RD Sharma XII Vol 2 2017 Solutions for Class 12 Science Math Chapter 6 Vector Or Cross Product are provided here with simple step-by-step explanations. These solutions for Vector Or Cross Product are extremely popular among class 12 Science students for Math Vector Or Cross Product Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 2 2017 Book of class 12 Science Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 2 2017 Solutions. All RD Sharma XII Vol 2 2017 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 25.29:

Question 1:

If a =i^+3j^-2k^ and b =-i^+3k^, find  a ×b .

Answer:

Given:a=i^+3j^-2k^ b=-i^+0j^+3k^ a×b=i^j^k^13-2-103               =9+0 i^-3-2 j^+0+3 k^                =9 i ^-j^+3k^a×b=92+-12+32                =91

Page No 25.29:

Question 2:

(i) If a =3i^+4j^ and b =i^+j^+k^, find the value of  a ×b .

(ii) If a =2i^+k^, b =i^+j^+k^, find the magnitude of a ×b .

Answer:

i Given: a=3i^+4j^+0k^ b=i^+j^+k^ a×b=i^j^k^340111         =4-0 i^ -3-0 j^ +3-4 k^          =4 i^-3j^-k^a×b=16+9+1                =26

ii Given:a=2i^+0j^+k^ b=i^+j^+k^a×b=i^j^k^201111         =0-1 i^ -2-1 j^ +2-0 k^          =- i^-j^+2k^a×b=-12+-12+22                =6

Page No 25.29:

Question 3:

(i) Find a unit vector perpendicular to both the vectors 4i^-j^+3k^ and -2i^+j^-2k^.

(ii) Find a unit vector perpendicular to the plane containing the vectors a =2i^+j^+k^ and b =i^+2j^+k^.

Answer:

i Given: a=4i^-j^+3k^  b=-2i^+j^-2k^ a×b=i^j^k^4-13-21-2          =2-3 i^ - -8+6 j^ +4-2 k^           =- i^+2j^+2k^a×b=1+22+22                =9                =3Unit vector perpendicular to a and b =a×ba×b=- i^+2j^+2k^3

ii Given:a=2i^+j^+k^ b=i^+2j^+k^ a×b=i^j^k^211121               =1-2 i^ -2-1 j^ +4-1k                 =-i^-j^+3k^a×b=1+1+9                 =11Unit vector perpendicular to the plane containing vectors a and b=±a×ba×bUnit vector perpendicular to the plane containing vectors a and b=±111-i^-j^+3k^

Page No 25.29:

Question 4:

Find the magnitude of a =3k^+4j^×i^+j^-k^.

Answer:

a=0i^+4j^+3k^×i^+j^-k^   =i^j^k^04311-1  =i^ -4-3-j^ 0-3+k^ 0-4  =-7i^+3j^-4k^a=-72+32+-42          =74

Page No 25.29:

Question 5:

If a =4i^+3j^+k^ and b =i^-2k^, then find  2b^×a .

Answer:

Given:a=4i^+3j^+k^ 2b=2i^+0j^-4k^2b×a=i^j^k^20-4431           =0+12 i^-2+16 j^+6-0k^            =12i^-18j^+6k^2b×a=122+-182+62                   =504

Page No 25.29:

Question 6:

If a =3i^-j^-2k^ and b =2i^+3j^+k^, find  a +2b × 2a -b .

Answer:

Given:a=3i^-j^-2k^ b=2i^+3j^+k^ a+2b=3i^-j^-2k^+2 2i^+3j^+k^                 =7i^+5j^+0k^ 2a-b=2 3i^-j^-2k^- 2i^+3j^+k^                 =4i^-5j^-5k^a+2b×2a-b=i^j^k^7504-5-5                                 =i^ -25+0-j^ -35+0+k^ -35-20                                 =-25i^+35j^-55k^

Page No 25.29:

Question 7:

(i) Find a vector of magnitude 49, which is perpendicular to both the vectors 2i^+3j^+6k^ and 3i^-6j^+2k^.

(ii) Find a vector whose length is 3 and which is perpendicular to the vector a =3i^+j^-4k^ and b =6i^+5j^-2k^.

Answer:

i Given:a=2i^+3j^+6k^ b=3i^-6j^+2k^a×b=i^j^k^2363-62          =6+36 i^ - 4-18 j^ +-12-9 k^            =42i^+14j^-21k^a×b=422+142+-212                =2401                =49Required vector =49×a×ba×b                              =49 ×42i^+14j^-21k^49                               =42i^+14j^-21k^

ii Given:a=3i^+j^-4k^ b=6i^+5j^-2k^a ×b=i^j^k^31-465-2               = -2+20 i^--6+24 j^ +15-6 k^                =18i^-18j^+9k^a×b=182+-182+92                =729                =27Required vector=3×a×ba×b                            =3×18i^-18j^+9k^27                             =32i^-2j^+k^3                             =2i^-2j^+k^

Page No 25.29:

Question 8:

Find the area of the parallelogram determined by the vectors:
(i) 2i^ and 3j^

(ii) 2i^+j^+3k^ and i^-j^

(iii) 3i^+j^-2k^ and i^-3j^+4k^

(iv) i^-3j^+k^ and i^+j^+k^.

