Examples (Revised) - Chapter 10 - Vector Algebra - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths: Chapter 10 - Vector Algebra Solutions

Example 1

Represent graphically a displacement of $40 \mathrm{~km}, 30^{\circ}$ west of south.

Solution

The vector $\overrightarrow{\mathrm{OP}}$ represents the required displacement (Fig).

Example 2

Classify the following measures as scalars and vectors.
(i) 5 seconds
(ii) $1000 \mathrm{~cm}^3$

(iii) 10 Newton
(iv) $30 \mathrm{~km} / \mathrm{hr}$
(v) $10 \mathrm{~g} / \mathrm{cm}^3$
(vi) $20 \mathrm{~m} / \mathrm{s}$ towards north

Solution
(i) Time-scalar
(ii) Volume-scalar
(iii) Force-vector
(iv) Speed-scalar
(v) Density-scalar
(vi) Velocity-vector

Example 3

In Fig 10.5, which of the vectors are:
(i) Collinear
(ii) Equal
(iii) Coinitial

Solution
(i) Collinear vectors : $\vec{a}, \vec{c}$ and $\vec{d}$.
(ii) Equal vectors : $\vec{a}$ and $\vec{c}$.
(iii) Coinitial vectors: $\vec{b}, \vec{c}$ and $\vec{d}$.

Example 4

Find the values of $x, y$ and $z$ so that the vectors $\vec{a}=x \hat{i}+2 \hat{j}+z \hat{k}$ and $\vec{b}=2 \hat{i}+y \hat{j}+\hat{k}$ are equal.

Solution

Note that two vectors are equal if and only if their corresponding components are equal. Thus, the given vectors $\vec{a}$ and $\vec{b}$ will be equal if and only if
$
x=2, y=2, z=1
$

Example 5

Let $\vec{a}=\hat{i}+2 \hat{j}$ and $\vec{b}=2 \hat{i}+\hat{j}$. Is $|\vec{a}|=|\vec{b}|$ ? Are the vectors $\vec{a}$ and $\vec{b}$ equal?
Solution

We have $|\vec{a}|=\sqrt{1^2+2^2}=\sqrt{5}$ and $|\vec{b}|=\sqrt{2^2+1^2}=\sqrt{5}$
So, $|\vec{a}|=|\vec{b}|$. But, the two vectors are not equal since their corresponding components are distinct.

Example 6

Find unit vector in the direction of vector $\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}$
Solution

The unit vector in the direction of a vector $\vec{a}$ is given by $\hat{a}=\frac{1}{|\vec{a}|} \vec{a}$.
Now
$
|\vec{a}|=\sqrt{2^2+3^2+1^2}=\sqrt{14}
$

Therefore $\quad \hat{a}=\frac{1}{\sqrt{14}}(2 \hat{i}+3 \hat{j}+\hat{k})=\frac{2}{\sqrt{14}} \hat{i}+\frac{3}{\sqrt{14}} \hat{j}+\frac{1}{\sqrt{14}} \hat{k}$
Example 7

Find a vector in the direction of vector $\vec{a}=\hat{i}-2 \hat{j}$ that has magnitude 7 units.

Solution

The unit vector in the direction of the given vector $\vec{a}$ is
$
\hat{a}=\frac{1}{|\vec{a}|} \vec{a}=\frac{1}{\sqrt{5}}(\hat{i}-2 \hat{j})=\frac{1}{\sqrt{5}} \hat{i}-\frac{2}{\sqrt{5}} \hat{j}
$

Therefore, the vector having magnitude equal to 7 and in the direction of $\vec{a}$ is
$
7 \hat{a}=7\left(\frac{1}{\sqrt{5}} \hat{i}-\frac{2}{\sqrt{5}} \hat{j}\right)=\frac{7}{\sqrt{5}} \hat{i}-\frac{14}{\sqrt{5}} \hat{j}
$

Example 8

Find the unit vector in the direction of the sum of the vectors, $\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+3 \hat{k}$.

