RD Sharma Class 12 Exercise 22.6 Algebra of Vectors Solutions Maths

RD Sharma Class 12 Exercise 22.6 Algebra of Vectors Solutions Maths - Download PDF Free Online

Edited By Satyajeet Kumar | Updated on Jan 24, 2022 06:19 PM IST

The RD Sharma solutions are a common and best source for the practice of mathematics. The Class 12 RD Sharma chapter 22 exercise 22.6 solution deals with the complex chapter of Algebra of vectors that is quite tough for the students to solve, but the RD Sharma class 12th exercise 22.6 is an effective solution for all the problems. It is highly recommended by thousands of students and teachers for the self-guidance of the students.

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RD Sharma Class 12 Solutions Chapter22 Algebra of vectors - Other Exercise
Algebra of Vectors Excercise: 22.6
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter22 Algebra of vectors - Other Exercise

Algebra of Vectors Excercise: 22.6

Algebra of Vectors Exercise 22.6 Question 1

Answer:
The magnitude of the vector is 7.
Hint:
Square and add using vector magnitude formula.
Given:
$\vec{a}=2 \hat{\imath}+3 \hat{\jmath}-6 \hat{k}$
Solution:
Given: $\vec{a}=2 \hat{\imath}+3 \hat{\jmath}-6 \hat{k}$
We know that
$\begin{aligned} |\vec{a}| &=\sqrt{\left[2^{2}+3^{2}+(-6)^{2}\right]} \\ &=\sqrt{4+9+36} \\ &=\sqrt{49}=7 \end{aligned}$

Algebra of Vectors Exercise 22.6 Question 2

Answer:

The unit vector will be $\frac{3}{13} \hat{\imath}+\frac{4}{13} \hat{\jmath}-\frac{12}{13} \hat{k}$

Hint:

Use vector magnitude formula.

Given:

$3 \hat{\imath}+4 \hat{\jmath}-12 \hat{k}$

Solution:

Let the unit vector in the direction of $\vec{A}=3 \hat{\imath}+4 \hat{\jmath}-12 \hat{k} \text { be } \vec{B}$

So, any unit vector in the direction of $\vec{A}=3 \hat{\imath}+4 \hat{j}-12 \hat{k}$ will be

$\vec{B}=\frac{\vec{A}}{|\vec{A}|}$

So, the magnitude of the vector

$|\vec{A}|=\sqrt{3^{2}+4^{2}+(-12)^{2}}$

$\begin{aligned} &|\vec{A}|=\sqrt{9+16+144}=\sqrt{169} \\ &|\vec{A}|=13 \end{aligned}$

So, the unit vector

$\vec{B}=\frac{3 \hat{\dagger}+4 \hat{\jmath}-12 \hat{k}}{13}=\frac{3}{13} \hat{i}+\frac{4}{13} \hat{\jmath}-\frac{12}{13} \hat{k}$

Algebra of Vectors Exercise 22.6 Question 3

Answer:
The unit vector will be $\frac{1}{\sqrt{21}}(4 \hat{\imath}+2 \hat{\jmath}-\hat{k})$
Hint:
Use vector magnitude formula.
Given:
$\vec{a}=\hat{\imath}-\hat{\jmath}+3 \hat{k}, \vec{b}=2 \hat{\imath}+\hat{\jmath}-2 \hat{k} \text { and } \overrightarrow{\mathrm{c}}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k}$
Solution:
To find the resultant vector we add all the vectors by vector addition.
$\vec{a}=\hat{\imath}-\hat{\jmath}+3 \hat{k}, \vec{b}=2 \hat{i}+\hat{\jmath}-2 \hat{k} \text { and } \overrightarrow{\mathrm{c}}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k}$
So, resultant vector is
$\begin{aligned} &\vec{P}=\vec{a}+\vec{b}+\vec{c} \\ &\vec{P}=(\hat{\imath}-\hat{\jmath}+3 \hat{k})+(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})+(\hat{\imath}+2 \hat{j}-2 \hat{k}) \\ &\vec{P}=4 \hat{\imath}+2 \hat{\jmath}-\hat{k} \end{aligned}$
So, the unit vector $\vec{P}=\frac{\vec{p}}{|\vec{P}|}$
Magnitude of vector
$\begin{aligned} &|\vec{P}|=\sqrt{4^{2}+2^{2}+(-1)^{2}} \\ &\quad=\sqrt{16+4+1}=\sqrt{21} \\ &\therefore \vec{P}=\frac{4 \hat{\imath}+2 \hat{\jmath}-\hat{k}}{\sqrt{21}}=\frac{1}{\sqrt{21}}(4 \hat{\imath}+2 \hat{j}-\hat{k}) \end{aligned}$