Answer:

i Let:a=2i^+0j^+0k^ b=0i^+3j^+0k^ a×b=i^j^k^200030              =0-0 i^ -0-0 j^ + 6-0 k^              =0 i^+0j^+6k^Area of the parallelogram=a×b                                                                                                      =0+0+62                                              =6 sq. units

ii Let: a =2i^+j^+3k^ b=i^-j^+0k^ a×b=i^j^k^2131-10              =0+3 i^ -0-3 j^ +-2-1 k^               = 3i^+3j^-3k^Area of the parallelogram =a×b                                             =32+32+32                                             =27                                             =33 sq. units

iii Let:a=3i^+j^-2k^b=1i^-3j^+4k^a×b=i^j^k^31-21-34         =i^ 4-6-j^ 12+2+k^ -9-1         =-2i^-14j^-10k^Area of the parallelogram=a×b                                             =-22+-142+-102                                             =300                                             =103 sq. units

iv Let: a=i^-3j^+k^ b=i^+j^+k^a×b=i^j^k^1-31111         =-3-1 i^ - 1-1 j^+1+3 k^          =-4i^+0j^+4k^Area of the parallelogram=a×b                                            =-42+0+42                                            =32                                            =42 sq. units.



Page No 25.30:

Question 9:

Find the area of the parallelogram whose diagonals are:
(i) 4i^-j^-3k^ and -2j^+j^-2k^

(ii) 2i^+k^ and i^+j^+k^

(iii) 3i^+4j^ and i^+j^+k^

(iv) 2i^+3j^+6k^ and 3i^-6j^+2k^

Answer:

i Let:a=4i^-j^-3k^b=-2i^+j^-2k^ a×b=i^j^k^4-1-3-21-2              =2+3 i^ --8-6j^ + 4-2k^              =5i^+14j^+2k^a×b=25+196+4                =225                =15Area of the parallelogram = 12a×b                                              =152 sq. units.

ii Let: a=2i^+0j^+k^ b=i^+j^+k^a×b=i^j^k^201111         =0-1 i^ -2-1 j^ +2-0 k^          =-i^-j^+2k^a×b=-12+-12+2                =6Area of the parallelogram = 12a×b                                              =62 sq. units

iii Let:a=3i^+4j^+0k^ b=i^+j^+k^ a×b=i^j^k^340111              =4-0 i^ -3-0 j^ +3-4 k^               =4i^-3j^-k^a×b=42+-32+-12                =26Area of the parallelogram = 12a×b                                             =262sq. units

iv Let:a=2i^+3j^+6k^b=3i^-6j^+2k^ a×b=i^j^k^2363-62              = 6+36i^-4-18j^ +-12-9k^               =42i^+14j^-21k^a×b=422+142+-212                 =2401                 =49Area of the parallelogram = 12a×b                                             =492 sq. units

Disclaimer: The answer given for (iii) and (iv) in the textbook is incorrect.

Page No 25.30:

Question 10:

If a =2i^+5j^-7k^, b =-3i^+4j^+k^ and c =i^-2j^-3k^, compute  a ×b ×c and a × b ×c  and verify that these are not equal.

Answer:

Given: a=2i^+5j^-7k^ b=-3i^+4j^+k^ c=i^-2j^-3k^ a×b=i^j^k^25-7-341              =5+28 i^ -2-21 j^ +8+15k^               =33i^+19j^+23k^a×b×c=i^j^k^3319231-2-3                        = -57+46i^- -99-23 j^+ -66-19 k^a×b×c =-11i^+122j^-85k^         ...(1) b×c=i^j^k^-3411-2-3               = -12+2 i^- 9-1 j^+6-4 k^               =-10i^-8j^+2k^ a×b×c=i^j^k^25-7-10-82                         =10-56 i^ -4-70 j^ +-16+50 k^  a×b×c=-46i^+66j^+34k^          ...(2)From (1) and (2), we geta×b×ca×b×c

Page No 25.30:

Question 11:

If  a =2, b =5 and  a ×b =8, find a ·b .

Answer:

We know  a.b2+a ×b2=a2b2a.b2+82=22×52          ( a ×b=8, a=2 and b=5)a.b2+64=100a.b2=36a.b=6

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Question 12:

Given a =172i^+3j^+6k^, b =173i^-6j^+2k^, c =176i^+2j^-3k^, i^, j^, k^ being a right handed orthogonal system of unit vectors in space, show that a , b , c is also another system.

Answer:

Given: a=17 2i^+3j^+6k^ b=17 3i^-6j^+2k^ c=176 i^+2j^-3k^a×b=17 17i^j^k^2363-62        =14942 i^+14j^-21k^        =1497 6 i^+2j^-3k^        =176 i^+2j^-3k^        =cb×c=17 17i^j^k^3-6262-3        =14914 i^+21j^+42k^        =1497 2i^+3j^+6k^        =17 2i^+3j^+6k^        =ac×a=17 17i^j^k^62-3236        =14921 i^-42j^+14k^        =1497 3i^-6j^+2k^        =17 3i^-6j^+2k^        =ba=174+9+36   =77   =1b=179+36+4    =77    =1c=1736+4+9   =77   =1Thus, ab and c form a right handed orthogonal system of unit vectors.

Page No 25.30:

Question 13:

If  a =13, b =5 and a . b =60, then find  a ×b .

Answer:

We know a.b2+a ×b2=a2b2602+a ×b2=132×52          ( a.b=60, a=13 and b =5)3600+a ×b2=4225a ×b2=625a ×b=25

Page No 25.30:

Question 14:

Find the angle between two vectors a and b , if  a ×b =a ·b .

Answer:

Let θ be the angle between aand b.Given:a×b=a. ba b sin θ=a b cos θsin θ=cos θtan θ = 1θ=π4

Page No 25.30:

Question 15:

If a ×b =b ×c 0, then show that a +c =mb, where m is any scalar.

