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#### Question 12:

If  are three mutually perpendicular vectors then $\left|\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}\right|$ = _______________.

Given:
are three mutually perpendicular vectors
then,

Hence, $\left|\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}\right|=\overline{)\sqrt{{\left|\stackrel{\to }{a}\right|}^{2}+{\left|\stackrel{\to }{b}\right|}^{2}+{\left|\stackrel{\to }{c}\right|}^{2}}}$.

#### Question 13:

If  are non-zero vectors such that  ,then the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is _________________.

Given:

Hence, the angle between  is $\overline{)\mathrm{\pi }}$.

#### Question 14:

For any vector  = ________________ .

Hence, ${\left(\stackrel{\to }{r}·\stackrel{^}{i}\right)}^{2}+{\left(\stackrel{\to }{r}·\stackrel{^}{j}\right)}^{2}+{\left(\stackrel{\to }{r}·\stackrel{^}{k}\right)}^{2}=\overline{){\left|\stackrel{\to }{r}\right|}^{2}}$.

#### Question 15:

If  are non-zero vectors of same magnitude such that the angle between

Given:
$\left|\stackrel{\to }{a}\right|=\left|\stackrel{\to }{b}\right|$        ...(1)
The angle between the vectors       ...(2)

$\stackrel{\to }{a}·\stackrel{\to }{b}=-8$       ....(3)

Hence, $\left|\stackrel{\to }{a}\right|=\overline{)4}$.

#### Question 16:

For any non-zero vectors $\stackrel{\to }{r},$ the expression  equals ___________.

Hence, the expression  equals $\overline{)\stackrel{\to }{r}}$.

#### Question 17:

The vectors  are the adjacent sides of a parallelogram. The acture angle between its diagonals is _____________.

Given:
vectors  are the adjacent sides of a parallelogram

Hence, the acute angle between its diagonals is $\overline{)\frac{\mathrm{\pi }}{4}}$.

Find when
(i)

(ii)

(iii)

#### Question 2:

For what value of λ are the vectors perpendicular to each other if
(i)

(ii)

(iii)

(iv)

#### Question 3:

If are two vectors such that , find the angle between

#### Question 5:

Find the angle between the vectors , where
(i)

(ii)

(iii)

(iv)

(v)

#### Question 6:

Find the angles which the vector makes with the coordinate axes.

#### Question 7:

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

#### Question 8:

If are unit vectors inclined at an angle θ, prove that
(i)

(ii)

#### Question 9:

If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is $\sqrt{3}$.

#### Question 10:

If are three mutually perpendicular unit vectors, then prove that .

If

#### Question 12:

Show that the vector $\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}$ is equally inclined to the coordinate axes.

#### Question 13:

Show that the vectors are mutually perpendicular unit vectors.

#### Question 14:

For any two vectors , show that .

#### Question 15:

If $\stackrel{\to }{a}=2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$$\stackrel{\to }{b}=\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}$ and $\stackrel{\to }{c}=\stackrel{^}{i}+3\stackrel{^}{j}-\stackrel{^}{k}$, find λ such that $\stackrel{\to }{a}$ is perpendicular to $\lambda \stackrel{\to }{b}+\stackrel{\to }{c}$.                 [NCERT EXEMPLAR]

The given vectors are $\stackrel{\to }{a}=2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$$\stackrel{\to }{b}=\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}$ and $\stackrel{\to }{c}=\stackrel{^}{i}+3\stackrel{^}{j}-\stackrel{^}{k}$.

Now,

$\lambda \stackrel{\to }{b}+\stackrel{\to }{c}=\lambda \left(\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}\right)+\left(\stackrel{^}{i}+3\stackrel{^}{j}-\stackrel{^}{k}\right)=\left(\lambda +1\right)\stackrel{^}{i}+\left(\lambda +3\right)\stackrel{^}{j}-\left(2\lambda +1\right)\stackrel{^}{k}$

It is given that

$\stackrel{\to }{a}\perp \left(\lambda \stackrel{\to }{b}+\stackrel{\to }{c}\right)\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{a}.\left(\lambda \stackrel{\to }{b}+\stackrel{\to }{c}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right).\left[\left(\lambda +1\right)\stackrel{^}{i}+\left(\lambda +3\right)\stackrel{^}{j}-\left(2\lambda +1\right)\stackrel{^}{k}\right]=0\phantom{\rule{0ex}{0ex}}⇒2\left(\lambda +1\right)-\left(\lambda +3\right)-\left(2\lambda +1\right)=0\phantom{\rule{0ex}{0ex}}⇒2\lambda +2-\lambda -3-2\lambda -1=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-2$

Thus, the value of λ is −2.

#### Question 16:

If then find the value of λ, so that and are perpendicular vectors.

#### Question 17:

If then express in the form of where is parallel to is perpendicular to .

#### Question 18:

If either , then But the converse need not be true. Justify your answer with an example.

#### Question 19:

Show that the vectors form a right-angled triangle.

#### Question 20:

If are such that is perpendicular to , then find the value of λ.

#### Question 21:

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

#### Question 22:

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

#### Question 23:

Show that the points whose position vectors are form a right triangle.

#### 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?

#### 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.

#### Question 26:

Find the projection of , where

#### Question 27:

If then show that the vectors are orthogonal.

#### Question 28:

A unit vector makes angles $\frac{\mathrm{\pi }}{4}\mathrm{and}\frac{\mathrm{\pi }}{3}$ with $\stackrel{^}{i}$ and $\stackrel{^}{j}$ respectively and an acute angle θ with $\stackrel{^}{k}$. Find the angle θ and components of .

