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Updated 6 August 2026

NCERT Solutions

Vector Algebra

CBSE · Class 12 · Mathematics

NCERT Solutions for Vector Algebra — CBSE Class 12 Mathematics.

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73 Questions Solved · 5 Sections

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EXERCISE 10.1

1Represent graphically a displacement of 40 km, 30° east of north.Show solution

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2Classify the following measures as scalars and vectors.Show solution

From the chapter’s example on scalars and vectors:
- Time has magnitude only, so it is a scalar.
- Volume has magnitude only, so it is a scalar.
- Force has magnitude and direction, so it is a vector.
- Speed has magnitude only, so it is a scalar.
- Density has magnitude only, so it is a scalar.
- Velocity has magnitude and direction, so it is a vector.

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3Classify the following as scalar and vector quantities.Show solution

Classify each quantity by whether it has magnitude only or magnitude and direction:
- Time period is a scalar.
- Distance is a scalar.
- Force is a vector.
- Velocity is a vector.
- Work done is a scalar.

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4In Fig 10.6 (a square), identify the following vectors.Show solution

From Fig. 10.5 given in the chapter:
- Collinear vectors are \vec{a}, \vec{c}, \vec{d}.
- Equal vectors are \vec{a} and \vec{c}.
- Coinitial vectors are \vec{b}, \vec{c}, \vec{d}.

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5Answer the following as true or false.Show solution

Use the chapter definitions:
- \vec{a} and -\vec{a} are along the same line, so they are collinear → True.
- Two collinear vectors need not have the same magnitude → False.
- Two vectors can have the same magnitude but different directions, so they need not be collinear → False.
- Two collinear vectors with the same magnitude and same direction are equal → True.

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EXERCISE 10.2

1Compute the magnitude of the following vectors:Show solution

Compute each magnitude using

|x\hat{i}+y\hat{j}+z\hat{k}|=\sqrt{x^2+y^2+z^2}

.

- For

\vec{a}=\hat{i}+\hat{j}+\hat{k}

|\vec{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}

But the question text shows

\hat{i}+\hat{j}+k

; the chapter uses the same vector as

\hat{i}+\hat{j}+\hat{k}

, so the magnitude is

\sqrt{3}

.
- For

\vec{b}=2\hat{i}-7\hat{j}-3\hat{k}

|\vec{b}|=\sqrt{2^2+(-7)^2+(-3)^2}=\sqrt{4+49+9}=\sqrt{62}

- For

\vec{c}=\frac1{\sqrt3}\hat{i}+\frac1{\sqrt3}\hat{j}-\frac1{\sqrt3}\hat{k}

|\vec{c}|=\sqrt{\frac13+\frac13+\frac13}=1

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2Write two different vectors having same magnitude.Show solution

We need two different vectors having the same magnitude. From the chapter’s example:

|\hat{i}+2\hat{j}|=\sqrt{1^2+2^2}=\sqrt5

|2\hat{i}+\hat{j}|=\sqrt{2^2+1^2}=\sqrt5

So these are different vectors with the same magnitude.

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3Write two different vectors having same direction.Show solution

Two vectors have the same direction if one is a scalar multiple of the other. For example,

2\hat{i}+2\hat{j}=2(\hat{i}+\hat{j})

So these two are different vectors but have the same direction.

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4Find the values of

x

and

y

so that the vectors

2\hat{i} + 3\hat{j}

and

x\hat{i} + y\hat{j}

are equal.Show solution

Two vectors are equal if and only if their corresponding components are equal. So,

2\hat{i}+3\hat{j}=x\hat{i}+y\hat{j}

implies

x=2,\quad y=3.

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5Find the scalar and vector components of the vector with initial point

(2, 1)

and terminal point

(-5, 7)

.Show solution

The vector from initial point

(2,1)

to terminal point

(-5,7)

(-5-2)\hat{i}+(7-1)\hat{j}=-7\hat{i}+6\hat{j}.

So the scalar components are

-7

and

6

, and the vector components are

-7\hat{i}

and

6\hat{j}

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6Find the sum of the vectors

\vec{a} = \hat{i} - 2\hat{j} + \hat{k}

\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k}

and

\vec{c} = \hat{i} - 6\hat{j} - 7\hat{k}

.Show solution

Add the vectors component-wise:

\vec{a}+\vec{b}+\vec{c}=(1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k}

=0\hat{i}-4\hat{j}-1\hat{k}

Hence the sum is

-4\hat{j}-\hat{k}.