Answer:

a×b=b×ca×b=-c×ba×b+c×b=0a+c×b=0           (Using right distributive property)Thus, a+c is parallel to b.a+c=mb, for some scalar m.

Page No 25.30:

Question 16:

If  a =2, b =7 and a ×b =3i^+2j^+6k^, find the angle between a and b .

Answer:

Let θ be the angle between a and b.a×b=3i^+2j^+6k^    (Given)a×b=9+4+36                =7We know a×b=a b sin θ7=2 7 sin θsin θ=12θ=π6

Page No 25.30:

Question 17:

What inference can you draw if a ×b =0  and a ·b =0.

Answer:

Given:a×b=0a=0      b=0  ab     Also,a.b=0a b cos θ=0 a=0 or b=0 or, ab      But a cannot be both perpendicular as well as parallel to b. a=0      b=0

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Question 18:

If a, b, c are three unit vectors such that a ×b =c , b ×c =a, c ×a =b . Show that a, b, c form an orthonormal right handed triad of unit vectors.

Answer:

Given:a×b=cb×c=ac×a=b          ...(1)Now,a×b=c=1      (∵ c is a unit vector)b×c=a=1      (∵ a is a unit vector)c×a=b=1      (∵ b is a unit vector) a×b=b×c=c×a=1        ...(2)From (1) and (2), we know a, b and c form an orthonormal right handed triad of unit vectors.

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Question 19:

Find a unit vector perpendicular to the plane ABC, where the coordinates of A, B and C are A (3, −1, 2), B (1, −1, −3) and C (4, −3, 1).

Answer:

The vector AB×AC is perpendicular to the vectors AB and AC. Required unit vector =AB×ACAB×ACNow,AB =Position vector of B- Position vector of A       =i^-j^-3k^-3i^-j^+2k^       =-2i^+0j^-5kAC  =Position vector of C- Position vector of A        =4i^-3j^+k^-3i^-j^+2k^        =i^-2j^-k^ AB×AC=i^j^k^-20-51-2-1                   =0-10 i-2+5 j+4-0 k^                   =-10i^-7j^+4k^AB×BC=-102+-72+42                 =165Unit vector perpendicular to the plane ABC =AB×ACAB×AC=-10i^-7j^+4k^165

Page No 25.30:

Question 20:

If a, b, c are the lengths of sides, BC, CA and AB of a triangle ABC, prove that BC+CA+AB=0 and deduce that asin A=bsin B=csin C.

Answer:

We haveBC=a CA=bAB=ca=ab=b       ( Length is always positive)c=c  Now,   BC+CA+AB=0           (Given)a+b+c=0a×a+b+c=a×0a×a+a×b+a×c=00+a×b-c×a=0a×b=c×aa bsin C=c a sin Bab sin C=ca sin BDividing both sides by abc, we getsin Cc=sin Bb        ...(1)Again,BC+CA+AB=0a+b+c=0b×a+b+c=b×0b×a+b×b+b×c=0-a×b+0+b×c=0a×b=b×ca b sin C=b c sin Aab sin C=bc sin ADividing both sides by abc, we getsin Cc=sin Aa        ...(2)From (1) and (2), we getsin Aa=sin Bb=sin Ccasin A=bsin B=csin C

Page No 25.30:

Question 21:

If a =i^-2j^+3k^, and b =2i^+3j^-5k^, then find a ×b . Verify that a and a ×b are perpendicular to each other.

Answer:

Given:a=i-2j+3k b=2i+3j-5k

a×b=i^j^k^1-2323-5         =i^+11j^+7k^Now, a. a×b=1-22+21                 =0Thus, a is perpendicular to a×b.

Page No 25.30:

Question 22:

If p and q are unit vectors forming an angle of 30°; find the area of the parallelogram having a =p +2q and b =2p +q as its diagonals.

Answer:

Given: a=p^+2q b= 2p+q^a×b=p+2q×2p+q=2 p×p+p×q+4q×p+2 q×q=20+p×q-4 p×q+2 0=-3p×qArea of the parallelogram=12a×b                                             =12-3p×q                                             =32p q sin 30o                                             =32 1 1 12         (∵ p and q are unit vectors)                                             =34sq. units

Page No 25.30:

Question 23:

For any two vectors a and b , prove that  a ×b 2=a .a a .b b .a b .b .

Answer:

RHS=a. aa. bb. ab. b         =a2a b cos θa b cos θb2        =a2 b2-a2 b2 cos2 θ        =a2 b2 1-cos2 θ        =a2 b2  sin2 θ        =ab sin θ2        =a×b2        =LHSHence proved.

Page No 25.30:

Question 24:

Define a ×b and prove that  a ×b = a .b  tan θ, where θ is the angle between a and b .

Answer:

If a and b are two non-zero non-parallel vectors, then the vector product denoted by a×b is defined as a×b=a b sin θ η^.Here, θ is the angle between a and b and η^ is the unit vector perpendicular to the plane of a and b such that a, b and η^ form a right handed system.LHS=a×b        =a b sin θ         =a b sin θ ×cos θcos θ        =a b cos θ sin θcos θ        =a b cos θ tan θ        =a. b tan θ        =RHSHence proved.

Page No 25.30:

Question 25:

If  a =26, b =7 and  a ×b =35, find a .b .

Answer:

We knowa.b2+a ×b2=a2b2a.b2+352=262×72          ( a×b=35, a=26 and b =7)a.b2+1225=1274a.b2=49a.b=7

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Question 26:

Find the area of the triangle formed by O, A, B when OA=i^+2j^+3k^, OB=-3i^-2j^+k^.