#### Question 29:

If two vectors are such that then find the value of

#### Question 30:

If is a unit vector, then find in each of the following.
(i)

(ii)

Find if
(i)

(ii)

(iii)

Find if
(i)

(ii)

(iii)

#### Question 33:

Find the angle between two vectors if
(i)

(ii)

#### Question 34:

Express the vector as the sum of two vectors such that one is parallel to the vector and other is perpendicular to .

#### Question 35:

If are two vectors of the same magnitude inclined at an angle of 30°, such that

#### Question 36:

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

#### Question 37:

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

#### Question 38:

Let Find λ such that is orthogonal to .

#### Question 39:

If what can you conclude about the vector ?

#### Question 40:

If is perpendicular to both , then prove that it is perpendicular to both .

If prove that

#### Question 42:

If are three non-coplanar vectors, such that then show that is the null vector.

Given that: $\stackrel{\to }{d}·\stackrel{\to }{a}=0$
so, either $\stackrel{\to }{d}$ = 0 or $\stackrel{\to }{d}\perp \stackrel{\to }{a}$

similarly, $\stackrel{\to }{d}·\stackrel{\to }{b}=0$
so, $\stackrel{\to }{d}$ = 0 or $\stackrel{\to }{d}\perp \stackrel{\to }{b}$

Also, $\stackrel{\to }{d}·\stackrel{\to }{c}=0$
so, $\stackrel{\to }{d}$ = 0 or $\stackrel{\to }{d}\perp \stackrel{\to }{c}$

But $\stackrel{\to }{d}$ cannot be perpendicular to  as   are non-coplanar.

so, $\stackrel{\to }{d}$=0. $\stackrel{\to }{d}$ is a null vector.

#### Question 43:

If a vector is perpendicular to two non-collinear vectors is perpendicular to every vector in the plane of

#### Question 44:

If show that the angle θ between the vectors is given by

cos θ =

#### Question 45:

Let be vectors such If then find

#### Question 46:

Let be three vectors. Find the values of x for which the angle between is acute and the angle between is obtuse.

#### Question 47:

Find the values of x and y if the vectors are mutually perpendicular vectors of equal magnitude.

#### Question 48:

If are two non-collinear unit vectors such that find

#### Question 49:

If , are two vectors such that , then prove that is perpendicular to .

#### Question 50:

Let a and b be unit vectors. If the vectors $\stackrel{\to }{c}=\stackrel{^}{a}+2\stackrel{^}{b}$ and $\stackrel{\to }{d}=5\stackrel{^}{a}-4\stackrel{^}{b}$ are perpendicular to each other, then find the angle between the vector a and b.

Given:
Since,  are perpendicular to each other
$\stackrel{\to }{c}·\stackrel{\to }{d}=0$
$⇒\left(\stackrel{^}{a}+2\stackrel{^}{b}\right)·\left(5\stackrel{^}{a}-4\stackrel{^}{b}\right)=0\phantom{\rule{0ex}{0ex}}⇒5-4\stackrel{^}{a}·\stackrel{^}{b}+10\stackrel{^}{a}·\stackrel{^}{b}-8=0\phantom{\rule{0ex}{0ex}}⇒6\stackrel{^}{a}·\stackrel{^}{b}=3\phantom{\rule{0ex}{0ex}}⇒\stackrel{^}{a}\stackrel{^}{b}=\frac{1}{2}$

$⇒\theta =\frac{\mathrm{\pi }}{3}$
Hence, the angle between $\stackrel{^}{a}$ and $\stackrel{^}{b}$ is $\frac{\mathrm{\pi }}{3}$.

#### Question 1:

In a triangle OAB, $\angle$AOB = 90º. If P and Q are points of trisection of AB, prove that ${\mathrm{OP}}^{2}+{\mathrm{OQ}}^{2}=\frac{5}{9}{\mathrm{AB}}^{2}$.

In triangle OAB, $\angle$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 $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$, 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, $\stackrel{\to }{\mathrm{OP}}=\frac{2\stackrel{\to }{a}+\stackrel{\to }{b}}{3}$                (Using section formula)
Position vector of Q, $\stackrel{\to }{\mathrm{OQ}}=\frac{\stackrel{\to }{a}+2\stackrel{\to }{b}}{3}$

Also, $\stackrel{\to }{\mathrm{OA}}\perp \stackrel{\to }{\mathrm{OB}}$.

$\therefore \stackrel{\to }{a}.\stackrel{\to }{b}=0$                .....(1)

Now,

${\mathrm{OP}}^{2}+{\mathrm{OQ}}^{2}\phantom{\rule{0ex}{0ex}}={\left|\stackrel{\to }{\mathrm{OP}}\right|}^{2}+{\left|\stackrel{\to }{\mathrm{OQ}}\right|}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{2\stackrel{\to }{a}+\stackrel{\to }{b}}{3}\right).\left(\frac{2\stackrel{\to }{a}+\stackrel{\to }{b}}{3}\right)+\left(\frac{\stackrel{\to }{a}+2\stackrel{\mathit{\to }}{b}}{3}\right).\left(\frac{\stackrel{\mathit{\to }}{a}+2\stackrel{\to }{b}}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{4{\left|\stackrel{\to }{a}\right|}^{2}+4\stackrel{\to }{a}.\stackrel{\mathit{\to }}{b}+{\left|\stackrel{\to }{b}\right|}^{2}+{\left|\stackrel{\mathit{\to }}{a}\right|}^{2}+4\stackrel{\to }{a}.\stackrel{\mathit{\to }}{b}+4{\left|\stackrel{\to }{b}\right|}^{2}}{9}$