Solution

The sum of the given vectors is
$
\vec{a}+\vec{b}(=\vec{c}, \text { say })=4 \hat{i}+3 \hat{j}-2 \hat{k}
$
and
$
|\vec{c}|=\sqrt{4^2+3^2+(-2)^2}=\sqrt{29}
$

Thus, the required unit vector is
$
\hat{c}=\frac{1}{|\vec{c}|} \vec{c}=\frac{1}{\sqrt{29}}(4 \hat{i}+3 \hat{j}-2 \hat{k})=\frac{4}{\sqrt{29}} \hat{i}+\frac{3}{\sqrt{29}} \hat{j}-\frac{2}{\sqrt{29}} \hat{k}
$

Example 9

Write the direction ratio's of the vector $\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$ and hence calculate its direction cosines.

Solution

Note that the direction ratio's $a, b, c$ of a vector $\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}$ are just the respective components $x, y$ and $z$ of the vector. So, for the given vector, we have $a=1, b=1$ and $c=-2$. Further, if $l, m$ and $n$ are the direction cosines of the given vector, then
$
1=\frac{a}{|\vec{r}|}=\frac{1}{\sqrt{6}}, \quad m=\frac{b}{|\vec{r}|}=\frac{1}{\sqrt{6}}, \quad n=\frac{c}{|\vec{r}|}=\frac{-2}{\sqrt{6}} \text { as }|\vec{r}|=\sqrt{6}
$

Thus, the direction cosines are $\left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right)$.

Example 10

Find the vector joining the points $P(2,3,0)$ and $Q(-1,-2,-4)$ directed from $\mathrm{P}$ to $\mathrm{Q}$.

Solution

Since the vector is to be directed from $\mathrm{P}$ to $\mathrm{Q}$, clearly $\mathrm{P}$ is the initial point and $\mathrm{Q}$ is the terminal point. So, the required vector joining $\mathrm{P}$ and $\mathrm{Q}$ is the vector $\overline{\mathrm{PQ}}$, given by
i.e.
$
\begin{aligned}
& \overline{\mathrm{PQ}}=(-1-2) \hat{i}+(-2-3) \hat{j}+(-4-0) \hat{k} \\
& \overline{\mathrm{PQ}}=-3 \hat{i}-5 \hat{j}-4 \hat{k} .
\end{aligned}
$

Example 11

Consider two points $\mathrm{P}$ and $\mathrm{Q}$ with position vectors $\overrightarrow{\mathrm{OP}}=3 \vec{a}-2 \vec{b}$ and $\overrightarrow{\mathrm{OQ}}=\vec{a}+\vec{b}$. Find the position vector of a point $\mathrm{R}$ which divides the line joining $\mathrm{P}$ and $\mathrm{Q}$ in the ratio $2: 1$, (i) internally, and (ii) externally.

Solution
(i) The position vector of the point $\mathrm{R}$ dividing the join of $\mathrm{P}$ and $\mathrm{Q}$ internally in the ratio $2: 1$ is
$
\overrightarrow{\mathrm{OR}}=\frac{2(\vec{a}+\vec{b})+(3 \vec{a}-2 \vec{b})}{2+1}=\frac{5 \vec{a}}{3}
$
(ii) The position vector of the point $\mathrm{R}$ dividing the join of $\mathrm{P}$ and $\mathrm{Q}$ externally in the ratio $2: 1$ is
$
\overrightarrow{\mathrm{OR}}=\frac{2(\vec{a}+\vec{b})-(3 \vec{a}-2 \vec{b})}{2-1}=4 \vec{b}-\vec{a}
$

Example 12

Show that the points $\mathrm{A}(2 \hat{i}-\hat{j}+\hat{k}), \mathrm{B}(\hat{i}-3 \hat{j}-5 \hat{k}), \mathrm{C}(3 \hat{i}-4 j-4 \hat{k})$ are the vertices of a right angled triangle.