Algebra of Vectors Exercise 22.6 Question 4

Answer:
The unit vector parallel to diagonals will be $\frac{1}{\sqrt{6}}(-\hat{\imath}+2 \hat{\jmath}+\hat{k}) \& \frac{-\hat{\imath}+\hat{k}}{\sqrt{2}}$
Hint:
Use vector magnitude formula.
Given:
$\hat{\imath}+\hat{\jmath}-\hat{k} \text { and }-2 \hat{\imath}+\hat{\jmath}+2 \hat{k}$
Solution:

Side BC parallel to: $A \vec{D}=\vec{b}, A \vec{B}=\vec{a}$
So, resultant vector $A \vec{C}=\vec{c}=\vec{a}+\vec{b}$
So, vector
$\begin{aligned} &\vec{c}=\vec{a}+\vec{b}=(\hat{\imath}+\hat{\jmath}-\hat{k})+(-2 \hat{\imath}+\hat{\jmath}+2 \hat{k}) \\\\ &\vec{c}=\hat{\imath}+\hat{\jmath}-\hat{k}-2 \hat{\imath}+\hat{\jmath}+2 \hat{k} \\\\ &\vec{c}=-\hat{\imath}+2 \hat{\jmath}+\hat{k} \end{aligned}$
So, unit vector along the diagonal of parallelogram is
$\begin{aligned} &\vec{c}=\frac{\vec{c}}{|\vec{c}|}=\frac{-\hat{\imath}+2 \hat{\jmath}+\hat{k}}{\sqrt{(-1)^{2}+2^{2}+1^{2}}} \\\\ &A \vec{C}=\vec{c}=\frac{-\hat{\imath}+2 \hat{\jmath}+\hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}}(-\hat{\imath}+2 \hat{\jmath}+\hat{k}) \end{aligned}$
Similarly $\begin{aligned} & B \vec{D}=\vec{b}-\vec{a}=-3 \hat{\imath}+3 \hat{k} \\\\ &\end{aligned}$
$\widehat{B D}=\frac{-3 \hat{\imath}+3 \hat{k}}{\sqrt{-3^{2}+3^{2}}}=\frac{-\hat{\imath}+\hat{k}}{\sqrt{2}}$

Algebra of Vectors Exercise 22.6 Question 5

Answer:
$|3 \vec{a}-2 \vec{b}+4 \vec{c}| \text { will be } \sqrt{398}$
Hint:
Use vector magnitude formula $|\vec{a}|=\sqrt{a^{2}+b^{2}+c^{2}}$
Given:
$\vec{a}=3 \hat{\imath}-\hat{\jmath}-4 \hat{k}, \vec{b}=-2 \hat{\imath}+4 \hat{\jmath}-3 \hat{k} \text { and } \overrightarrow{\mathrm{c}}=\hat{\imath}+2 \hat{\jmath}-\hat{k}$
Solution:
We want to find the magnitude of vector $3 \vec{a}-2 \vec{b}+4 \vec{c}$
So,
$\begin{aligned} &3 \vec{a}-2 \vec{b}+4 \vec{c}=3(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})-2(-2 \hat{\imath}+4 \hat{\jmath}-3 \hat{k})+4(\hat{\imath}+2 \hat{\jmath}-\hat{k}) \\ &3 \vec{a}-2 \vec{b}+4 \vec{c}=9 \hat{i}-3 \hat{j}-12 \hat{k}+4 \hat{\imath}-8 \hat{\jmath}+6 \hat{k}+4 \hat{\imath}+8 \hat{j}-4 \hat{k} \\ &3 \vec{a}-2 \vec{b}+4 \vec{c}=17 \hat{\imath}-3 \hat{\jmath}-10 \hat{k} \end{aligned}$
If a vector is given by:
$\vec{A}=a \hat{i}-b \hat{j}-c \hat{k}$ then the magnitude of vector is generally by $|\vec{a}|$ which is equal to $\sqrt{a^{2}+b^{2}+c^{2}}$
$\begin{array}{r} |3 \vec{a}-2 \vec{b}+4 \vec{c}|=\sqrt{17^{2}+(-3)^{2}+(-10)^{2}} \\\\ =\sqrt{289+9+100}=\sqrt{398} \end{array}$