#### Question 2:

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

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 $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$, respectively. Then,

$\stackrel{\to }{\mathrm{OA}}=\stackrel{\to }{a}$ and $\stackrel{\to }{\mathrm{OB}}=\stackrel{\to }{b}$
Position vector of mid-point of AB, $\stackrel{\to }{\mathrm{OE}}=\frac{\stackrel{\to }{a}+\stackrel{\to }{b}}{2}$

∴ Position vector of C, $\stackrel{\to }{\mathrm{OC}}=\stackrel{\to }{a}+\stackrel{\to }{b}$

By the triangle law of vector addition, we have

$\stackrel{\to }{\mathrm{OA}}+\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{\mathrm{OB}}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{\mathrm{OB}}-\stackrel{\to }{\mathrm{OA}}=\stackrel{\to }{b}-\stackrel{\to }{a}$

Since $\stackrel{\to }{\mathrm{AB}}\perp \stackrel{\to }{\mathrm{OC}}$,

$⇒\stackrel{\to }{\mathrm{AB}}.\stackrel{\to }{\mathrm{OC}}=0\phantom{\rule{0ex}{0ex}}⇒\left(\stackrel{\to }{b}-\stackrel{\to }{a}\right).\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)=0\phantom{\rule{0ex}{0ex}}⇒{\left|\stackrel{\to }{b}\right|}^{2}-{\left|\stackrel{\to }{a}\right|}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{\left|\stackrel{\to }{a}\right|}^{2}={\left|\stackrel{\to }{b}\right|}^{2}\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{a}\right|=\left|\stackrel{\to }{b}\right|\phantom{\rule{0ex}{0ex}}⇒\mathrm{OA}=\mathrm{OB}$

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

#### 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.

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

$\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{b}$ and $\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{c}$

Since $\stackrel{\to }{\mathrm{AB}}\perp \stackrel{\to }{\mathrm{AC}}$ $⇒\stackrel{\to }{b}.\stackrel{\to }{c}=0$          .....(1)

Now,

${\left|\stackrel{\to }{\mathrm{AB}}\right|}^{2}+{\left|\stackrel{\to }{\mathrm{AC}}\right|}^{2}={\left|\stackrel{\to }{b}\right|}^{2}+{\left|\stackrel{\to }{c}\right|}^{2}$              .....(2)

Also,

From (2) and (3), we have

${\left|\stackrel{\to }{\mathrm{AB}}\right|}^{2}+{\left|\stackrel{\to }{\mathrm{AC}}\right|}^{2}={\left|\stackrel{\to }{\mathrm{BC}}\right|}^{2}$

#### 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.

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 $\stackrel{\to }{b}$ and $\stackrel{\to }{d}$, respectively. Then,

$\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{b}$ and $\stackrel{\to }{\mathrm{AD}}=\stackrel{\to }{d}$

Using triangle law of vector addition, we have

$\stackrel{\to }{\mathrm{AD}}+\stackrel{\to }{\mathrm{DB}}=\stackrel{\to }{\mathrm{AB}}\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{\mathrm{DB}}=\stackrel{\to }{b}-\stackrel{\to }{d}$

In ∆ABC,

$\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}=\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{AD}}=\stackrel{\to }{b}+\stackrel{\to }{d}$

Now,

Also,

From (1) and (2), we have

${\left|\stackrel{\to }{\mathrm{AB}}\right|}^{2}+{\left|\stackrel{\to }{\mathrm{BC}}\right|}^{2}+{\left|\stackrel{\to }{\mathrm{CD}}\right|}^{2}+{\left|\stackrel{\to }{\mathrm{DA}}\right|}^{2}={\left|\stackrel{\to }{\mathrm{DB}}\right|}^{2}+{\left|\stackrel{\to }{\mathrm{AC}}\right|}^{2}$

#### Question 5:

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

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

Now,

$\stackrel{\to }{\mathrm{PQ}}=\stackrel{\to }{\mathrm{PB}}+\stackrel{\to }{\mathrm{BQ}}=\frac{1}{2}\stackrel{\to }{\mathrm{AB}}+\frac{1}{2}\stackrel{\to }{\mathrm{BC}}=\frac{1}{2}\left(\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}\right)=\frac{1}{2}\stackrel{\to }{\mathrm{AC}}$              .....(1)

$\stackrel{\to }{\mathrm{SR}}=\stackrel{\to }{\mathrm{SD}}+\stackrel{\to }{\mathrm{DR}}=\frac{1}{2}\stackrel{\to }{\mathrm{AD}}+\frac{1}{2}\stackrel{\to }{\mathrm{DC}}=\frac{1}{2}\left(\stackrel{\to }{\mathrm{AD}}+\stackrel{\to }{\mathrm{DC}}\right)=\frac{1}{2}\stackrel{\to }{\mathrm{AC}}$            .....(2)

From (1) and (2), we have

$\stackrel{\to }{\mathrm{PQ}}=\stackrel{\to }{\mathrm{SR}}$

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

Now,

Also,

From (3) and (4), we have

${\left|\stackrel{\to }{\mathrm{PQ}}\right|}^{2}={\left|\stackrel{\to }{\mathrm{PS}}\right|}^{2}\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{\mathrm{PQ}}\right|=\left|\stackrel{\to }{\mathrm{PS}}\right|$

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

#### Question 6:

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

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 $\stackrel{\to }{a}$ and $\stackrel{\to }{c}$, respectively. Then,