Solution

We have
$
\text { and } \quad \begin{aligned}
& \overrightarrow{\mathrm{AB}}=(1-2) \hat{i}+(-3+1) \hat{j}+(-5-1) \hat{k}=-\hat{i}-2 \hat{j}-6 \hat{k} \\
& \overrightarrow{\mathrm{BC}}=(3-1) \hat{i}+(-4+3) \hat{j}+(-4+5) \hat{k}=2 \hat{i}-\hat{j}+\hat{k} \\
& \overrightarrow{\mathrm{CA}}=(2-3) \hat{i}+(-1+4) \hat{j}+(1+4) \hat{k}=-\hat{i}+3 \hat{j}+5 \hat{k}
\end{aligned}
$

Further, note that
$
|\overrightarrow{\mathrm{AB}}|^2=41=6+35=|\overrightarrow{\mathrm{BC}}|^2+|\overrightarrow{\mathrm{CA}}|^2
$

Hence, the triangle is a right angled triangle.

Example 13

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes 1 and 2 respectively and when $\vec{a} \cdot \vec{b}=1$.
Solution

Given $\vec{a} \cdot \vec{b}=1,|\vec{a}|=1$ and $|\vec{b}|=2$. We have
$
\theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}
$

Example 14

Find angle ' $\theta$ ' between the vectors $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$.
Solution

The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by
$
\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}
$

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

Therefore, we have
$
\cos \theta=\frac{-1}{3}
$
hence the required angle is
$
\theta=\cos ^{-1}\left(-\frac{1}{3}\right)
$

Example 15

If $\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$, then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

Solution

We know that two nonzero vectors are perpendicular if their scalar product is zero.
Here $\quad \vec{a}+\vec{b}=(5 \hat{i}-\hat{j}-3 \hat{k})+(\hat{i}+3 \hat{j}-5 \hat{k})=6 \hat{i}+2 \hat{j}-8 \hat{k}$
and $\quad \vec{a}-\vec{b}=(5 \hat{i}-\hat{j}-3 \hat{k})-(\hat{i}+3 \hat{j}-5 \hat{k})=4 \hat{i}-4 \hat{j}+2 \hat{k}$
So $\quad(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(6 \hat{i}+2 \hat{j}-8 \hat{k}) \cdot(4 \hat{i}-4 \hat{j}+2 \hat{k})=24-8-16=0$.
Hence $\quad \vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular vectors.
Example 16

Find the projection of the vector $\vec{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}$ on the vector $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$
Solution

The projection of vector $\vec{a}$ on the vector $\vec{b}$ is given by
$
\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b})=\frac{(2 \times 1+3 \times 2+2 \times 1)}{\sqrt{(1)^2+(2)^2+(1)^2}}=\frac{10}{\sqrt{6}}=\frac{5}{3} \sqrt{6}
$

Example 17

Find $|\vec{a}-\vec{b}|$, if two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}|=2,|\vec{b}|=3$ and $\vec{a} \cdot \vec{b}=4$.

Solution

We have
$
\begin{aligned}
|\vec{a}-\vec{b}|^2 & =(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b}) \\
= & \vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}
\end{aligned}
$

$\begin{aligned}
& =|\vec{a}|^2-2(\vec{a} \cdot \vec{b})+|\vec{b}|^2 \\
& =(2)^2-2(4)+(3)^2 \\
|\vec{a}-\vec{b}| & =\sqrt{5}
\end{aligned}$

Example 18

If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8$, then find $|\vec{x}|$.
Solution

Since $\vec{a}$ is a unit vector, $|\vec{a}|=1$. Also,
$
(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8
$
or
$
\vec{x} \cdot \vec{x}+\vec{x} \cdot \vec{a}-\vec{a} \cdot \vec{x}-\vec{a} \cdot \vec{a}=8
$
or
$
|\vec{x}|^2-1=8 \text { i.e. }|\vec{x}|^2=9
$

Therefore
$|\vec{x}|=3$ (as magnitude of a vector is non negative).

