thata. b=a b cos θ-89=103×103 cos θ-89=109cos θcos θ=-89×910=-45

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Question 2:

If a ·i^=a ·i^+j^=a ·i^+j^+k^=1, then a =
(a) 0 

(b) i^

(c) j^

(d) i^+j^+k^

Answer:


(b) i^

Let a=a1i^+a2j^+a3k^a. i^=a1and a. i^+j^=a1+a2and a. i^+j^+k^=a1+a2+a3Given,a. i^=a. i^+j^=a. i^+j^+k^=1a1=a1+a2=a1+a2+a3=1a1=1, a2=0, a3=0So, a=a1i^+a2j^+a3k^=1 i^+0j^+0k^=i^

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Question 3:

If a +b +c =0 , a =3, b =5, c =7, then the angle between a and b is
(a) π6

(b) 2π3

(c) 5π3

(d) π3

Answer:

(d) π3

Given, a =3, b=5 and c=7   ...iLet θ be the angle between aand bGiven thata+b+c=0a+b=-ca+b=-c2a2+b2+2a. b=c22a. b=c2-a2-b22a. b=72-32-52     Using i2a. b=152 a b cos θ=152 3 5 cos θ=15     Using icos θ=12 θ=π3

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Question 4:

Let a and b be two unit vectors and α be the angle between them. Then, a +b is a unit vector if
(a) α=π4

(b) α=π3

(c) α=2π3

(d) α=π2

Answer:


(c) α=2π3

a and b are unit vectors.a=b=1  ... 1Now,a. b=a b cos αa. b=cos α ... 2               Using 1Given that a+b=1Squaring both sides, we geta+b2=1a2+b2+2 a. b=11+1+2 cos α=1        Using 1 and 22+ 2 cos α=12 cos α=-12 cos α=-1cos α=-12α=2π3

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Question 5:

The vector (cos α cos β)i^ + (cos α sin β)j^ + (sin α)k^ is a
(a) null vector
(b) unit vector
(c) constant vector
(d) None of these

Answer:

(b) unit vector

Let a=cos α cos β i^+cos α sin β j^+sin α k^a=cos2 α cos2 β+cos2 α sin2 β+sin2 α=cos2 α cos2 β+sin2 β+sin2 α=cos2 α1+sin2 α=cos2 α+sin2 α=1=1So, a is a unit vector.

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Question 6:

If the position vectors of P and Q are i^+3j^-7k^ and 5i^-2j^+4k^ then the cosine of the angle between PQ and y-axis is
(a) 5162

(b) 4162

(c) -5162

(d) 11162

Answer:

(c) -5162

PQ=OQ-OP=5i^-2j^+4k^-i^+3j^-7k^=4i^-5j^+11k^The unit vector along y-axis is j^.Let θ be the required angle.cos θ=PQ. j^PQj^=-516+25+1211=-5162

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Question 7:

If a and b are unit vectors, then which of the following values of a .b is not possible?
(a) 3

(b) 3/2

(c) 1/2

(d) −1/2

Answer:

(a) 3

It is given that a and b are unit vectors.a=b=1Now,a. b=a b cos θ=1 1 cos θ=cos θThe range of cos θ is -1, 1 3 is not a possible value of cosθ as it is greater than 1.

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Question 8:

If the vectors i^-2xj^+3yk^ and i^+2xj^-3yk^ are perpendicular, then the locus of (x, y) is
(a) a circle
(b) an ellipse
(c) a hyperbola
(d) None of these

Answer:

(b) an ellipse

Let, a=i^-2x j^+3y k^ and b=i^+2x j^-3y k^It is given that the vectors are perpendicular. So, their dot product is zero.a.b=0i^-2x j^+3y k^.i^+2x j^-3y k^=01-4x2-9y2=04x2+9y2=1Dividing both sides by 36, we getx29+y24=1This is an ellipse.

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Question 9:

The vector component of b perpendicular to a is
(a)  b .c a 

(b) a × b ×a  a 2

(c) a × b ×a 

(d) None of these

Answer:

(b) a × b ×a  a 2

The vector component of b perpendicular to a isa×b×a a2

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Question 10:

What is the length of the longer diagonal of the parallelogram constructed on 5a +2b and a -3b if it is given that  a =22, b =3 and the angle between a and b is π/4?
(a) 15
(b) 113
(c) 593
(d) 369

Answer:

(c) 593


Let ABCD be a parallelogram in which side AB = DC = 5a+2band AD = BC=a-3band diagonals are AC and BD.Now, AC=AB+BC             =5a+2b+a-3b             =6a-b  AC=6a-b           =6a2+b2-2×6a×bcosθ           =36a2+b2-12×a×bcosπ4           =36222+32-12×22×3×12           =288+9-72           =225=15 units BD= BA+BD             =-AB+BD             =-5a+2b+a-3b             =-4a-5b  BD=-4a-5b           =4a+5b           =4a2+5b2+2×4a×5bcosθ           =16a2+25b2+40×a×bcosπ4           =16222+2532+40×22×3×12           =128+225+240           =593 units Therefore, the larger diagonal=593



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Question 11:

If a is a non-zero vector of magnitude 'a' and λ is a non-zero scalar, then λa  is a unit vector if
(a) λ = 1
(b) λ = −1
(c) a = |λ|
(d) a=1λ

Answer:

(d) a=1λ

Given thata=a;Now,λa=1λ a=1λa=1a=1λ

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Question 12:

If θ is the angle between two vectors a and b , then a ·b 0 only when
(a) 0<θ<π2

(b) 0θπ2

(c) 0 < θ < π

(d) 0 ≤ θ ≤ π

Answer:

(b) 0θπ2

a. b0a b cos θ0cos θ00θπ2  

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Question 13:

The values of x for which the angle between a =2x2i^+4xj^+k^, b =7i^-2j^+xk^ is obtuse and the angle between b and the z-axis is acute and less than π6 are
(a) x>12 or x<0

(b) 0<x<12

(c) 12<x<15

(d) Ï•

Answer:

(b) 0<x<12

a =2x2i^+4xj^+k^, b =7i^-2j^+xk^

Let the angle between vector a and vector b be A.

cosA=a .b a b =2x2i^+4xj^+k^.7i^-2j^+xk^2x2i^+4xj^+k^ 7i^-2j^+xk^                         =14x2-8x+x4x4+16x2+149+4+x2                         =14x2-7x4x4+16x2+153+x2Now, A is an obtuse angle.cosA<014x2-7x4x4+16x2+153+x2<014x2-7x<02x2-x<0x2x-1<0x<0 & 2x-1>0  or    x>0 & 2x-1<0x<0 & x>12  or    x>0 & x<12x>0 & x<12         As there cannot be any number less than zero and greater than 1/2x0, 12      ...(i)

Let the equation of the z-axis be zk^.And let the angle between b and z-axis be B.cosB=7i^-2j^+xk^.zk^7i^-2j^+xk^ zk^          =xzz49+4+x2          =x53+x2Now, angle B is acute and less than π/6.0<x53+x2<cosπ60<x<3253+x2  ...(ii)From (i) and (ii) we get0<x<12

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Question 14:

If a , b , c are any three mutually perpendicular vectors of equal magnitude a, then  a +b +c  is equal to
(a) a
(b) 2a
(c) 3a
(d) 2a
(e) None of these

Answer:

(c) 3a

Given thatSo, ab=c=a     ...iSince they are mutually perpendicular,a.b=b.c=c.a=0     ...iiNow,a+b+c2 =a2+b2+c2+2 a. b+2 b.c+2 c.a                  =a2+a2+a2+0+0+0      Using i and ii                  =3a2 a+b+c=3a

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Question 15:

If the vectors 3i^+λj^+k^ and 2i^-j^+8k^ are perpendicular, then λ is equal to
(a) −14
(b) 7
(c) 14
(d) 17

Answer:

(c) 14

It is given that vectors 3i^+λj^+k^ and 2i^-j^+8k^ are perpendicular. So, their dot product is zero.3i^+λj^+k^. 2i^-j^+8k^=06-λ+8=014-λ=0 λ=14

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Question 16:

The projection of the vector i^+j^+k^ along the vector of j^ is
(a) 1
(b) 0
(c) 2
(d) −1
(e) −2

Answer:

(a) 1

Let a=i^+j^+k^ and b= j^The projection of a on b isa. bb=i^+j^+k^.j^j^=0+1+01=1

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Question 17:

The vectors 2i^+3j^-4k^ and ai^+bj^+ck^ are perpendicular if
(a) a = 2, b = 3, c = −4
(b) a = 4, b = 4, c = 5
(c) a = 4, b = 4, c = −5
(d) a = −4, b = 4, c = −5

Answer:

(b) a = 4, b = 4, c = 5
 
It is given that vectors 2i^+3j^-4k^ and ai^+bj^+ck^ are perpendicular. So, their dot product is zero.2a+3b-4c=0b a=4; b=4; c=524+34-45=08+12-20=00=0, which is true.