$\stackrel{\to }{\mathrm{OA}}=\stackrel{\to }{a}$ and $\stackrel{\to }{\mathrm{OC}}=\stackrel{\to }{c}$

In ∆OAB,

$\stackrel{\to }{\mathrm{OB}}=\stackrel{\to }{\mathrm{OA}}+\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{\mathrm{OA}}+\stackrel{\to }{\mathrm{OC}}=\stackrel{\to }{a}+\stackrel{\to }{c}$                  $\left(\stackrel{\to }{\mathrm{AB}}=\stackrel{\to }{\mathrm{OC}}\right)$

Position vector of mid-point of $\stackrel{\to }{\mathrm{OB}}=\frac{1}{2}\left(\stackrel{\to }{a}+\stackrel{\to }{c}\right)$

Position vector of mid-point of $\stackrel{\to }{\mathrm{AC}}=\frac{1}{2}\left(\stackrel{\to }{a}+\stackrel{\to }{c}\right)$                (Mid-point formula)

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

Now,

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

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

#### Question 7:

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

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

Suppose the position vectors of points B and D be $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$, respectively.

Now,

$\stackrel{\to }{\mathrm{AC}}=\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{BC}}=\stackrel{\to }{\mathrm{AB}}+\stackrel{\to }{\mathrm{AD}}=\stackrel{\to }{a}+\stackrel{\to }{b}$

Also, $\stackrel{\to }{\mathrm{BD}}=\stackrel{\to }{a}-\stackrel{\to }{b}$

Since ABCD is rectangle, so $\stackrel{\to }{\mathrm{AB}}\perp \stackrel{\to }{\mathrm{AD}}$.

$\therefore \stackrel{\to }{a}.\stackrel{\to }{b}=0$               .....(1)

Now, diagonals AC and BD are perpendicular

iff $\stackrel{\to }{\mathrm{AC}}.\stackrel{\to }{\mathrm{BD}}=0$

iff ABCD is a square

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

#### Question 8:

If AD is the median of ∆ABC, using vectors, prove that ${\mathrm{AB}}^{2}+{\mathrm{AC}}^{2}=2\left({\mathrm{AD}}^{2}+{\mathrm{CD}}^{2}\right)$.

Taking A as the origin, let the position vectors of B and C be $\stackrel{\to }{b}$ and $\stackrel{\to }{c}$, respectively.

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

∴ Position vector of mid-point of BC = $\stackrel{\to }{\mathrm{AD}}=\frac{\stackrel{\to }{b}+\stackrel{\to }{c}}{2}$             (Mid-point formula)

Now,

${\mathrm{AB}}^{2}+{\mathrm{AC}}^{2}={\left|\stackrel{\to }{\mathrm{AB}}\right|}^{2}+{\left|\stackrel{\to }{\mathrm{AC}}\right|}^{2}={\left|\stackrel{\to }{b}\right|}^{2}+{\left|\stackrel{\to }{c}\right|}^{2}$         .....(1)

Also,

From (1) and (2), we have

${\mathrm{AB}}^{2}+{\mathrm{AC}}^{2}=2\left({\mathrm{AD}}^{2}+{\mathrm{CD}}^{2}\right)$

#### Question 9:

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

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 $\stackrel{\to }{b}$ and $\stackrel{\to }{c}$, respectively. Then,

Position vector of D = $\frac{\stackrel{\to }{b}+\stackrel{\to }{c}}{2}$               (Mid-point formula)

Now,

$\stackrel{\to }{\mathrm{AD}}$ = Position vector of D − Position vector of A = $\frac{\stackrel{\to }{b}+\stackrel{\to }{c}}{2}$

$\stackrel{\to }{\mathrm{BC}}$ = Position vector of C − Position vector of B = $\stackrel{\to }{c}-\stackrel{\to }{b}$

Since $\stackrel{\to }{\mathrm{AD}}\perp \stackrel{\to }{\mathrm{BC}}$,

$\therefore \stackrel{\to }{\mathrm{AD}}.\stackrel{\to }{\mathrm{BC}}=0\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left(\stackrel{\to }{b}+\stackrel{\to }{c}\right).\left(\stackrel{\to }{c}-\stackrel{\to }{b}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\stackrel{\to }{c}+\stackrel{\to }{b}\right).\left(\stackrel{\to }{c}-\stackrel{\to }{b}\right)=0\phantom{\rule{0ex}{0ex}}⇒{\left|\stackrel{\to }{c}\right|}^{2}-{\left|\stackrel{\to }{b}\right|}^{2}=0\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{c}\right|=\left|\stackrel{\to }{b}\right|\phantom{\rule{0ex}{0ex}}⇒\mathrm{AC}=\mathrm{AB}$

Hence, the ∆ABC is an isosceles triangle.

#### Question 10:

In a quadrilateral ABCD, prove that ${\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}+{\mathrm{CD}}^{2}+{\mathrm{DA}}^{2}={\mathrm{AC}}^{2}+{\mathrm{BD}}^{2}+4{\mathrm{PQ}}^{2}$, where P and Q are middle points of diagonals AC and BD.