Algebra of Vectors Exercise 22.6 Question 6

Answer:
The position vector Q has co - ordinate $\left ( 4,1,1 \right )$
Hint:
Find the vector equation of Q.
Given:
$P \vec{Q}=3 \hat{\imath}+2 \hat{\jmath}-\hat{k}$ and the co - ordinates of P is $\left ( 1,-1,2 \right )$
Solution:
Let the position vector of point Q is $\overrightarrow{a}$
So, we need to find the value of $\overrightarrow{a}$
$P \vec{Q}=$ Position vector of Q – Position vector of P
$\begin{gathered} 3 \hat{\imath}+2 \hat{\jmath}-\hat{k}=\vec{a}-(\hat{\imath}-\hat{\jmath}+2 \hat{k}) \\ 3 \hat{\imath}+2 \hat{\jmath}-\hat{k}=\vec{a}-\hat{i}+\hat{j}-2 \hat{k} \\ \vec{a}=3 \hat{\imath}+2 \hat{\jmath}-\hat{k}+\hat{i}-\hat{j}+2 \hat{k} \\ \end{gathered}$
$\vec{a}=4 \hat{\imath}+\hat{\jmath}+\hat{k}$
So, the coordinates of position vector Q is $\left ( 4,1,1 \right )$

Algebra of Vectors Exercise 22.6 Question 7

Answer:
Proved L H S = R H S
Hint:
Use vector magnitude formula for both L.H.S and R.H.S.
Given:
$\hat{\imath}-\hat{j}, 4 \hat{\imath}-3 \hat{j}+\hat{k} \text { and } 2 \hat{i}-4 \hat{\jmath}+5 \hat{k}$
Solution:
In a right triangle $C A^{2}=A B^{2}+B C^{2}$ , where CA is the hypotenuse.
BC is the perpendicular and AB is the base.
Vertices of triangle are given below
$A=(1,-1,0), B=(4,-3,1), C=(2,-4,5)$
So,
$\begin{gathered} A \vec{B}=\vec{B}-\vec{A}=(4 \hat{\imath}-3 \hat{j}+\hat{k})-(\hat{i}-\hat{j}) \\ \\A \vec{B}=4 \hat{\imath}-3 \hat{\jmath}+\hat{k}-\hat{\imath}+\hat{\jmath}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \\\\ |A \vec{B}|=\sqrt{3^{2}+(-2)^{2}+1^{2}}=\sqrt{9+4+1} \\ \end{gathered}$
$|A \vec{B}|=\sqrt{14}$ Eq.(i)
Similarly,
$\begin{aligned} & B \vec{C}=\vec{C}-\vec{B}=(2 \hat{i}-4 \hat{\jmath}+5 \hat{k})-(4 \hat{\imath}-3 \hat{\jmath}+\hat{k}) \\\\ &B \vec{C}=-2 \hat{\imath}-\hat{\jmath}+4 \hat{k} \\ & \begin{array}{l} \end{array} \end{aligned}$