Answer:

Given:OA =i^+2j^+3k^OB=-3i^-2j^+k^ OA×OB=i^j^k^123-3-21               =8i^-10j^+4k^OA×OB=64+100+16                      =180                      =65Area of the triangle=12OA×OB                                 =12 65                                 =3 5 sq. units

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Question 27:

Let a =i^+4j^+2k^, b =3i^-2j^+7k^ and c =2i^-j^+4k^. Find a vector d which is perpendicular to both a and d and c ·d =15.

Answer:

Given:a=i^+4j^+2k^ b=3i^-2j^+7k^c=2i^-j^+4k^Since d is perpendicular to both a and b, it is parallel to a×b.Suppose d=λa×b for some scalar λ.d=λ i^j^k^1423-27   =λ 28+4i^-7-6 j^+-2-12k^   =λ 32i^- j^-14k^c.d=15              (Given)2i^-j^+4k^.λ 32i^- j^-14k^=15λ64+1-56=15λ=53 d=5332i^- j^-14k^d=13160i^- 5j^-70k^

Disclaimer: The question should contain
"which is perpendicular to both a and binstead of "which is perpendicular to both a and d



Page No 25.31:

Question 28:

Find a unit vector perpendicular to each of the vectors a +b and a -b , where a =3i^+2j^+2k^ and b =i^+2j^-2k^.

Answer:

Given: a=3i^+2j^+2k^b=i^+2j^-2k^ a+b=4i^+4j^+0k^     a-b=2i^+0j^+4k^a+b×a-b=i^j^k^440204                            =16i^-16j^-8k^ a+b×a-b=256+256+64                                  =576                                  =24Unit vector perpendicular to both a+b and a-b=a+b×a-ba+b×a-b=16i^-16j^-8k^24=8 2i^-2j^-k^24=132i^-2j^-k^

Page No 25.31:

Question 29:

Using vectors find the area of the triangle with vertices, A (2, 3, 5), B (3, 5, 8) and C (2, 7, 8).

Answer:

Let a, b and c be the position vectors of A, B and C, respectively. Then,a=2i^+3j^+5k^b=3i^+5j^+8k^c=2i^+7j^+8k^Now, AB=b-a     =i^+2j^+3k^AC=c-a      =0i^+4j^+3k^ AB×AC=i^j^k^123043                    =-6i^-3j^+4k^AB×AC=36+9+16                      =61Area of triangle ABC=12AB×AC                                    =612sq. units

Page No 25.31:

Question 30:

If a=2i^-3j^+k^, b=-i^+k^, c=2j^-k^ are three vectors, find the area of the parallelogram having diagonals a+b and b+c.     [CBSE 2014]

Answer:


It is given that a=2i^-3j^+k^, b=-i^+k^, c=2j^-k^.

a+b=2i^-3j^+k^+-i^+k^=i^-3j^+2k^

 b+c=-i^+k^+2j^-k^=-i^+2j^

We know that the area of parallelogram is 12d1×d2, where d1 and d2 are the diagonal vectors.

Now,

a+b×b+c=i^j^k^1-32-120=-4i^-2j^-k^

∴ Area of the parallelogram having diagonals a+b and b+c

 =12a+b×b+c=12-4i^-2j^-k^=12-42+-22+-12=212 square units

Thus, the required area of the parallelogram is 212 square units.

Page No 25.31:

Question 31:

The two adjacent sides of a parallelogram are 2i^-4j^+5k^ and i^-2j^-3k^. Find the unit vector parallel to one of its diagonals. Also, find its area.

Answer:

Suppose  ABCD is the given parallelogram and AC is its diagonal.Let:AB=2i^-4j^+5k^BC=i^-2j^-3k^ Diagonal AC=AB+BC                        =3i^-6j^+2k^AC=9+36+4            =7Unit vector parallel to AC=ACAC                                            =3i^-6j^+2k^7Now,AB×BC=i^j^k^2-451-2-3               =22i^+11j^+0k^AB×AC=484+121                      =605                      =115Area of triangle ABC=12AB×AC                                    =1152sq. units

Page No 25.31:

Question 32:

If either a =0 or b =0 , then a ×b =0 . Is the converse true? Justify your answer with an example.

Answer:

If a=0 or b=0, then a b sin θ n^=0.a×b=0But the converse is not true as whenever a×b=0, we cannot be sure that either a=0 or b=0.For example:a=i^+2j^+3k^b=i^+2j^+3k^Here,a≠0b≠0But a×b=i^j^k^123123                  =0i^+0j^+0k^                  =0

Page No 25.31:

Question 33:

If a =a1i^+a2j^+a3k^, b =b1i^+b2j^+b3k^ and c =c1i^+c2j^+c3k^, then verify that a × b +c =a ×b +a ×c .