Example 19

For any two vectors $\vec{a}$ and $\vec{b}$, we always have $|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}|$ (CauchySchwartz inequality).
Solution

The inequality holds trivially when either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}$. Actually, in such a situation we have $|\vec{a} \cdot \vec{b}|=0=|\vec{a}||\vec{b}|$. So, let us assume that $|\vec{a}| \neq 0 \neq|\vec{b}|$.
Then, we have
$
\frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|}=|\cos \theta| \leq 1
$

Therefore
$
|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}|
$

Example 20

For any two vectors $\vec{a}$ and $\vec{b}$, we always have $|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$ (triangle inequality).

Solution

The inequality holds trivially in case either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}$ (How?). So, let $|\vec{a}| \neq \overrightarrow{0} \neq|\vec{b}|$. Then,

$
\begin{aligned}
|\vec{a}+\vec{b}|^2 & =(\vec{a}+\vec{b})^2=(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b}) \\
& =\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}
\end{aligned}
$

$
\begin{aligned}
& =|\vec{a}|^2+2 \vec{a} \cdot \vec{b}+|\vec{b}|^2 \\
& \leq|\vec{a}|^2+2|\vec{a} \cdot \vec{b}|+|\vec{b}|^2 \\
& \leq|\vec{a}|^2+2|\vec{a} \| \vec{b}|+|\vec{b}|^2 \\
& =(|\vec{a}|+|\vec{b}|)^2
\end{aligned}
$
(scalar product is commutative)
(since $x \leq|x| \forall x \in \mathbf{R}$ )
(from Example 19)

Hence
$
|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|
$

Remark If the equality holds in triangle inequality (in the above Example 20), i.e.
then
$
\begin{aligned}
|\vec{a}+\vec{b}| & =|\vec{a}|+|\vec{b}|, \\
|\overline{\mathrm{AC}}| & =|\overline{\mathrm{AB}}|+|\overline{\mathrm{BC}}|
\end{aligned}
$
showing that the points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear.
Example 21

Show that the points $\mathrm{A}(-2 \hat{i}+3 \hat{j}+5 \hat{k}), \mathrm{B}(\hat{i}+2 \hat{j}+3 \hat{k})$ and $\mathrm{C}(7 \hat{i}-\hat{k})$ are collinear.

Solution

We have
$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =(1+2) \hat{i}+(2-3) \hat{j}+(3-5) \hat{k}=3 \hat{i}-\hat{j}-2 \hat{k}, \\
\overrightarrow{\mathrm{BC}} & =(7-1) \hat{i}+(0-2) \hat{j}+(-1-3) \hat{k}=6 \hat{i}-2 \hat{j}-4 \hat{k}, \\
\overrightarrow{\mathrm{AC}} & =(7+2) \hat{i}+(0-3) \hat{j}+(-1-5) \hat{k}=9 \hat{i}-3 \hat{j}-6 \hat{k} \\
|\overrightarrow{\mathrm{AB}}| & =\sqrt{14},|\overrightarrow{\mathrm{BC}}|=2 \sqrt{14} \text { and }|\overrightarrow{\mathrm{AC}}|=3 \sqrt{14} \\
|\overrightarrow{\mathrm{AC}}| & =|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|
\end{aligned}
$

Therefore
Hence the points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear.

Example 22

Find $|\vec{a} \times \vec{b}|$, if $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}$
Solution

We have
$
\begin{aligned}
\vec{a} \times \vec{b} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & 3 \\
3 & 5 & -2
\end{array}\right| \\
& =\hat{i}(-2-15)-(-4-9) \hat{j}+(10-3) \hat{k}=-17 \hat{i}+13 \hat{j}+7 \hat{k} \\
\text { Hence } \quad|\vec{a} \times \vec{b}| & =\sqrt{(-17)^2+(13)^2+(7)^2}=\sqrt{507}
\end{aligned}
$

Example 23

Find a unit vector perpendicular to each of the vectors $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$, where $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \quad \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$.
Solution

We have $\vec{a}+\vec{b}=2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $\vec{a}-\vec{b}=-\hat{j}-2 \hat{k}$
A vector which is perpendicular to both $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ is given by
$
(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
0 & -1 & -2
\end{array}\right|=-2 \hat{i}+4 \hat{j}-2 \hat{k} \quad(=\vec{c}, \text { say })
$