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

Position vector of P = $\frac{\stackrel{\to }{c}}{2}$                       (Mid-point formula)

Position vector of Q = $\frac{\stackrel{\to }{b}+\stackrel{\to }{d}}{2}$                (Mid-point formula)

Now,

Also,

From (1) and (2), we have

${\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}+{\mathrm{CD}}^{2}+{\mathrm{DA}}^{2}={\mathrm{AC}}^{2}+{\mathrm{BD}}^{2}+4{\mathrm{PQ}}^{2}$

#### Question 1:

The vectors satisfy the equations If θ is the angle between , then
(a)

(b)

(c)

(d)

(c)

#### Question 2:

If then
(a)

(b) $\stackrel{^}{i}$

(c) $\stackrel{^}{j}$

(d) $\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}$

(b) $\stackrel{^}{i}$

#### Question 3:

If then the angle between is
(a) $\frac{\mathrm{\pi }}{6}$

(b) $\frac{2\mathrm{\pi }}{3}$

(c) $\frac{5\mathrm{\pi }}{3}$

(d) $\frac{\mathrm{\pi }}{3}$

(d) $\frac{\mathrm{\pi }}{3}$

#### Question 4:

Let be two unit vectors and α be the angle between them. Then, is a unit vector if
(a) $\mathrm{\alpha }=\frac{\mathrm{\pi }}{4}$

(b) $\mathrm{\alpha }=\frac{\mathrm{\pi }}{3}$

(c) $\mathrm{\alpha }=\frac{2\mathrm{\pi }}{3}$

(d) $\mathrm{\alpha }=\frac{\mathrm{\pi }}{2}$

(c) $\mathrm{\alpha }=\frac{2\mathrm{\pi }}{3}$

#### Question 5:

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

(b) unit vector

#### Question 6:

If the position vectors of P and Q are then the cosine of the angle between $\stackrel{\to }{PQ}$ and y-axis is
(a) $\frac{5}{\sqrt{162}}$

(b) $\frac{4}{\sqrt{162}}$

(c) $-\frac{5}{\sqrt{162}}$

(d) $\frac{11}{\sqrt{162}}$

(c) $-\frac{5}{\sqrt{162}}$

#### Question 7:

If are unit vectors, then which of the following values of is not possible?
(a) $\sqrt{3}$

(b) $\sqrt{3}/2$

(c) $1/\sqrt{2}$

(d) −1/2

(a) $\sqrt{3}$

#### Question 8:

If the vectors are perpendicular, then the locus of (x, y) is
(a) a circle
(b) an ellipse
(c) a hyperbola
(d) None of these

(b) an ellipse

#### Question 9:

The vector component of perpendicular to is
(a)

(b)

(c)

(d) None of these

(b)

#### Question 10:

What is the length of the longer diagonal of the parallelogram constructed on if it is given that and the angle between is π/4?
(a) 15
(b) $\sqrt{113}$
(c) $\sqrt{593}$
(d) $\sqrt{369}$

(c) $\sqrt{593}$

#### Question 11:

If is a non-zero vector of magnitude 'a' and λ is a non-zero scalar, then λ is a unit vector if
(a) λ = 1
(b) λ = −1
(c) a = |λ|
(d) $a=\frac{1}{\left|\lambda \right|}$

(d) $a=\frac{1}{\left|\lambda \right|}$

#### Question 12:

If θ is the angle between two vectors only when
(a) $0<\mathrm{\theta }<\frac{\mathrm{\pi }}{2}$

(b) $0\le \mathrm{\theta }\le \frac{\mathrm{\pi }}{2}$

(c) 0 < θ < π

(d) 0 ≤ θ ≤ π

(b) $0\le \mathrm{\theta }\le \frac{\mathrm{\pi }}{2}$

#### Question 13:

The values of x for which the angle between is obtuse and the angle between and the z-axis is acute and less than $\frac{\mathrm{\pi }}{6}$ are
(a)

(b) $0

(c) $\frac{1}{2}

(d) ϕ

(b) $0

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

#### Question 14:

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

(c) $\sqrt{3}a$

#### Question 15:

If the vectors are perpendicular, then λ is equal to
(a) −14
(b) 7
(c) 14
(d) $\frac{1}{7}$

(c) 14

#### Question 16:

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

(a) 1

#### Question 17:

The vectors $2\stackrel{^}{i}+3\stackrel{^}{j}-4\stackrel{^}{k}$ and $a\stackrel{^}{i}+b\stackrel{^}{j}+c\stackrel{^}{k}$ 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

(b) a = 4, b = 4, c = 5

#### Question 18:

If
(a) positive
(b) negative
(c) 0
(d) None of these

(c) 0

#### Question 19:

If are unit vectors inclined at an angle θ, then the value of is
(a)

(b) 2 sin θ

(c)

(d) 2 cos θ

(a)

#### Question 20:

If are unit vectors, then the greatest value of is
(a) 2
(b) $2\sqrt{2}$
(c) 4
(d) None of these

(c) 4

#### Question 21:

If the angle between the vectors is acute, then x lies in the interval
(a) (−4, 7)
(b) [−4, 7]
(c) R −[−4, 7]
(d) R −(4, 7)

(c) R −[−4, 7]

#### Question 22:

If are two unit vectors inclined at an angle θ, such that then
(a) $\mathrm{\theta }<\frac{\mathrm{\pi }}{3}$

(b) $\mathrm{\theta }>\frac{2\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}$

(c) $\frac{\mathrm{\pi }}{3}<\mathrm{\theta }<\frac{2\mathrm{\pi }}{3}$

(d) $\frac{2\mathrm{\pi }}{3}<\mathrm{\theta }<\mathrm{\pi }$

(d) $\frac{2\mathrm{\pi }}{3}<\mathrm{\theta }<\mathrm{\pi }$

#### Question 23:

Let be three unit vectors, such that =1 and is perpendicular to . If makes angles α and β with respectively, then cos α + cos β =
(a) $-\frac{3}{2}$