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7Find the unit vector in the direction of the vector

\vec{a} = \hat{i} + \hat{j} + 2\hat{k}

.Show solution

For

\vec{a}=\hat{i}+\hat{j}+2\hat{k}

|\vec{a}|=\sqrt{1^2+1^2+2^2}=\sqrt6.

The unit vector in its direction is

\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{1}{\sqrt6}(\hat{i}+\hat{j}+2\hat{k}).

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8Find the unit vector in the direction of vector

\overline{PQ}

, where

P

and

Q

are the points

(1, 2, 3)

and

(4, 5, 6)

, respectively.Show solution

First find

\overrightarrow{PQ}

\overrightarrow{PQ}=(4-1)\hat{i}+(5-2)\hat{j}+(6-3)\hat{k}=3\hat{i}+3\hat{j}+3\hat{k}.

Its magnitude is

|\overrightarrow{PQ}|=\sqrt{3^2+3^2+3^2}=3\sqrt3.

Hence the unit vector in the direction of

\overrightarrow{PQ}

\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{1}{\sqrt3}(\hat{i}+\hat{j}+\hat{k}).

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9For given vectors,

\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}

and

\vec{b} = -\hat{i} + \hat{j} - \hat{k}

, find the unit vector in the direction of the vector

\vec{a} + \vec{b}

.Show solution

Add the vectors:

\vec{a}+\vec{b}=(2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k}=\hat{i}+\hat{k}.

Its magnitude is

|\vec{a}+\vec{b}|=\sqrt{1^2+0^2+1^2}=\sqrt2.

So the unit vector is

\frac{1}{\sqrt2}(\hat{i}+\hat{k}).

The computed answer is therefore

\frac{1}{\sqrt2}(\hat{i}+\hat{k})

, which is not among the standard options if different ones were printed.

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10Find a vector in the direction of vector

5\hat{i} - \hat{j} + 2\hat{k}

which has magnitude 8 units.Show solution

First find the magnitude of the given direction vector:

|5\hat{i}-\hat{j}+2\hat{k}|=\sqrt{5^2+(-1)^2+2^2}=\sqrt{30}.

Unit vector in its direction is

\frac{1}{\sqrt{30}}(5\hat{i}-\hat{j}+2\hat{k}).

A vector of magnitude 8 in this direction is

8\cdot \frac{1}{\sqrt{30}}(5\hat{i}-\hat{j}+2\hat{k})=\frac{8}{\sqrt{30}}(5\hat{i}-\hat{j}+2\hat{k}).

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11Show that the vectors

2\hat{i} - 3\hat{j} + 4\hat{k}

and

-4\hat{i} + 6\hat{j} - 8\hat{k}

are collinear.Show solution

Let

\vec{u}=2\hat{i}-3\hat{j}+4\hat{k}, \qquad \vec{v}=-4\hat{i}+6\hat{j}-8\hat{k}.

Now,

\vec{v}=-2(2\hat{i}-3\hat{j}+4\hat{k})=-2\vec{u}.

Since one vector is a nonzero scalar multiple of the other, they are collinear.

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12Find the direction cosines of the vector

\hat{i} + 2\hat{j} + 3\hat{k}

.Show solution

For

\vec{a}=\hat{i}+2\hat{j}+3\hat{k}

|\vec{a}|=\sqrt{1^2+2^2+3^2}=\sqrt{14}.

Direction cosines are components divided by magnitude:

\left(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right).

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13Find the direction cosines of the vector joining the points

A(1, 2, -3)

and

B(-1, -2, 1)

, directed from

A

B

.Show solution

The vector from

A(1,2,-3)

B(-1,-2,1)

\overrightarrow{AB}=(-1-1)\hat{i}+(-2-2)\hat{j}+(1+3)\hat{k}=-2\hat{i}-4\hat{j}+4\hat{k}.

Its magnitude is

|\overrightarrow{AB}|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{36}=6.

So the direction cosines are

\left(-\frac{2}{6},-\frac{4}{6},\frac{4}{6}\right)=\left(-\frac{1}{3},-\frac{2}{3},\frac{2}{3}\right).

This is the computed value; if matching printed options, choose the one equivalent to these.