Now
$
|\vec{c}|=\sqrt{4+16+4}=\sqrt{24}=2 \sqrt{6}
$

Therefore, the required unit vector is
$
\frac{\vec{c}}{|\vec{c}|}=\frac{-1}{\sqrt{6}} \hat{i}+\frac{2}{\sqrt{6}} \hat{j}-\frac{1}{\sqrt{6}} \hat{k}
$

Example 24

Find the area of a triangle having the points $\mathrm{A}(1,1,1), \mathrm{B}(1,2,3)$ and $C(2,3,1)$ as its vertices.

Solution

We have $\overrightarrow{\mathrm{AB}}=\hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{AC}}=\hat{i}+2 \hat{j}$. The area of the given triangle is $\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$.

Now
$
\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 1 & 2 \\
1 & 2 & 0
\end{array}\right|=-4 \hat{i}+2 \hat{j}-\hat{k}
$

Therefore
$
|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{16+4+1}=\sqrt{21}
$

Thus, the required area is $\frac{1}{2} \sqrt{21}$

Example 25

Find the area of a parallelogram whose adjacent sides are given by the vectors $\vec{a}=3 \hat{i}+\hat{j}+4 \hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$
Solution

The area of a parallelogram with $\vec{a}$ and $\vec{b}$ as its adjacent sides is given by $|\vec{a} \times \vec{b}|$.

Now
$
\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & 4 \\
1 & -1 & 1
\end{array}\right|=5 \hat{i}+\hat{j}-4 \hat{k}
$

Therefore
$
|\vec{a} \times \vec{b}|=\sqrt{25+1+16}=\sqrt{42}
$
and hence, the required area is $\sqrt{42}$.

Example 26

Write all the unit vectors in XY-plane.
Solution

Let $\vec{r}=x \hat{i}+y \hat{j}$ be a unit vector in XY-plane (Fig 10.28). Then, from the figure, we have $x=\cos \theta$ and $y=\sin \theta($ since $|\vec{r}|=1)$. So, we may write the vector $\vec{r}$ as
$
\vec{r}(=\overrightarrow{\mathrm{OP}})=\cos \theta \hat{i}+\sin \theta \hat{j}
$

Clearly,
$
|\vec{r}|=\sqrt{\cos ^2 \theta+\sin ^2 \theta}=1
$

Also, as $\theta$ varies from 0 to $2 \pi$, the point $\mathrm{P}$ (Fig 10.28) traces the circle $x^2+y^2=1$ counterclockwise, and this covers all possible directions. So, (1) gives every unit vector in the XY-plane.

Example 27

If $\hat{i}+\hat{j}+\hat{k}, 2 \hat{i}+5 \hat{j}, 3 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\hat{i}-6 \hat{j}-\hat{k}$ are the position vectors of points $A, B, C$ and $D$ respectively, then find the angle between $\overline{A B}$ and $\overline{C D}$. Deduce that $\overline{\mathrm{AB}}$ and $\overline{\mathrm{CD}}$ are collinear.

Solution

Note that if $\theta$ is the angle between $A B$ and $C D$, then $\theta$ is also the angle between $\overrightarrow{\mathrm{AB}}$ and $\overline{\mathrm{CD}}$.
Now

Therefore
$
\begin{aligned}
\overline{\mathrm{AB}} & =\text { Position vector of } \mathrm{B}-\text { Position vector of } \mathrm{A} \\
& =(2 \hat{i}+5 \hat{j})-(\hat{i}+\hat{j}+\hat{k})=\hat{i}+4 \hat{j}-\hat{k}
\end{aligned}
$
$
|\overline{\mathrm{AB}}|=\sqrt{(1)^2+(4)^2+(-1)^2}=3 \sqrt{2}
$