(b) $\frac{3}{2}$

(c) 1

(d) −1

(d) −1

#### Question 24:

The orthogonal projection of is
(a)

(b)

(c)

(d)

(b)

#### Question 25:

If θ is an acute angle and the vector (sin θ)$\stackrel{^}{i}$ + (cos θ)$\stackrel{^}{j}$ is perpendicular to the vector $\stackrel{^}{i}-\sqrt{3}\stackrel{^}{j},$ then θ =
(a) $\frac{\mathrm{\pi }}{6}$

(b) $\frac{\mathrm{\pi }}{5}$

(c) $\frac{\mathrm{\pi }}{4}$

(d) $\frac{\mathrm{\pi }}{3}$

(d) $\frac{\mathrm{\pi }}{3}$

#### Question 26:

The angle between two vectors  with magnitudes $\sqrt{3}$ and 4 respectively and

Given:
$\left|\stackrel{\to }{a}\right|=\sqrt{3}\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{b}\right|=4\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}·\stackrel{\to }{b}=2\sqrt{3}$

$\stackrel{\to }{a}·\stackrel{\to }{b}=2\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|\mathrm{cos}\theta =2\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}×4×\mathrm{cos}\theta =2\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{2\sqrt{3}}{4\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{\mathrm{\pi }}{3}$

Hence, the correct option is (b).

#### Question 27:

If  are unit vectors such that $\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}=\stackrel{\to }{0},$ then the value of
(a) 1
(b) 3
(c) $-\frac{3}{2}$
(d) none of these

Given:
are unit vectors
$\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}=\stackrel{\to }{0}$

Hence, the correct option is (c).

#### Question 28:

If $\stackrel{\to }{a},\stackrel{\to }{b,}\stackrel{\to }{c}$ are three vectors such that  then the value of
(a) 0
(b) 1
(c) -19
(d) 38

Given:
$\left|\stackrel{\to }{a}\right|=2,\left|\stackrel{\to }{b}\right|=3,\left|\stackrel{\to }{c}\right|=5$
$\stackrel{\to }{a}+\stackrel{\to }{b}+\stackrel{\to }{c}=\stackrel{\to }{0}$

Hence, the correct option is (c).

#### Question 29:

If  are unit vectors, then what is the angle between  to be a unit vector?

Given:
are unit vectors
$\sqrt{3}\stackrel{\to }{a}-\stackrel{\to }{b}$ is a unit vector

Hence, the correct option is (a).

#### Question 30:

If the projection of $\stackrel{\to }{a}=\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k}$ on $\stackrel{\to }{b}=2\stackrel{^}{i}+\lambda \stackrel{^}{k}$ is zero, then the value of $\lambda$ is
(a) 0
(b) 1
(c) $-\frac{2}{3}$
(d) $-\frac{3}{2}$

Given:
Now,
Projection of $\stackrel{\to }{a}$ on $\stackrel{\to }{b}$ = $\frac{\stackrel{\to }{a}·\stackrel{\to }{b}}{\left|\stackrel{\to }{b}\right|}$
Here, $\frac{\stackrel{\to }{a}·\stackrel{\to }{b}}{\left|\stackrel{\to }{b}\right|}=0$
$⇒\stackrel{\to }{a}·\stackrel{\to }{b}=0\phantom{\rule{0ex}{0ex}}⇒\left(\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k}\right)·\left(2\stackrel{^}{i}+\lambda \stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}⇒2+0+3\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =\frac{-2}{3}$
Hence, the correct answer is option C.

#### Question 1:

If  are two unit vectors such that $\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|$ is also a unit vector, then

Given:
are unit vectors
$\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|$ is also a unit vector

Hence, $\left|\stackrel{\to }{a}-\stackrel{\to }{b}\right|=\overline{)\sqrt{3}}$.

#### Question 2:

is perpendicular to

Given:
$\left|\stackrel{\to }{a}\right|=3,\left|\stackrel{\to }{b}\right|=4$
$\stackrel{\to }{a}+\lambda \stackrel{\to }{b}$ is perpendicular to $\stackrel{\to }{a}-\lambda \stackrel{\to }{b}$

Hence, $\lambda =\overline{)±\frac{3}{4}}$.

#### Question 3:

Given:
$\stackrel{\to }{a}=2\stackrel{^}{i}-7\stackrel{^}{j}+\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{b}=\stackrel{^}{i}+3\stackrel{^}{j}-5\stackrel{^}{k}$
$\stackrel{\to }{a}·m\stackrel{\to }{b}=120$

$\stackrel{\to }{a}·m\stackrel{\to }{b}=120\phantom{\rule{0ex}{0ex}}⇒m\left(\stackrel{\to }{a}·\stackrel{\to }{b}\right)=120\phantom{\rule{0ex}{0ex}}⇒m\left(\left(2\right)\left(1\right)+\left(-7\right)\left(3\right)+\left(1\right)\left(-5\right)\right)=120\phantom{\rule{0ex}{0ex}}⇒m\left(2-21-5\right)=120\phantom{\rule{0ex}{0ex}}⇒m\left(-24\right)=120\phantom{\rule{0ex}{0ex}}⇒m=-\frac{120}{24}\phantom{\rule{0ex}{0ex}}⇒m=-\frac{10}{2}\phantom{\rule{0ex}{0ex}}⇒m=-5$

Hence, $m=\overline{)-5}$.

#### Question 4:

The non-zero vectors , then the angle between  is ____________.