$|B \vec{C} |=\sqrt{(-2)^{2}+(-1)^{2}+4^{2}}=\sqrt{4+1+16} \\$

$|B \vec{C}|=\sqrt{21}$ Eq.(ii)

$\begin{aligned} &C \vec{A}=\vec{A}-\vec{C}=(\hat{\imath}-\hat{j})-(2 \hat{i}-4 \hat{\jmath}+5 \hat{k}) \\\\ &C \vec{A}=-\hat{\imath}+3 \hat{\jmath}-5 \hat{k} \\\\ &|C \vec{A}|=\sqrt{(-1)^{2}+3^{2}+(-5)^{2}}=\sqrt{1+9+25} \\\\ &|C \vec{A}|=\sqrt{35} \\ \end{aligned}$ Eq.(iii)
$\begin{aligned} &C A^{2}=A B^{2}+B C^{2} \\\\ &(\sqrt{35})^{2}=(\sqrt{14})^{2}+(\sqrt{21})^{2} \\\\ &35=14+21 \\\\ &35=35 \end{aligned}$
So, three points form a right angle triangle.

Algebra of Vectors Exercise 22.6 question 8

Answer:
$\begin{aligned} &|A \vec{B}|=\sqrt{\left(b_{1}-a_{1}\right)^{2}+\left(b_{2}-a_{2}\right)^{2}+\left(b_{3}-a_{3}\right)^{2}} \\\\ &|B \vec{C}|=\sqrt{\left(c_{1}-b_{1}\right)^{2}+\left(c_{2}-b_{2}\right)^{2}+\left(c_{3}-b_{3}\right)^{2}} \\\\ &|C \vec{A}|=\sqrt{\left(a_{1}-c_{1}\right)^{2}+\left(a_{2}-c_{2}\right)^{2}+\left(a_{3}-c_{3}\right)^{2}} \end{aligned}$
Hint:
Use position vector formula.
Given:
$a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}, b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k} \text { and } c_{1} \hat{\imath}+c_{2} \hat{\jmath}+c_{3} \hat{k}$
Solution:
Let the position vector of the vertex A is $a_{1} \hat{\imath}+a_{2} \hat{j}+a_{3} \hat{k}$
And similarly $B=b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k} \text { and } C=c_{1} \hat{\imath}+c_{2} \hat{\jmath}+c_{3} \hat{k}$
Side AB is
$\begin{aligned} &A \vec{B}=\vec{B}-\vec{A}=\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)-\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right)\\\\ &A \vec{B}=\left(b_{1}-a_{1}\right) \hat{\imath}+\left(b_{2}-a_{2}\right) \hat{\jmath}+\left(b_{3}-a_{3}\right) \hat{k} \end{aligned}$ Eq.(i)

Equation ( i ) vector representation of the side AB
Magnitude of side AB,
$|A \vec{B}|=\sqrt{\left(b_{1}-a_{1}\right)^{2}+\left(b_{2}-a_{2}\right)^{2}+\left(b_{3}-a_{3}\right)^{2}}$
And similarly for side BC and CA
$\begin{aligned} &B \vec{C}=\vec{C}-\vec{B}=\left(c_{1} \hat{\imath}+c_{2} \hat{\jmath}+c_{3} \hat{k}\right)-\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)\\\\ &B \vec{C}=\left(c_{1}-b_{1}\right) \hat{\imath}+\left(c_{2}-b_{2}\right) \hat{\jmath}+\left(c_{3}-b_{3}\right) \hat{k} \end{aligned}$ Eq.(ii)

$\begin{aligned} &C \vec{A}=\vec{A}-\vec{C}=\left(a_{1} \hat{\imath}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right)-\left(c_{1} \hat{\imath}+c_{2} \hat{\jmath}+c_{3} \hat{k}\right) \\\\ &C \vec{A}=\left(a_{1}-c_{1}\right) \hat{\imath}+\left(a_{2}-c_{2}\right) \hat{\jmath}+\left(a_{3}-c_{3}\right) \hat{k} \end{aligned}$