Similarly
$
\overrightarrow{\mathrm{CD}}=-2 \hat{i}-8 \hat{j}+2 \hat{k} \text { and }|\overrightarrow{\mathrm{CD}}|=6 \sqrt{2}
$

Thus
$
\begin{aligned}
\cos \theta & =\frac{\overline{\mathrm{AB}} \cdot \overline{\mathrm{CD}}}{|\overline{\mathrm{AB}}||\overline{\mathrm{CD}}|} \\
& =\frac{1(-2)+4(-8)+(-1)(2)}{(3 \sqrt{2})(6 \sqrt{2})}=\frac{-36}{36}=-1
\end{aligned}
$

Since $0 \leq \theta \leq \pi$, it follows that $\theta=\pi$. This shows that $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{CD}}$ are collinear,
Alternatively, $\overline{\mathrm{AB}}=-\frac{1}{2} \overline{\mathrm{CD}}$ which implies that $\overline{\mathrm{AB}}$ and $\overrightarrow{\mathrm{CD}}$ are collinear vectors.
Example 28

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three vectors such that $|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5$ and each one of them being perpendicular to the sum of the other two, find $|\vec{a}+\vec{b}+\vec{c}|$.

Solution

Given $\vec{a} \cdot(\vec{b}+\vec{c})=0, \vec{b} \cdot(\vec{c}+\vec{a})=0, \vec{c} \cdot(\vec{a}+\vec{b})=0$.
Now
$
\begin{aligned}
|\vec{a}+\vec{b}+\vec{c}|^2= & (\vec{a}+\vec{b}+\vec{c})^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c}) \\
= & \vec{a} \cdot \vec{a}+\vec{a} \cdot(\vec{b}+\vec{c})+\vec{b} \cdot \vec{b}+\vec{b} \cdot(\vec{a}+\vec{c}) \\
& +\vec{c} \cdot(\vec{a}+\vec{b})+\vec{c} \cdot \vec{c} \\
= & |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 \\
= & 9+16+25=50
\end{aligned}
$

Therefore
$
|\vec{a}+\vec{b}+\vec{c}|=\sqrt{50}=5 \sqrt{2}
$

Example 29

Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$. Evaluate the quantity $\mu=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$, if $|\vec{a}|=3,|\vec{b}|=4$ and $|\vec{c}|=2$.

Solution

Since $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, we have
$
\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}=0
$
or
$
\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=0
$

Therefore
$
\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=-|\vec{a}|^2=-9
$

Again,
$
\vec{b} \cdot(\vec{a}+\vec{b}+\vec{c})=0
$
or
$
\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}=-|\vec{b}|^2=-16
$

Similarly
$
\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{c}=-4 .
$

Adding (1), (2) and (3), we have
$
2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{a} \cdot \vec{c})=-29
$
or
$
2 \mu=-29 \text {, i.e., } \mu=\frac{-29}{2}
$

Example 30

If with reference to the right handed system of mutually perpendicular unit vectors $\hat{i}, \hat{j}$ and $\hat{k}, \vec{\alpha}=3 \hat{i}-\hat{j}, \vec{\beta}=2 \hat{i}+\hat{j}-3 \hat{k}$, then express $\vec{\beta}$ in the form $\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2$, where $\vec{\beta}_1$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$.
Solution

Let $\vec{\beta}_1=\lambda \vec{\alpha}, \lambda$ is a scalar, i.e., $\vec{\beta}_1=3 \lambda \hat{i}-\lambda \hat{j}$.
Now
$
\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1=(2-3 \lambda) \hat{i}+(1+\lambda) \hat{j}-3 \hat{k} .
$

Now, since $\vec{\beta}_2$ is to be perpendicular to $\vec{\alpha}$, we should have $\vec{\alpha} \cdot \vec{\beta}_2=0$. i.e.,
$
3(2-3 \lambda)-(1+\lambda)=0
$
or
$
\lambda=\frac{1}{2}
$

Therefore
$
\vec{\beta}_1=\frac{3}{2} \hat{i}-\frac{1}{2} \hat{j} \text { and } \vec{\beta}_2=\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k}
$