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14Show that the vector

\hat{i} + \hat{j} + \hat{k}

is equally inclined to the axes

OX

OY

and

OZ

.Show solution

For the vector

\vec{a}=\hat{i}+\hat{j}+\hat{k},

its components are equal:

1,1,1

. Hence its direction cosines are

l=m=n=\frac{1}{\sqrt{3}}.

So the angles it makes with the positive

x

y

- and

z

-axes are equal. Therefore,

\vec{a}

is equally inclined to the axes

OX

OY

and

OZ

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15Find the position vector of a point

R

which divides the line joining two points

P

and

Q

whose position vectors are

\hat{i} + 2\hat{j} - \hat{k}

and

-\hat{i} + \hat{j} + \hat{k}

respectively, in the ratio

2 : 1

Show solution

Let the position vectors of

P

and

Q

\vec{a}=\hat{i}+2\hat{j}-\hat{k}, \qquad \vec{b}=-\hat{i}+\hat{j}+\hat{k}.

Since

R

divides

PQ

internally in the ratio

2:1

, by the section formula,

\overrightarrow{OR}=\frac{1\cdot \vec{a}+2\cdot \vec{b}}{2+1}.

So,

\overrightarrow{OR}=\frac{(\hat{i}+2\hat{j}-\hat{k})+2(-\hat{i}+\hat{j}+\hat{k})}{3}

=\frac{-\hat{i}+4\hat{j}+\hat{k}}{3}.

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16Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, -2).Show solution

For points

P(2,3,4)

and

Q(4,1,-2)

, the midpoint is

\left(\frac{2+4}{2},\frac{3+1}{2},\frac{4+(-2)}{2}\right)=(3,2,1).

Therefore the position vector of the midpoint is

3\hat{i}+2\hat{j}+\hat{k}.

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17Show that the points A, B and C with position vectors,

\vec{a} = 3\hat{i} - 4\hat{j} - 4\hat{k}

\vec{b} = 2\hat{i} - \hat{j} + \hat{k}

and

\vec{c} = \hat{i} - 3\hat{j} - 5\hat{k}

, respectively form the vertices of a right angled triangle.Show solution

Let the position vectors of

A,B,C

\vec{a}=3\hat{i}-4\hat{j}-4\hat{k},\quad \vec{b}=2\hat{i}-\hat{j}+\hat{k},\quad \vec{c}=\hat{i}-3\hat{j}-5\hat{k}.

Then

\overrightarrow{AB}=\vec{b}-\vec{a}=(-1)\hat{i}+3\hat{j}+5\hat{k},

\overrightarrow{BC}=\vec{c}-\vec{b}=(-1)\hat{i}-2\hat{j}-6\hat{k},

\overrightarrow{CA}=\vec{a}-\vec{c}=2\hat{i}-\hat{j}+\hat{k}.

Now compute squares of magnitudes:

|\overrightarrow{AB}|^2=1+9+25=35,

|\overrightarrow{BC}|^2=1+4+36=41,

|\overrightarrow{CA}|^2=4+1+1=6.

Since

|\overrightarrow{CA}|^2+|\overrightarrow{BC}|^2=6+35=41=|\overrightarrow{AB}|^2,

triangle

ABC

is right angled.

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18In triangle ABC (Fig 10.18), which of the following is not true:Show solution

From triangle law,

\overline{AB}+\overline{BC}=\overline{AC}.

Hence

\overline{AB}+\overline{BC}-\overline{AC}=\vec{0},

so option (B) is true. Option (C) is the same expression as (B), so it is also true. Also,

\overline{CB}=-\overline{BC}

\overline{AB}-\overline{CB}+\overline{CA}=\overline{AB}+\overline{BC}+\big(-\overline{AC}\big)=\vec{0},

which is true as well. Therefore, as printed, there is no incorrect option among them.

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19If

\vec{a}

and

\vec{b}

are two collinear vectors, then which of the following are incorrect:Show solution

For collinear vectors:
- (A)

\vec{b}=\lambda\vec{a}

for some scalar

\lambda

is correct.
- (B)

\vec{a}=\pm\vec{b}

is not always true, because collinear vectors can have any nonzero scalar multiple, not only

\pm1

.
- (C) “respective components are not proportional” is incorrect, because for collinear vectors components are proportional.
- (D) “same direction, but different magnitudes” is not always true; collinear vectors may have opposite directions too.
So the incorrect statements are (B), (C), and (D), but if the exercise expects only the clearly false statements, then (C) and (D) are definitely incorrect.