Length of side BC and CA
$\begin{aligned} &|B \vec{C}|=\sqrt{\left(c_{1}-b_{1}\right)^{2}+\left(c_{2}-b_{2}\right)^{2}+\left(c_{3}-b_{3}\right)^{2}} \\\\ &|C \vec{A}|=\sqrt{\left(a_{1}-c_{1}\right)^{2}+\left(a_{2}-c_{2}\right)^{2}+\left(a_{3}-c_{3}\right)^{2}} \end{aligned}$

Algebra of Vectors Exercise 22.6 question 9

Answer:
Vector $O \vec{C}$ will be $\frac{2}{3} \hat{\imath}+\frac{2}{3} \hat{j}+\frac{4}{3} \hat{k}$
Hint
Use position vector formula.
Given:
Vertices of the triangle are $(1,-1,2),(2,1,3) \text { and }(-1,2,-1)$
Solution:
Centroid of the triangle with vertices $(1,-1,2),(2,1,3)$ and $(-1,2,-1)$ is given by:
$(x, y, z)=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right)$
In vector algebra, x is consider as a co – efficient of $\hat{i}\text { and } \mathrm{y}$ as a co – efficient of $\hat{j}\text { and } \mathrm{z}$ as a co – efficient of $\hat{k}$
So, the position vector of the centroid
$\begin{aligned} &(\hat{i}, \hat{\jmath}, \hat{k})=\frac{1+2-1}{3}, \frac{-1+1+2}{3}, \frac{2+3-1}{3} \\ &(\hat{i}, \hat{\jmath}, \hat{k})=\frac{2}{3}, \frac{2}{3}, \frac{4}{3} \end{aligned}$
So, the location of the centroid is $\left(\frac{2}{3}, \frac{2}{3}, \frac{4}{3}\right)$
And the vector is,
$O \vec{C}=\frac{2}{3} \hat{\imath}+\frac{2}{3} \hat{j}+\frac{4}{3} \hat{k}$

Algebra of Vectors Exercise 22.6 question 10 (i)

Answer:
$\frac{-\hat{\imath}+4 \hat{\jmath}+3 \hat{k}}{3}$
Hint:
Use position vector formula.
Given:
$P(\hat{\imath}+2 \hat{\jmath}+\hat{k}) \text { and } Q(-\hat{\imath}+\hat{\jmath}+\hat{k})$ in the ratio 2 : 1.
Solution:
A vector divides an line Internally as in
Position vector of P and Q are given as: $\mathrm{m}: \mathrm{n}=\frac{m Q+n P}{m+n}$
$O \vec{P}=(\hat{\imath}+2 \hat{\jmath}+\hat{k}) \text { and } O \vec{Q}=(-\hat{\imath}+\hat{j}+\hat{k})$
$\frac{m}{n}=\frac{2}{1}$
The position vector pf point R which divides the line joining two points P and Q internally in the ratio 2 : 1 is given by:
$\begin{aligned} &O \vec{R}=\frac{(\hat{\imath}+2 \hat{\jmath}+\hat{k})+2(-\hat{\imath}+\hat{\jmath}+\hat{k})}{2+1}=\frac{-\hat{\imath}+4 \hat{\jmath}+3 \hat{k}}{3} \\\\ &O \vec{R}=-\frac{1}{3} \hat{\imath}+\frac{4}{3} \hat{\jmath}+\hat{k} \end{aligned}$

Algebra of Vectors Exercise 22.6 question 10 (ii)

Answer:
$-3 \hat{\imath}+\hat{k}$
Hint:
Use position vector formula.
Given:
$P(\hat{\imath}+2 \hat{\jmath}+\hat{k}) \text { and } Q(-\hat{\imath}+\hat{\jmath}+\hat{k})$ i the ratio 2:1
Solution:
Position vector of P and Q are given as:
$\begin{aligned} &O \vec{P}=(\hat{\imath}+2 \hat{\jmath}+\hat{k}) \text { and } O \vec{Q}=(-\hat{\imath}+\hat{\jmath}+\hat{k}) \\\\ &\frac{m}{n}=\frac{2}{1} \end{aligned}$