Answer:

Given:a=a1i^+a2j^+a3k^b=b1i^+b2j^+b3k^ c=c1i^+c2j^+c3k^b+c=b1+c1 i^+b2+c2 j^+b3+c3 k^ a×b+c=i^j^k^a1a2a3b1+c1b2+c2b3+c3                       =a2b3+a2c3-a3b2-a3c2i^-a1b3+a1c3-a3b1-a3c1j^+ a1b2+a1c2-a2b1-a2c1k^        ...(1)Now, a×b=i^j^k^a1a2a3b1b2b3          = a2b3-a3b2i^- a1b3-a3b1j^+ a1b2-a2b1k^a×c=i^j^k^a1a2a3c1c2c3          =a2c3-a3c2i^ -a1c3-a3c1j^ +a1c2-a2c1k^ a×b+b×c= a2b3+a2c3-a3b2-a3c2i^-a1b3+a1c3-a3b1-a3c1 j^ + a1b2+a1c2-a2b1-a2c1k^      ...(2)From (1) and (2), we geta×b+c=a×b+b×c

Page No 25.31:

Question 34:

Using vectors, find the area of the triangle with vertices:
(i) A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)
(ii) A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1)                                   [CBSE 2011, NCERT EXEMPLAR]
                                                 

Answer:


(i) The vertices of the triangle are A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Position vector of A = i^+j^+2k^

Position vector of B = 2i^+3j^+5k^

Position vector of C = i^+5j^+5k^

AB=2i^+3j^+5k^-i^+j^+2k^=i^+2j^+3k^

AC=i^+5j^+5k^-i^+j^+2k^=4j^+3k^

Now,

AB×AC=i^j^k^123043=-6i^-3j^+4k^

∴ Area of ∆ABC = 12AB×AC

                          =12-6i^-3j^+4k^=12-62+-32+42=1236+9+16=612 square units

(ii) The vertices of the triangle are A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1).

Position vector of A = i^+2j^+3k^

Position vector of B = 2i^-j^+4k^

Position vector of C = 4i^+5j^-k^

AB=2i^-j^+4k^-i^+2j^+3k^=i^-3j^+k^

AC=4i^+5j^-k^-i^+2j^+3k^=3i^+3j^-4k^

Now,

AB×AC=i^j^k^1-3133-4=9i^+7j^+12k^

∴ Area of ∆ABC = 12AB×AC

                          =129i^+7j^+12k^=1292+72+122=1281+49+144=2742 square units

Page No 25.31:

Question 35:

Find all vectors of magnitude 103 that are perpendicular to the plane of i^+2j^+k^ and -i^+3j^+4k^.                  [NCERT EXEMPLAR]

Answer:


Let a=i^+2j^+k^ and b=-i^+3j^+4k^.

Unit vectors perpendicular to both a and b = ±a×ba×b

Now,
a×b=i^j^k^121-134=5i^-5j^+5k^a×b=5i^-5j^+5k^=52+-52+52=75=53

Unit vectors perpendicular to both a and b = ±5i^-5j^+5k^53=±i^-j^+k^3

∴ Required vectors = 103±i^-j^+k^3=±10i^-j^+k^

Thus, the vectors of magnitude 103 that are perpendicular to the plane of i^+2j^+k^ and -i^+3j^+4k^ are ±10i^-j^+k^.



Page No 25.33:

Question 1:

Define vector product of two vectors.

Answer:

If a and b are two non-zero non-parallel vectors, then the vector product denoted by a×b is defined as a×b=a b sin θ η^.Here, θ is the angle between a and b and η^ is the unit vector perpendicular to the plane of a and b such that a, b and η^ form a right handed system.

Page No 25.33:

Question 2:

Write the value i^×j^·k^+i^·j^.

Answer:

i^×j^. k^+i^. j^=k^. k^ +0=k^2+0=12+0           ( k=1)=1

Page No 25.33:

Question 3:

Write the value of i^.j^×k^+j^.k^×i^+k^.j^×i^.

Answer:

i^. j^×k^+j^. k^×i^+k^. j^×i^=i^. i^+j^. j+k^. -k^=i^2+j^2-k^2 =1+1-1                   ( i^=1, j^=1 and k^=1)=1

Page No 25.33:

Question 4:

Write the value of i^.j^×k^+j^.k^×i^+k^.i^×j^.

Answer:

i^. j^×k^+j^. k^×i^+k^. i^×j^=i^. i^+j^. j+k^. k^=i^2+j^2+k^2=1+1+1=3

Page No 25.33:

Question 5:

Write the value of i^×j^+k^+j^×k^+i^+k^×i^+j^.

Answer:

i^× j^+k^+j^× k^+i^+k^× i^+j^=i^× j^+i^× k^+j^× k^+j^× i^+k^× i^+k^× j^=k^-j^+i^-k^+j^-i^=0

Page No 25.33:

Question 6:

Write the expression for the area of the parallelogram having a and b as its diagonals.

Answer:

Given: a and b are diagonals of a parallelogram.Area of the parallelogram = 12 a×b

Page No 25.33:

Question 7:

For any two vectors a and b write the value of  a .b 2+ a ×b 2 in terms of their magnitudes.

Answer:

a. b2+a×b2=a b cos θ2+a b sin θ2=a2 b2 cos2 θ+sin2 θ=a2 b2 1=a2 b2 

Page No 25.33:

Question 8:

If a and b are two vectors of magnitudes 3 and 23 respectively such that a ×b is a unit vector. Write the angle between a and b .

Answer:

Let θ be the angle between aand b.It is given that a×b is a unit vector.a×b=1We know a×b=a b sin θ1=3 23 sin θsin θ=12θ=45o, 135o

Page No 25.33:

Question 9:

If  a =10, b =2 and  a ×b =16, find a .b .

Answer:

We know a.b2+a ×b2=a2b2a.b2+162=102×22          ( a×b=16, a=10 and b=2)a.b2+256=400a.b2=144a.b=±12

Page No 25.33:

Question 10:

For any two vectors a and b , find a . b ×a .