Given:

Hence, the angle between  is $\overline{)\mathrm{\pi }}$.

#### Question 5:

If  are mutually perpendicular unit vectors, then $\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|=____________.$

Given:
are unit vectors
are mutually perpendicular vectors

Hence, $\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|=\overline{)\sqrt{2}}$.

#### Question 6:

If the angle between the vectors

Given:
The angle between the vectors

Hence, .

#### Question 7:

If  are unit vectors such that $\stackrel{\to }{a}+\stackrel{\to }{b}-\stackrel{\to }{c}=\stackrel{\to }{0}$, then the angle between  is ___________.

Given:
are unit vectors
$\stackrel{\to }{a}+\stackrel{\to }{b}-\stackrel{\to }{c}=\stackrel{\to }{0}$

Hence, the angle between  is $\overline{)\frac{2\mathrm{\pi }}{3}}$.

#### Question 8:

If  are unit vectors such that $\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|=\sqrt{3},$ then the angle between  is ___________.

Given:
are unit vectors
$\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|=\sqrt{3}$

Hence, the angle between  is $\overline{)\frac{\mathrm{\pi }}{3}}$.

#### Question 9:

If  are two non-zero vectors, then the projection of  is _____________.

The projection of  is given by $\left(\stackrel{\to }{a}·\stackrel{^}{b}\right)=\frac{1}{\left|\stackrel{\to }{b}\right|}\left(\stackrel{\to }{a}·\stackrel{\to }{b}\right)$.

Hence, the projection of  is $\overline{)\frac{1}{\left|\stackrel{\to }{b}\right|}\left(\stackrel{\to }{a}·\stackrel{\to }{b}\right)}$.

#### Question 10:

Let  be unit vectors such that $\stackrel{\to }{a}-\sqrt{2}\stackrel{\to }{b}$ is also a unit vector. Then the angle between  is _____________.

Given:
are unit vectors
$\stackrel{\to }{a}-\sqrt{2}\stackrel{\to }{b}$ is also a unit vector

Hence, the angle between  is $\overline{)\frac{\mathrm{\pi }}{4}}$.

#### Question 11:

then the angle between  is ______________.

Given:
$\left|\stackrel{\to }{a}\right|=1\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{b}\right|=3\phantom{\rule{0ex}{0ex}}\left|\stackrel{\to }{a}-\stackrel{\to }{b}\right|=\sqrt{7}$

Hence, the angle between  is $\overline{)\frac{\mathrm{\pi }}{3}}$.

#### Question 18:

If $\stackrel{\to }{a}=2\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}$ and $\stackrel{\to }{b}=5\stackrel{^}{i}-3\stackrel{^}{j}-4\stackrel{^}{k}$, then  = ........................

Given:
Now,

#### Question 1:

What is the angle between vectors with magnitudes 2 and $\sqrt{3}$ respectively? Given

#### Question 2:

are two vectors such that Write the projection of .

#### Question 3:

Find the cosine of the angle between the vectors

#### Question 4:

If the vectors are orthogonal, find m.

#### Question 5:

If the vectors are parallel, find the value of m.

#### Question 6:

If are vectors of equal magnitude, write the value of

#### Question 7:

If are two vectors such that find the relation between the magnitudes of .

#### Question 8:

For any two vectors , write when holds.

#### Question 9:

For any two vectors , write when holds.

#### Question 10:

If are two vectors of the same magnitude inclined at an angle of 60° such that write the value of their magnitude.

#### Question 11:

If what can you conclude about the vector ?

#### Question 12:

If  is a unit vector such that

#### Question 13:

If are unit vectors such that is a unit vector, write the value of

If

#### Question 15:

If find the projection of .

#### Question 16:

For any two non-zero vectors, write the value of

#### Question 17:

Write the projections of on the coordinate axes.

#### Question 18:

Write the component of along .

#### Question 19:

Write the value of where is any vector.

#### Question 20:

Find the value of θ ∈(0, π/2) for which vectors are perpendicular.

#### Question 21:

Write the projection of $\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}$ along the vector $\stackrel{^}{j}$.

#### Question 22:

Write a vector satisfying

#### Question 23:

If are unit vectors, find the angle between

#### Question 24:

If are mutually perpendicular unit vectors, write the value of

#### Question 25:

If are mutually perpendicular unit vectors, write the value of

#### Question 26:

Find the angle between the vectors

#### Question 27:

For what value of λ are the vectors perpendicular to each other?

#### Question 28:

Find the projection of

#### Question 29:

Write the value of p for which are parallel vectors.

#### Question 30:

Find the value of λ if the vectors are perpendicular to each other.

#### Question 31:

If find the projection of .

#### Question 32:

Write the angle between two vectors with magnitudes $\sqrt{3}$ and 2 respectively if

#### Question 33:

Write the projection of the vector $\stackrel{^}{i}+3\stackrel{^}{j}+7\stackrel{^}{k}$ on the vector $2\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}$.

We know that projection of $\stackrel{\to }{a}$ on $\stackrel{\to }{b}$ = $\frac{\stackrel{\to }{a}.\stackrel{\to }{b}}{\left|\stackrel{\to }{b}\right|}$.

Let $\stackrel{\to }{a}=\stackrel{^}{i}+3\stackrel{^}{j}+7\stackrel{^}{k}$ and $\stackrel{\to }{b}=2\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}$.