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EXERCISE 10.3

1Find the angle between two vectors

\vec{a}

and

\vec{b}

with magnitudes

\sqrt{3}

and 2, respectively having

\vec{a} \cdot \vec{b} = \sqrt{6}

.Show solution

Use

\cos\theta=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}.

Given

|\vec{a}|=\sqrt3

|\vec{b}|=2

, and

\vec{a}\cdot\vec{b}=\sqrt6

\cos\theta=\frac{\sqrt6}{2\sqrt3}=\frac{\sqrt2}{2}.

So,

\theta=\cos^{-1}\left(\frac{\sqrt2}{2}\right)=\frac{\pi}{4}.

The computed value is

\frac{\pi}{4}

, not

\frac{pi}{3}

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2Find the angle between the vectors

\hat{i} - 2\hat{j} + 3\hat{k}

and

3\hat{i} - 2\hat{j} + \hat{k}

Show solution

Let

\vec{a}=\hat{i}-2\hat{j}+3\hat{k},\quad \vec{b}=3\hat{i}-2\hat{j}+\hat{k}.

Then

\vec{a}\cdot\vec{b}=1\cdot3+(-2)(-2)+3\cdot1=3+4+3=10.

Also,

|\vec{a}|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14},

|\vec{b}|=\sqrt{3^2+(-2)^2+1^2}=\sqrt{14}.

Hence

\cos\theta=\frac{10}{\sqrt{14}\cdot\sqrt{14}}=\frac{10}{14}=\frac{5}{7}.

So,

\theta=\cos^{-1}\left(\frac{5}{7}\right).

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3Find the projection of the vector

\hat{i} - \hat{j}

on the vector

\hat{i} + \hat{j}

.Show solution

The projection of **

\vec a

\vec b

** is

\vec a \cdot \frac{\vec b}{|\vec b|}.

Here,

\vec a=\hat{i}-\hat{j},\qquad \vec b=\hat{i}+\hat{j}.

First find

|\vec b|

|\vec b|=\sqrt{1^2+1^2}=\sqrt{2}.

So the projection vector is

(\hat{i}-\hat{j})\cdot \frac{\hat{i}+\hat{j}}{\sqrt{2}} =\frac{1-1}{\sqrt{2}}=0.

Thus the projection is zero. Since the printed book question refers to the projection along the vector, the scalar projection is 0, and the projection vector is the zero vector.

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4Find the projection of the vector

\hat{i} + 3\hat{j} + 7\hat{k}

on the vector

7\hat{i} - \hat{j} + 8\hat{k}

.Show solution

The projection of **

\vec a

\vec b

** is

\frac{\vec a\cdot \vec b}{|\vec b|}.

Given

\vec a=\hat{i}+3\hat{j}+7\hat{k},\qquad \vec b=7\hat{i}-\hat{j}+8\hat{k}.

Compute the dot product:

\vec a\cdot \vec b=(1)(7)+(3)(-1)+(7)(8)=7-3+56=60.

Now

|\vec b|=\sqrt{7^2+(-1)^2+8^2}=\sqrt{49+1+64}=\sqrt{114}.

So the projection is

\frac{60}{\sqrt{114}}=\frac{60\sqrt{114}}{114}=\frac{10\sqrt{114}}{19}.

If the question means the scalar projection, this is the value. The chapter’s worked method gives projection as

(\vec a\cdot \vec b)/|\vec b|

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5Show that each of the given three vectors is a unit vector:Show solution

For each vector, find its magnitude using

|\vec r|=\sqrt{x^2+y^2+z^2}.

The three given vectors are

\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}),\quad \frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k}),\quad \frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k}).

### First vector

\left|\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})\right| =\frac{1}{7}\sqrt{2^2+3^2+6^2} =\frac{1}{7}\sqrt{4+9+36} =\frac{1}{7}\sqrt{49}=1.

### Second vector

\left|\frac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})\right| =\frac{1}{7}\sqrt{3^2+(-6)^2+2^2} =\frac{1}{7}\sqrt{9+36+4} =\frac{1}{7}\sqrt{49}=1.