Answer:

Let: a=a1i^+a2j^+a3k^b=b1i^+b2j^+b3k^b×a=i^j^k^b1b2b3a1a2a3         =i^ b2a3-b3a2-j^ b1a3-b3a1+k^ b1a2-b2a1Now,a. b×a=a1i^+a2j^+a3k^. i^ b2a3-b3a2-j^ b1a3-b3a1+k^ b1a2-b2a1=a1b2a3-b3a2-a2 b1a3-b3a1+a3 b1a2-b2a1=a1b2a3-a1b3a2-a2b1a3+a2b3a1+a3b1a2-a3b2a1=0

Page No 25.33:

Question 11:

If a and b are two vectors such that  a ×b =3and a .b =1, find the angle between.

Answer:

a×b=3a b sin θ=3        ...(1)a. b=1a b cos θ=1           ...(2)Dividing (1) by (2), we geta b sin θa b cos θ=3tan θ=3θ=60o

Page No 25.33:

Question 12:

For any three vectors a, b and c write the value of a × b +c +b × c +a +c × a +b .

Answer:

a×b+c+b×c+a+c×a+b=a×b+a×c+b×c+b×a+c×a+c×b=a×b+a×c+b×c-a×b-a×c-b×c=0

Page No 25.33:

Question 13:

For any two vectors a and b , find  a ×b  . b .

Answer:

Let:a=a1i^+a2j^+a3k^b=b1i^+b2j^+b3k^a×b=i^j^k^a1a2a3b1b2b3=i^ a2b3-a3b2-j^ a1b3-a3b1+k^ a1b2-a2b1a×b. b=i^ a2b3-a3b2-j^ a1b3-a3b1+k^ a1b2-a2b1 . b1i^+b2j^+b3k^=b1 a2b3-a3b2 -b2 a1b3-a3b1+b3 a1b2-a2b1=a2b1b3-a3b1b2-a1b2b3+a3b1b2+a1b2b3-a2b1b3=0

Page No 25.33:

Question 14:

Write the value of i^×j^×k^.

Answer:

i^×j^×k^=i^×i^=0

Page No 25.33:

Question 15:

If a =3i^-j^+2k^ and b =2i^+j^-k^, then find  a ×b  a .

Answer:

Since a×b is a vector, a×b a without any dot or cross product in between is meaningless.

Page No 25.33:

Question 16:

Write a unit vector perpendicular to i^+j^ and j^+k^.

Answer:

Let a=i^+j^+0k^; b=0i^+j^+k^a×b=i^j^k^110011         = i^-j^+k^a×b=1+1+1                =3Unit vector perpendicular to a and b is,a×ba×b=13i^-j^+k^

Page No 25.33:

Question 17:

If  a ×b 2+ a .b 2=144 and  a =4, find  b .

Answer:

We know a×b2+a. b2=a2 b2144=42 b2144=16 b2b2=9b=3

Page No 25.33:

Question 18:

If r =xi^+yj^+zk^, then write the value of r ×i^2.

Answer:

Given:r=xi^+yj^+zk^Now,i=i^+0j^+0k^r×i=i^j^k^xyz100         =0 i^+zj^-yk^r×i=z2+y2r×i2=z2+y2

Page No 25.33:

Question 19:

If a and b are unit vectors such that a ×b is also a unit vector, find the angle between a and b .

Answer:

Let θ be the angle between a and b.Given:a×b=1a=1b=1We knowa×b=a b sin θ1=1 1 sin θsin θ=1 θ=π2



Page No 25.34:

Question 20:

If a and b are two vectors such that  a .b = a ×b , write the angle between a and b .

Answer:

Let θ be the angle between a and b.We know a×b=a b sin θa. b=a b cos θNow,a×b=a. b            (Given)a b sin θ=a b cos θsin θ=cos θθ=π4

Page No 25.34:

Question 21:

If a and b are unit vectors, then write the value of  a ×b 2+ a .b 2.

Answer:

It is given that a and b are unit vectors.a=b=1         ...(1)Now,a. b2+a×b2=a b cos θ2+a b sin θ2=a2 b2 cos2 θ+sin2 θ=a2 b2 1=a2 b2 =12 12                 [From (1)]=1

Page No 25.34:

Question 22:

If a is a unit vector such that a ×i^=j^, find a .i^.

Answer:

We know k^×i^=j^                      ...(1)Given: a×i^=j^         ...(2)               Comparing (1) and (2), we geta=k^Now,a. i^=k^. i^      =0

Page No 25.34:

Question 23:

If c is a unit vector perpendicular to the vectors a and b ,write another unit vector perpendicular to a and b .

Answer:

c is a unit vector perpendicular to both a and b.c=a×ba×b⇒-c=b×aa×bTherefore, -c is perpendicular to b and a.Thus,-c is another unit vector perpendicular to a and b.

Page No 25.34:

Question 24:

Find the angle between two vectors a and b with magnitudes 1 and 2 respectively and when  a ×b =3.

Answer:

Let θ be the angle between a and b.We know a×b=a b sin θ3=1 2 sin θsin θ=32θ=π3

Page No 25.34:

Question 25:

Vectors a and b are such that  a =3, b =23and a ×b  is a unit vector. Write the angle between a and b .

Answer:

Given: a×b is a unit vector.a×b=1              ...(1)Let θ be the angle between aand b.We know a×b=a b sin θFrom (1), we get1=3 23 sin θ    sin θ=32θ=π3

Page No 25.34:

Question 26:

Find λ, if 2i^+6j^+14k^×i^-λj^+7k^=0 .