The position vector of point R which divides the line joining two points P and Q externally in the ratio 2 : 1 is given by:

A vector divides an line Internally as in m:n Externally $=\frac{m \vec{Q}-n \vec{p}}{m-n}$
$\begin{aligned} &O \vec{R}=\frac{(\hat{\imath}+2 \hat{\jmath}+\hat{k})-2(-\hat{\imath}+\hat{\jmath}+\hat{k})}{1-2}=\frac{3 \hat{\imath}-\hat{k}}{-1} \\\\ &O \vec{R}=-3 \hat{\imath}+\hat{k} \end{aligned}$

Algebra of Vectors Exercise 22.6 Question 11

Answer:
$3 \hat{\imath}-\hat{j}+\hat{k}$
Hint:
Mid – point of the vector joining the point $\vec{P} \text { and } \vec{Q}=\frac{\vec{p}+\vec{Q}}{2}$
Given:
Two points are $P(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}) \text { and } Q(4 \hat{\imath}+\hat{\jmath}-2 \hat{k})$
Solution:
Let $\overrightarrow {R}$ be the mid – point of the vector joining the point $\overrightarrow {P}$ and $\overrightarrow {Q}$
Then, $\vec{R} =\frac{\vec{p}+\vec{Q}}{2}$
$\begin{aligned} \\ &=\frac{(2 \hat{t}-3 \hat{\jmath}+4 \hat{k})+(4 \hat{\imath}+\hat{j}-2 \hat{k})}{2} \end{aligned}$ [ By mid – point formula ]
$\begin{aligned} =\frac{6 \hat{\imath}-2 \hat{\jmath}+2 \hat{k}}{2} \\ \therefore \vec{R}=3 \hat{\imath}-\hat{j}+\hat{k} \end{aligned}$

Algebra of Vectors Exercise 22.6 Question 12

Answer:
$\frac{1}{\sqrt{3}}(\hat{\imath}+\hat{\jmath}+\hat{k})$
Hint:
Unit vector $=\frac{P \vec{Q}}{|P \vec{Q}|}$
Given:
P and Q are the points $(1,2,3) \text { and }(4,5,6)$
Solution:
$\begin{aligned} P \vec{Q} &=\vec{Q}-\vec{P} \\ &=(4 \hat{i}+5 \hat{\jmath}+6 \hat{k})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \\ P \vec{Q} &=((3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k})) \end{aligned}$
Any unit vector on the direction of $P \vec{Q}$ is
$\begin{aligned} &\frac{P \vec{Q}}{|P \vec{Q}|}=\frac{3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}}{\sqrt{3^{2}+3^{2}+3^{2}}}=\frac{(3,3,3)}{\sqrt{27}} \\ &=\frac{3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}}{3 \sqrt{3}} \\ &=\frac{\hat{\imath}+\hat{\jmath}+\hat{k}}{\sqrt{3}} \end{aligned}$
Required unit vector $=\frac{1}{\sqrt{3}}(\hat{\imath}+\hat{\jmath}+\hat{k})$

Algebra of Vectors Exercise 22.6 Question 13

Answer:
Hence proved
Hint:
Use distance formula $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$
Given:
$A(2 \hat{\imath}-\hat{\jmath}+\hat{k}), B(\hat{\imath}-3 \hat{\jmath}-5 \hat{k}), C(3 \hat{\imath}-4 \hat{\jmath}-4 \hat{k})$
Solution:

Let BC length be a, AC length be b and AB length be c.
$\begin{aligned} \text { Then, } a &=C-B=\sqrt{(3-1)^{2}+(-4+3)^{2}+(-4+5)^{2}} \\ &=\sqrt{4+1+1} \\ &=\sqrt{6} \end{aligned}$ [ Use distance formula of two points ]
$\begin{aligned} b &=C-A=\sqrt{(3-2)^{2}+(-4+1)^{2}+(-4-1)^{2}} \\ &=\sqrt{1+9+25} \\ &=\sqrt{35} \end{aligned}$ [ Use distance formula of two points ]