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Question 18:

If  a = b , then  a +b · a -b =
(a) positive
(b) negative
(c) 0
(d) None of these

Answer:

(c) 0

Given thata=aa+b.a-b=a2-b2                           =a2-a2                           =0

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Question 19:

If a and b are unit vectors inclined at an angle θ, then the value of  a -b  is
(a) 2 sinθ2

(b) 2 sin θ

(c) 2 cosθ2

(d) 2 cos θ

Answer:

(a) 2 sinθ2

a. b=a b cos θ       =1 ×1 cos θ(Because a and b are unit vectors)       =cos θ  ...ia-b2=a2+b2-2 a. b           =1+1-2 cos θ     Using i           =2-2 cos θ            =2 1-cos θ           =2  2 sin2 θ2           =4 sin2 θ2 a-b=2 sin θ2

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Question 20:

If a and b are unit vectors, then the greatest value of 3 a +b + a -b  is
(a) 2
(b) 22
(c) 4
(d) None of these

Answer:

(c) 4

We have3 a +b + a -b =3×  a 2+b 2+2a b cosθ+ a 2+b 2-2a b cosθ=3 × 12+12+2×1×1cosθ+ 12+12-2×1×1 cosθ       As a and b  unit vectors=3 × 2+2 cosθ+ 2-2 cosθ=3 × 21+cosθ+ 21- cosθ=3 × 2×2cos2θ2+ 2×2sin2θ2=23 cosθ2+2sin θ2=23 cosθ2+sin θ2=2×232 cosθ2+12sin θ2=2×2sinπ3 cosθ2+ cosπ3sin θ2=4 sinπ3+ θ2Now, maximum value of sinα=1Maximum value of sinπ3+ θ2=1Maximum value of 4sinπ3+ θ2=4Maximum value of 3 a +b + a -b =4

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Question 21:

If the angle between the vectors xi^+3j^-7k^ and xi^-xj^+4k^ is acute, then x lies in the interval
(a) (−4, 7)
(b) [−4, 7]
(c) R −[−4, 7]
(d) R −(4, 7)

Answer:

(c) R −[−4, 7]

Let θ be the angle between a and b.cos θ=a. ba b=x2-3x-28x2+32+49 x2+x2+42For θ to be acute,cos θ>0x2-3x-28>0x-7x+4>0x-, -4 7, xR--4, 7

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Question 22:

If a and b are two unit vectors inclined at an angle θ, such that  a +b <1, then
(a) θ<π3

(b) θ>2π3

(c) π3<θ<2π3

(d) 2π3<θ<π

Answer:

(d) 2π3<θ<π

We have a +b <1a 2+b 2+2a ×b  cosθ<112+12+2×1×1× cosθ<12+2  cosθ<121+ cosθ<12×2cos2θ2<12cosθ2<1cosθ2<12π3<θ2<2π32π3<θ<4π3But here θ cannot be more than π.2π3<θ<π



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Question 23:

Let a , b , c  be three unit vectors, such that  a +b +c  =1 and a is perpendicular to b . If c makes angles α and β with a and b respectively, then cos α + cos β =
(a) -32

(b) 32

(c) 1

(d) −1

Answer:

(d) −1

Given that ab and c are unit vectors.So, a=1, b=1and c=1.Sincea and b are mutually perpendicular,a.b=0Now,a+b+c=1a+b+c2=1a2+b2+c2+2 a. b+2 b.c+2 c.a=11+1+1+20+2 a b cos β+2 c a cos α=13+2 cos α+cos β=12 cos α+cos β=-2cos α+cos β=-1

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Question 24:

The orthogonal projection of a on b is
(a)  a ·b  a  a 2

(b)  a ·b  b  b 2

(c) a  a 

(d) b  b 

Answer:

(b)  a ·b  b  b 2

The orthogonal projection of a on b isa. b bb2

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Question 25:

If θ is an acute angle and the vector (sin θ)i^ + (cos θ)j^ is perpendicular to the vector i^-3j^, then θ =
(a) π6

(b) π5

(c) π4

(d) π3

Answer:

(d) π3

The given vectors are perpendicular. So, their dot product is zero.sin θ i^+ cos θ j^.i^-3 j^=0sin θ-3 cos θ=0sin θ=3 cos θtan θ=3θ=π3 (Because θ is acute)

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Question 26:

If a and b be two unit vectors and θ the angle between them, then a +b is a unit vector if θ =
(a) π4

(b) π3

(c) π2

(d) 2π3

Answer:

(d) 2π3

We havea=1 and b=1Now,  a+b=1a2+b2+2 a. b=11+1+2 a b cos θ=12+2 cos θ=12 cos θ=-1cos θ=-12θ=2π3



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