Answer:

Given: i^j^k^26141-λ7=0i^ 42+14λ-0j^+k^ -2λ-6=0i^+0j^+0k^42+14λ=0; -2λ-6=0λ=-3      (This satisfies the above equations) 

Page No 25.34:

Question 27:

Write the value of the area of the parallelogram determined by the vectors 2i^ and 3j^.

Answer:

Let:a=2i^b=3j^a×b=6 i^×j^          =6k^Area of the parallelogram=a×b=6 k^=61=6 sq. units

Page No 25.34:

Question 28:

Write the value of i^×j^·k^+j^+k^·j^

Answer:

i^×j^. k^+j^+k^. j^=k^. k^ +j^. j^+k^. j^     ( i^×j^= k^) =k^2+j^2+0           ( k^. j^=0)=12+12=2

Page No 25.34:

Question 29:

Find a vector of magnitude 171 which is perpendicular to both of the vectors a=i^+2j^-3k^ and b=3i^-j^+2k^.

Answer:


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

Unit vectors perpendicular to both a and b = ±a×ba×b

Now,
a×b=i^j^k^12-33-12=i^-11j^-7k^a×b=i^-11j^-7k^=12+-112+-72=1+121+49=171

Unit vectors perpendicular to both a and b = ±i^-11j^-7k^171

∴ Required vectors = 171±i^-11j^-7k^171=±i^-11j^-7k^

Thus, the vectors of magnitude 171 which are perpendicular to both the given vectors are ±i^-11j^-7k^.

Page No 25.34:

Question 1:

If a is any vector, then a ×i^2+a ×j^2+a ×k^2=
(a) a2

(b) 2a2

(c) 3a2

(d) 4a2

Answer:

(b) 2a2

Let a=a1 i^+a2j^+a3 k^a×i^=i^j^k^a1a2a3100        =a3j^-a2k^ a×i^2=a3j^-a2k^ 2                 =a32j^2+a22k^2-2 a3a2 j^. k^                 =a32+a22                (∵ j^. k^=0)          ...(1) a×j^=i^j^k^a1a2a3010              =-a3i^+a1k^a×j^2=-a3i^+a1k^ 2                  =a32i^2+a12k^2-2 a3a2 i^. k^                  =a32+a12             (∵ i^. k^=0)         ...(2)  a×k^=i^j^k^a1a2a3001              =a2i^-a1j^a×k2=a2i^-a1j^2                  =a22i^2+a12j2+2 a1a2 i^. j^                  =a22+a12             (∵ i^. j^=0)      ...(3)Adding (1), (2) and (3), we geta×i^2+a×j^2+a×k2=a32+a22+a32+a12+a22+a12                                               =2 a12+a22+a32                                              =2 a2                    (∵ a=a12+a22+a32 ) 



Page No 25.35:

Question 2:

If a ·b =a ·c and a ×b =a ×c, a 0, then
(a) b =c 

(b) b =0 

(c) b +c =0 

(d) none of these

Answer:

(a) b =c 

a ·b =a ·c  a  ·b -a ·c =0 a .b -c =0   Let θ be the angle between a and b-c     ab -c cos θ      ...(1)

and  a ×b =a ×c  a ×b -a ×c =0a  ×b -c =0Then , a b -c  sin θ=0      ...(2)Here, it is given that a 0Therefore, for eq (1) and eq (2) to be 0We have , b -c  cos θ=0 For   b -c  cos θ=0 , one of   b -c  or cos θ must be 0Case 1:Let cos  θ=0θ=90°sin θ=1& if  b -c  sin θ=0 and sin θ=1 Then  b -c =0b =cCase 2:Let b -c =0b =cHence, b =c 

Page No 25.35:

Question 3:

The vector b =3i^+4k^ is to be written as the sum of a vector α parallel to a =i^+j^ and a vector β perpendicular to a . Then α =
(a) 32i^+j^

(b) 23i^+j^

(c) 12i^+j^

(d) 13i^+j^

Answer:

(a) 32i^+j^

Let:α=a1i^+a2j^+a3k^β=b1i^+b2j^+b3k^Now,b=3i^+4k^=α+β                                           (Given)3i^+0j^+4k^=a1+b1 i^+a2+b2 j^+a3+b3 k^a1+b1=3; a2+b2=0; a3+b3=4a1+b1=3; a2=-b2; a3+b3=4              ...(1)a=i^+j^                                                           (Given) Also, α is parallel to a.α×a=0i^j^k^a1a2a3110=0-a3i^+a3j^+a1-a2k^=0i^+0j^+0k^a3=0; a1-a2=0 a3=0; a1=a2                             ...(2)Since β is perpendicular to a, we getβ. a=0b1i^+b2j^+b3k^. i^+j^=0b1+b2=0b1=-b2                                             ...(3)Solving (1), (2) and (3), we geta1=32; a2=32; a3=0∴ α=a1i^+a2j^+a3k^      =32i^+32j^+0k^      =32 i^+j^

Page No 25.35:

Question 4:

The unit vector perpendicular to the plane passing through points Pi^-j^+2k^, Q2i^-k^ and R2j^+k^ is
(a) 2i^+j^+k^

(b) 62i^+j^+k^

(c) 162i^+j^+k^

(d) 162i^+j^+k^

Answer:

(c) 162i^+j^+k^


The vector  PQ×PR is perpendicular to the vectors PQ  and PR. Required unit vector==PQ×PRPQ×PRNow, PQ=P.V. of Q-P.V. of P                  =i^+j^-3k^PR=P.V. of R-P.V. of P      =-i^+3j^-k^ PQ×PR=i^j^k^11-3-13-1                    =8i^+4j^+4k^                    =4 2i^+j^+k^PQ×PR=64+16+16                      =96                     =46Required unit vector=PQ×PRPQ×PR                                   =4 2i^+j^+k^46                                   = 162i^+j^+k^