### Third vector

\left|\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})\right| =\frac{1}{7}\sqrt{6^2+2^2+(-3)^2} =\frac{1}{7}\sqrt{36+4+9} =\frac{1}{7}\sqrt{49}=1.

Hence, each vector has magnitude 1, so each is a unit vector.

To show they are mutually perpendicular, take dot products:

(2,3,6)\cdot(3,-6,2)=6-18+12=0,

(3,-6,2)\cdot(6,2,-3)=18-12-6=0,

(6,2,-3)\cdot(2,3,6)=12+6-18=0.

Thus the vectors are also mutually perpendicular.

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6Find

|\vec{a}|

and

|\vec{b}|

, if

(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 8

and

|\vec{a}| = 8|\vec{b}|

.Show solution

From

(\vec a+\vec b)\cdot(\vec a-\vec b)=8,

expand:

\vec a\cdot\vec a-\vec a\cdot\vec b+\vec b\cdot\vec a-\vec b\cdot\vec b=8.

Since dot product is commutative,

|\vec a|^2-|\vec b|^2=8.

Also given

|\vec a|=8|\vec b|.

Let

|\vec b|=x

. Then

|\vec a|=8x

. So

(8x)^2-x^2=8

64x^2-x^2=8

63x^2=8

x^2=\frac{8}{63}.

Hence

|\vec b|=\sqrt{\frac{8}{63}}, \qquad |\vec a|=8\sqrt{\frac{8}{63}}.

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7Evaluate the product

(3\vec{a} - 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})

.Show solution

Expand using distributive property:

(3\vec a-5\vec b)\cdot(2\vec a+7\vec b)

=3\vec a\cdot 2\vec a+3\vec a\cdot 7\vec b-5\vec b\cdot 2\vec a-5\vec b\cdot 7\vec b.

=6\vec a\cdot\vec a+21\vec a\cdot\vec b-10\vec b\cdot\vec a-35\vec b\cdot\vec b.

Since

\vec a\cdot\vec b=\vec b\cdot\vec a

=6\vec a\cdot\vec a+11\vec a\cdot\vec b-35\vec b\cdot\vec b.

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8Find the magnitude of two vectors

\vec{a}

and

\vec{b}

, having the same magnitude and such that the angle between them is

60^\circ

and their scalar product is

\frac{1}{2}

.Show solution

If the vectors have the same magnitude, let that magnitude be

r

. The angle between them is

60^\circ

, so

\vec a\cdot\vec b=|\vec a||\vec b|\cos 60^\circ=r\cdot r\cdot \frac12=\frac{r^2}{2}.

Given that their scalar product is

\frac12

, we get

\frac{r^2}{2}=\frac12.

Thus

r^2=1 \implies r=1.

So the magnitude of each vector is 1.

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9Find

|\vec{x}|

, if for a unit vector

\vec{a}

(\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12

.Show solution

Since

\vec a

is a unit vector,

|\vec a|=1

.

Now

(\vec x-\vec a)\cdot(\vec x+\vec a)=12.

Using distributive property,

\vec x\cdot\vec x+\vec x\cdot\vec a-\vec a\cdot\vec x-\vec a\cdot\vec a=12.

Because dot product is commutative,

\vec x\cdot\vec a=\vec a\cdot\vec x

, so the middle terms cancel:

|\vec x|^2-| \vec a|^2=12.

Since

|\vec a|^2=1

|\vec x|^2-1=12

|\vec x|^2=13.

Therefore

|\vec x|=\sqrt{13}.

The computed answer from the chapter text is

\sqrt{13}

; if a different printed answer is seen, it is not among the chapter’s working. However, the standard evaluation gives

\sqrt{13}

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10If

\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}

\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}

and

\vec{c} = 3\hat{i} + \hat{j}

are such that

\vec{a} + \lambda\vec{b}

is perpendicular to

\vec{c}

, then find the value of

\lambda

.Show solution

Since

\vec a+\lambda\vec b

is perpendicular to

\vec c

, their dot product is zero:

(\vec a+\lambda\vec b)\cdot\vec c=0.

Given

\vec a=2\hat i+2\hat j+3\hat k, \quad \vec b=-\hat i+2\hat j+\hat k, \quad \vec c=3\hat i+\hat j.

Now

\vec a+\lambda\vec b=(2-\lambda)\hat i+(2+2\lambda)\hat j+(3+\lambda)\hat k.