∴ Projection of $\stackrel{\to }{a}$ on $\stackrel{\to }{b}$

$=\frac{\left(\stackrel{^}{i}+3\stackrel{^}{j}+7\stackrel{^}{k}\right).\left(2\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)}{\left|2\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right|}\phantom{\rule{0ex}{0ex}}=\frac{1×2+3×\left(-3\right)+7×6}{\sqrt{{2}^{2}+{\left(-3\right)}^{2}+{6}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{2-9+42}{\sqrt{49}}\phantom{\rule{0ex}{0ex}}=\frac{35}{7}\phantom{\rule{0ex}{0ex}}=5$

#### Question 34:

Find λ when the projection of is 4 units.

#### Question 35:

For what value of λ are the vectors perpendicular to each other?

#### Question 36:

Write the projection of the vector $7\stackrel{^}{i}+\stackrel{^}{j}-4\stackrel{^}{k}$ on the vector $2\stackrel{^}{i}+6\stackrel{^}{j}+3\stackrel{^}{k}.$

#### Question 37:

Write the value of λ so that the vectors are perpendicular to each other.

#### Question 38:

Write the projection of

#### Question 39:

If $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are perpendicular vectors, $\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|=3$ and $\left|\stackrel{\to }{a}\right|=5$, find the value of $\left|\stackrel{\to }{b}\right|$.                           [CBSE 2014]

Disclaimer: $\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|=13$ has been taken in order to solve the question.

It is given that $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are perpendicular vectors.

$\therefore \stackrel{\to }{a}.\stackrel{\to }{b}=0$           .....(1)

$⇒{\left|\stackrel{\to }{b}\right|}^{2}=169-25=144\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{b}\right|=12$

Thus, the value of $\left|\stackrel{\to }{b}\right|$ is 12.

#### Question 40:

If the vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are such that $\left|\stackrel{\to }{a}\right|=3,\left|\stackrel{\to }{b}\right|=\frac{2}{3}$ and $\stackrel{\to }{a}×\stackrel{\to }{b}$ is a unit vector, then write the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$.          [CBSE 2014]

Let the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ be $\theta$.

It is given that $\stackrel{\to }{a}×\stackrel{\to }{b}$ is a unit vector.

$\therefore \left|\stackrel{\to }{a}×\stackrel{\to }{b}\right|=1\phantom{\rule{0ex}{0ex}}⇒\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|\mathrm{sin}\theta =1\phantom{\rule{0ex}{0ex}}⇒3×\frac{2}{3}×\mathrm{sin}\theta =1\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin}\theta =\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{\mathrm{\pi }}{6}$

Thus, the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is $\frac{\mathrm{\pi }}{6}$.

#### Question 41:

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

Let the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ be $\theta$.

It is given that $\left|\stackrel{\to }{a}\right|=\left|\stackrel{\to }{b}\right|=\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|=1$.

$\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|=1\phantom{\rule{0ex}{0ex}}⇒{\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{\left|\stackrel{\to }{a}\right|}^{2}+2\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|\mathrm{cos}\theta +{\left|\stackrel{\to }{b}\right|}^{2}=1\phantom{\rule{0ex}{0ex}}⇒1+2×1×1×\mathrm{cos}\theta +1=1\phantom{\rule{0ex}{0ex}}⇒2\mathrm{cos}\theta =-1\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =-\frac{1}{2}=\mathrm{cos}\frac{2\mathrm{\pi }}{3}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{2\mathrm{\pi }}{3}$

Thus, the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is $\frac{2\mathrm{\pi }}{3}$.

#### Question 42:

If $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are unit vectors, then find the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$, given that $\left(\sqrt{3}\stackrel{\to }{a}-\stackrel{\to }{b}\right)$ is a unit vector.                [CBSE 2014]

Let the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ be $\theta$.

It is given that $\left|\stackrel{\to }{a}\right|=\left|\stackrel{\to }{b}\right|=\left|\sqrt{3}\stackrel{\to }{a}-\stackrel{\to }{b}\right|=1$.

$\left|\sqrt{3}\stackrel{\to }{a}-\stackrel{\to }{b}\right|=1\phantom{\rule{0ex}{0ex}}⇒{\left|\sqrt{3}\stackrel{\to }{a}-\stackrel{\to }{b}\right|}^{2}=1\phantom{\rule{0ex}{0ex}}⇒{\left|\sqrt{3}\stackrel{\to }{a}\right|}^{2}-2\sqrt{3}\stackrel{\to }{a}.\stackrel{\to }{b}+{\left|\stackrel{\to }{b}\right|}^{2}=1\phantom{\rule{0ex}{0ex}}⇒3{\left|\stackrel{\to }{a}\right|}^{2}-2\sqrt{3}\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|\mathrm{cos}\theta +{\left|\stackrel{\to }{b}\right|}^{2}=1$
$⇒3×1-2\sqrt{3}×1×1×\mathrm{cos}\theta +1=1\phantom{\rule{0ex}{0ex}}⇒2\sqrt{3}\mathrm{cos}\theta =3\phantom{\rule{0ex}{0ex}}⇒\mathrm{cos}\theta =\frac{\sqrt{3}}{2}=\mathrm{cos}\frac{\mathrm{\pi }}{6}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{\mathrm{\pi }}{6}$

Thus, the angle between $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is $\frac{\mathrm{\pi }}{6}$.

#### Question 43:

Find the magnitude of each of the two vectors , having the same magnitude such that the angle between them is 60° and their scalar product is $\frac{9}{2}.$

$\left|\stackrel{\to }{a}\right|=\left|\stackrel{\to }{b}\right|$
Scalar product of two vectors = $\stackrel{\to }{a}·\stackrel{\to }{b}=\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|\mathrm{cos}\theta$