Page No 25.35:

Question 5:

If a, b represent the diagonals of a rhombus, then
(a) a ×b =0 

(b) a ·b =0

(c) a ·b =1

(d) a ×b =a 

Answer:

(b) a ·b =0

We know that the diagonals in a rhombus (a and b) are perpendicular.Therefore, their dot product is zero.a. b=0

Page No 25.35:

Question 6:

Vectors a and b are inclined at angle θ = 120°. If  a =1, b =2, then a +3b ×3a -b 2 is equal to
(a) 300
(b) 325
(c) 275
(d) 225

Answer:

(a) 300

a+3b×3a-b=3 a×a-a×b+9 b×a-3 b×b=3 0-a×b-9 a×b-3 0=-10 a×bNow,a×3b×3a-b 2=-10 a×b2=100 a×b2=100 a2 b2 sin2120=100 12 22 322=400 ×34=300

Page No 25.35:

Question 7:

If a =i^+j^-k^, b =-i^+2j^+2k^ and c =-i^+2j^-k^, then a unit vector normal to the vectors a +b and b -c  is
(a) i^

(b) j^

(c) k^

(d) none of these

Answer:

(a) i^

a+b=0i^+3j^+k^b-c=0i^-0j^+3k^a+b×b-c=i^j^k^031003                            =9i^a+b×b-c=9  i^                              =91                              =9Unit vector perpendicular to both a+b and b-c=a+b×b-ca+b×b-c=9i^9=i^

Page No 25.35:

Question 8:

A unit vector perpendicular to both i^+j^ and j^+k^ is

(a) i^-j^+k^

(b) i^+j^+k^

(c) 13i^+j^+k^

(d) 13i^-j^+k^

Answer:

(d) 13i^-j^+k^

Let:a=i^+j^+0k^ b=0i^+j^+k^  a×b=i^j^k^110011              = i^-j^+k^a×b=1+1+1                =3Unit vector perpendicular to a and b=a×ba×b=i^-j^+k^3


Disclaimer: The answer given for this question in the textbook is incorrect.

Page No 25.35:

Question 9:

If a =2i^-3j^-k^ and b =i^+4j^-2k^, then a ×b is
(a) 10i^+2j^+11k^

(b) 10i^+3j^+11k^

(c) 10i^-3j^+11k^

(d) 10i^-2j^-10k^

Answer:

(b) 10i^+3j^+11k^

a×b=i^j^k^2-3-114-2         =10 i ^+3j^+11k^

Page No 25.35:

Question 10:

If i^, j^, k^ are unit vectors, then
(a) i^·j^=1

(b) i^·i^=1

(c) i^×j^=1

(d) i^×j^×k^=1

Answer:

(b)  i^·i^=1

Let us check each option one by one.(a) We knowi^. j^=0    1(b) We knowi^. i^=i^2     =12     =1(c) i^×j^ =k^              1(d) i^ ×j^×k^= i^ ×i^                      =0                      1

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Question 11:

If θ is the angle between the vectors 2i^-2j^+4k^ and 3i^+j^+2k^, then sin θ =
(a) 23

(b) 27

(c) 27

(d) 27

Answer:

(b) 27

Let:a=2i^-2j^+4k^b=3i^+j^+2k^a=22+-22+42         =4+4+16         =24         =26 b=32+12+22        =9+1+4        =14a×b=i^j^k^2-24312          =-8i^+8j^+8k^a×b=64+64+64           =192           =8 3Let θ be the angle between a and b.a×b=a b sin θ 8 3=2614 sin θ sin θ= 8 3421               =27θ=sin-127

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Question 12:

If  a ×b =4, a ·b =2, then  a 2 b 2=

(a) 6
(b) 2
(c) 20
(d) 8

Answer:

(c) 20

We know a. b2+a×b2=a2 b2           ...(1)a. b=2                                         (Given)a. b2=a. b2From (1), we get22+42=a2 b2 a2 b2 =20



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Question 13:

The value of  a ×b 2 is

(a)  a 2+ b 2- a ·b 2

(b)  a 2  b 2- a ·b 2

(c)  a 2+ b 2-2 a ·b 

(d)  a 2+ b 2-a ·b 

Answer:

(b)  a 2  b 2- a ·b 2

a. b2+a×b2=a b cos θ2+a b sin θ2=a2 b2 cos2 θ+sin2 θ=a2 b2 1=a2 b2  a×b2=a2 b2 -a. b2Thus, the value of a×b2 is a2 b2 -a. b2.

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Question 14:

The value of i^·j^×k^+j^·i^×k^+k^·i^×j^, is
(a) 0
(b) −1
(c) 1
(d) 3

Answer:

 (c) 1

i^. j^×k^+j^. i^×k^+k^. i^×j^=i^. i^+j^. -j^+k^. k^=i^2-j^2+k^2=1-1+1=1

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Question 15:

If θ is the angle between any two vectors a and b , then  a · b = a ×b  when θ is equal to

(a) 0
(b) π/4
(c) π/2
(d) π

Answer:

(b)  π/4

Let θ be the angle between aand b.We know a×b=a b sin θa. b=a b cos θa. b=a b cos θ=a' b cos θGiven: a. b=a×ba b cos θ=a b sin θcos θ=sin θθ=π4



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