Dot with

\vec c

(2-\lambda,\ 2+2\lambda,\ 3+\lambda)\cdot(3,1,0)=0

3(2-\lambda)+(2+2\lambda)=0

6-3\lambda+2+2\lambda=0

8-\lambda=0

\lambda=8.

But the chapter’s worked example gives the perpendicularity condition as above; using it directly, the correct value is 8.

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11Show that

|\vec{a}|\vec{b} + |\vec{b}|\vec{a}

is perpendicular to

|\vec{a}|\vec{b} - |\vec{b}|\vec{a}

, for any two nonzero vectors

\vec{a}

and

\vec{b}

.Show solution

Let

\vec u=|\vec a|\vec b+|\vec b|\vec a, \qquad \vec v=|\vec a|\vec b-|\vec b|\vec a.

To show they are perpendicular, compute the dot product:

\vec u\cdot\vec v =(|\vec a|\vec b+|\vec b|\vec a)\cdot(|\vec a|\vec b-|\vec b|\vec a).

Expanding,

=|\vec a|^2(\vec b\cdot\vec b)-|\vec a||\vec b|(\vec b\cdot\vec a)+|\vec a||\vec b|(\vec a\cdot\vec b)-|\vec b|^2(\vec a\cdot\vec a).

Since

\vec a\cdot\vec b=\vec b\cdot\vec a

, the middle terms cancel:

=|\vec a|^2|\vec b|^2-|\vec b|^2|\vec a|^2=0.

Hence

(|\vec a|\vec b+|\vec b|\vec a)\perp(|\vec a|\vec b-|\vec b|\vec a).

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12If

\vec{a} \cdot \vec{a} = 0

and

\vec{a} \cdot \vec{b} = 0

, then what can be concluded about the vector

\vec{b}

?Show solution

From

\vec a\cdot\vec a=0

, we get

|\vec a|^2=0 \implies \vec a=\vec 0.

Then

\vec a\cdot\vec b=0

gives no further restriction beyond this, but the conclusion from the chapter is that **

\vec b

** can be any vector if only one vector is zero. Here, since

\vec a=\vec 0

, the condition

\vec a\cdot\vec b=0

is automatically true.

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13If

\vec{a}, \vec{b}, \vec{c}

are unit vectors such that

\vec{a} + \vec{b} + \vec{c} = \vec{0}

, find the value of

\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}

.Show solution

Since

\vec a+\vec b+\vec c=\vec 0,

take square of both sides:

(\vec a+\vec b+\vec c)^2=0.

Expanding,

|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0.

Since each vector is a unit vector,

1+1+1+2S=0,

where

S=\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a.

3+2S=0

S=-\frac32.

Hence the required value is

-\frac32

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14If either vector

\vec{a} = \vec{0}

\vec{b} = \vec{0}

, then

\vec{a} \cdot \vec{b} = 0

. But the converse need not be true. Justify your answer with an example.

15If the vertices A, B, C of a triangle ABC are

(1, 2, 3)

(-1, 0, 0)

(0, 1, 2)

, respectively, then find

\angle ABC

. [

\angle ABC

is the angle between the vectors

\overline{BA}

and

\overline{BC}

16Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear.

17Show that the vectors

2\hat{i} - \hat{j} + \hat{k}

\hat{i} - 3\hat{j} - 5\hat{k}

and

3\hat{i} - 4\hat{j} - 4\hat{k}

form the vertices of a right angled triangle.

18If

\vec{a}

is a nonzero vector of magnitude '

a

' and

\lambda

a nonzero scalar, then

\lambda\vec{a}

is unit vector if

EXERCISE 10.4

1Find

|\vec{a} \times \vec{b}|

, if

\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}

and

\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}

2Find a unit vector perpendicular to each of the vector

\vec{a} + \vec{b}

and

\vec{a} - \vec{b}

, where

\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}

and

\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}

3If a unit vector

\vec{a}

makes angles

\frac{\pi}{3}

with

\hat{i}

\frac{\pi}{4}

with

\hat{j}

and an acute angle

\theta

with

\hat{k}

, then find

\theta

and hence, the components of

\vec{a}

4Show that

5Find

\lambda

and

\mu

(2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}

6Given that

\vec{a} \cdot \vec{b} = 0

and

\vec{a} \times \vec{b} = \vec{0}

. What can you conclude about the vectors

\vec{a}

and

\vec{b}

7Let the vectors

\vec{a}, \vec{b}, \vec{c}

be given as

a_1\hat{i} + a_2\hat{j} + a_3\hat{k}

b_1\hat{i} + b_2\hat{j} + b_3\hat{k}

c_1\hat{i} + c_2\hat{j} + c_3\hat{k}

. Then show that

\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}

8If either

\vec{a} = \vec{0}

\vec{b} = \vec{0}

, then

\vec{a} \times \vec{b} = \vec{0}

. Is the converse true? Justify your answer with an example.

9Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

10Find the area of the parallelogram whose adjacent sides are determined by the vectors

\vec{a} = \hat{i} - \hat{j} + 3\hat{k}

and

\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}

11Let the vectors

\vec{a}

and

\vec{b}

be such that

|\vec{a}| = 3

and

|\vec{b}| = \frac{\sqrt{2}}{3}

, then

\vec{a} \times \vec{b}

is a unit vector, if the angle between

\vec{a}

and

\vec{b}

12Area of a rectangle having vertices A, B, C and D with position vectors

-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}

\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}

\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}

and

-\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}

, respectively is

Miscellaneous Exercise on Chapter 10

1Write down a unit vector in XY-plane, making an angle of

30^\circ

with the positive direction of

x

-axis.

2Find the scalar components and magnitude of the vector joining the points

P(x_1, y_1, z_1)

and

Q(x_2, y_2, z_2)

3A girl walks 4 km towards west, then she walks 3 km in a direction

30^\circ

east of north and stops. Determine the girl's displacement from her initial point of departure.

4If

\vec{a} = \vec{b} + \vec{c}

, then is it true that

|\vec{a}| = |\vec{b}| + |\vec{c}|

? Justify your answer.

5Find the value of

x

for which

x(\hat{i} + \hat{j} + \hat{k})

is a unit vector.

6Find a vector of magnitude 5 units, and parallel to the resultant of the vectors

\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}

and

\vec{b} = \hat{i} - 2\hat{j} + \hat{k}

7If

\vec{a} = \hat{i} + \hat{j} + \hat{k}

\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}

and

\vec{c} = \hat{i} - 2\hat{j} + \hat{k}

, find a unit vector parallel to the vector

2\vec{a} - \vec{b} + 3\vec{c}

8Show that the points A (1, -2, -8), B (5, 0, -2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.

9Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are

(2\vec{a} + \vec{b})

and

(\vec{a} - 3\vec{b})

externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.

10The two adjacent sides of a parallelogram are

2\hat{i} - 4\hat{j} + 5\hat{k}

and

\hat{i} - 2\hat{j} - 3\hat{k}

. Find the unit vector parallel to its diagonal. Also, find its area.

11Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are

\pm \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)

12Let

\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}

\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}

and

\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}

. Find a vector

\vec{d}

which is perpendicular to both

\vec{a}

and

\vec{b}

, and

\vec{c} \cdot \vec{d} = 15

13The scalar product of the vector

\hat{i} + \hat{j} + \hat{k}

with a unit vector along the sum of vectors

2\hat{i} + 4\hat{j} - 5\hat{k}

and

\hat{i} + 2\hat{j} + 3\hat{k}

is equal to one. Find the value of

\lambda

14If

\vec{a}

\vec{b}

\vec{c}

are mutually perpendicular vectors of equal magnitudes, show that the vector

\vec{c} \cdot \vec{d} = 15

is equally inclined to

\vec{a}

\vec{b}

and

\vec{c}

15Prove that

(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2

, if and only if

\vec{a}, \vec{b}

are perpendicular, given

\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}

16If

\theta

is the angle between two vectors

\vec{a}

and

\vec{b}

, then

\vec{a} \cdot \vec{b} \geq 0

only when

17Let

\vec{a}

and

\vec{b}

be two unit vectors and

\theta

is the angle between them. Then

\vec{a} + \vec{b}

is a unit vector if

18The value of

\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})

19If

\theta

is the angle between any two vectors

\vec{a}

and

\vec{b}

, then

|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|

when

\theta

is equal to

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EXERCISE 10.1

EXERCISE 10.2

EXERCISE 10.3

EXERCISE 10.4

Miscellaneous Exercise on Chapter 10

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