$\begin{aligned} c &=A-B=\sqrt{(2-1)^{2}+(-1+3)^{2}+(1+5)^{2}} \\ &=\sqrt{1+4+36} \\ &=\sqrt{41} \end{aligned}$ [ Use distance formula of two points ]
$\begin{aligned} &a^{2}+b^{2}=(\sqrt{6})^{2}+(\sqrt{35})^{2} \\ &=6+35 \\ &=41 \\ &=(\sqrt{41})^{2}=c^{2} \\ &a^{2}+b^{2}=c^{2} \end{aligned}$
ABC is right angle triangle.

Algebra of Vectors Excercise 22.6 Question 14

Answer:
$(3,2,1)$
Hint:
Mid – point $=\frac{ \vec{OP}+ \vec{OQ}}{2}$
Given:
The points are $P(2,3,4) \text { and } Q(4,1,-2)$
Solution:
$\begin{aligned} &O \vec{P}=2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k} \\ &\vec{OQ}=4 \hat{\imath}+\hat{\jmath}-2 \hat{k} \end{aligned}$
Let R the mid - point of the line joining the point P and Q.
$\begin{aligned} &\therefore O \vec{R}=\frac{O \vec{P}+O \vec{Q}}{2} \\ &=\frac{(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})+(4 \hat{\imath}+\hat{\jmath}-2 \hat{k})}{2} \end{aligned}$
$\begin{aligned} &=\frac{6 \hat{\imath}+4 \hat{j}+2 \hat{k}}{2} \\ &=3 \hat{i}+2 \hat{j}+\hat{k} \end{aligned}$
∴The mid – point are $\left ( 3,2,1 \right )$

Algebra of Vectors Excercise 22.6 Question 15

Answer:
$\frac{1}{\sqrt{3}}$
Hint:
Unit vector of $\vec{a}=\frac{\vec{a}}{|\vec{a}|}$
Given:
$x(\hat{\imath}+\hat{\jmath}+\hat{k})$ is a unit vector.
Solution:
$\begin{aligned} &\text { Let } \vec{a}=\hat{\imath}+\hat{\jmath}+\hat{k} \\ &\hat{a}=\frac{\vec{a}}{|\vec{a}|}=x \vec{a} \\ &\Rightarrow x=\frac{1}{|\vec{a}|} \\ &\Rightarrow|\vec{a}|=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3} \\ &x=\frac{1}{\sqrt{3}} \end{aligned}$

Algebra of Vectors Exercise 22.6 Question 16

Answer:
$\frac{3}{\sqrt{22}} \hat{\imath}-\frac{3}{\sqrt{22}} \hat{\jmath}+\frac{2}{\sqrt{22}} \hat{k}$
Hint:
Use vector of a line $\mathrm{AB}=\frac{A \vec{B}}{|A \vec{B}|}$
Given:
$\vec{a}=(\hat{\imath}+\hat{\jmath}+\hat{k}), \vec{b}=(2 \hat{\imath}-\hat{\jmath}+3 \hat{k}), \vec{c}=(\hat{\imath}-2 \hat{\jmath}+\hat{k})$
Solution:
$\begin{aligned} &\text { Let } \vec{d}=2 \vec{a}-\vec{b}+3 \vec{c} \\\\ &=2(\hat{\imath}+\hat{\jmath}+\hat{k})+(2 \hat{\imath}-\hat{\jmath}+3 \hat{k})+3(\hat{\imath}-2 \hat{\jmath}+\hat{k}) \\\\ &=3 \hat{\imath}-3 \hat{\jmath}+2 \hat{k} \end{aligned}$

$\begin{aligned} &|\vec{d}|=\sqrt{(3)^{2}+(3)^{2}+(2)^{2}}=\sqrt{22} \\\\ &\hat{d}=\frac{\vec{d}}{|\vec{d}|} \Rightarrow \hat{d}=\frac{3 \hat{\imath}-3 \hat{\jmath}+2 \hat{k}}{\sqrt{22}} \end{aligned}$
$\hat{d}=\frac{3}{\sqrt{22}} \hat{\imath}-\frac{3}{\sqrt{22}} \hat{\jmath}+\frac{2}{\sqrt{22}} \hat{k}$

Algebra of Vectors Exercise 22.6 Question 17

Answer:
$2 \hat{\imath}-4 \hat{\jmath}+4 \hat{k}$
Hint:
Unit vector $=\frac{\vec{A}}{|\vec{A}|}$ , new vector = $6\times$ unit vector of $2 \vec{a}-\vec{b}+3 \vec{c}$
Given:
$\vec{a}=(\hat{\imath}+\hat{\jmath}+\hat{k}), \vec{b}=(4 \hat{\imath}-2 \hat{\jmath}+3 \hat{k}), \vec{c}=(\hat{\imath}-2 \hat{\jmath}+\hat{k})$
Solution:
$\begin{aligned} &\text { Let } \vec{A}=2 \vec{a}-\vec{b}+3 \vec{c} \\\\ &=2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k}-4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}+3 \hat{\imath}-6 \hat{\jmath}+3 \hat{k} \\\\ &=\hat{\imath}-2 \hat{\jmath}+2 \hat{k} \end{aligned}$

$\begin{aligned} &|2 \vec{a}-\vec{b}+3 \vec{c}|=\sqrt{1^{2}+2^{2}+2^{2}} \\\\ &=\sqrt{9}=3 \end{aligned}$

Let $\vec{B}$ be the new vector
$\begin{aligned} &\vec{B}=6 \hat{A}=6 \frac{\vec{A}}{|\vec{A}|} \\\\ &=6 \times \frac{\vec{A}}{3}=2 \vec{A} \\\\ &=2(\hat{\imath}-2 \hat{\jmath}+2 \hat{k}) \\\\ &=2 \hat{\imath}-4 \hat{\jmath}+4 \hat{k} \end{aligned}$

Algebra of Vectors Exercise 22.6 Question 18

Answer:
$\frac{15}{\sqrt{10}} \hat{\imath}+\frac{5}{\sqrt{10}} \hat{\jmath}$
Hint:
A vector having magnitude 1 and parallel to $\vec{a}=1 \frac{\vec{a}}{|\vec{a}|}$
Given:
$\vec{a}=(2 \hat{\imath}+3 \hat{\jmath}-\hat{k}), \vec{b}=(\hat{\imath}-2 \hat{\jmath}+\hat{k})$
Solution:
If $\vec{c}$ is the resultant of $\vec{a} \text { and } \vec{b}$
$\begin{aligned} &\vec{c}=(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})+(\hat{\imath}-2 \hat{\jmath}+\hat{k}) \\\\ &\vec{c}=3 \hat{\imath}+\hat{\jmath}+0 \hat{k} \end{aligned}$
Now, a vector having magnitude 5 and parallel to $\vec{c}$
$\begin{aligned} &\frac{5 \vec{c}}{|\vec{c}|}=\frac{5(3 \hat{\imath}+\hat{j})}{\sqrt{3^{2}+1^{2}}} \\\\ &=\frac{15}{\sqrt{10}} \hat{\imath}+\frac{5}{\sqrt{10}} \hat{\jmath} \end{aligned}$

Algebra of Vectors Exercise 22.6 Question 19

Answer:
$2\sqrt{2}$ units
Hint:
By triangle law of addition.
Given:
$\begin{aligned} &A \vec{B}=\hat{\jmath}+\hat{\imath} \\\\ &A \vec{C}=3 \hat{\imath}-\hat{\jmath}+4 \hat{k} \end{aligned}$
Solution:

$\begin{aligned} B \vec{C}=A \vec{C}-A \vec{B} &=(3 \hat{\imath}-\hat{\jmath}+4 \hat{k})-(\hat{\jmath}+\hat{\imath}) \\ &=2 \hat{\imath}-2 \hat{\jmath}+4 \hat{k} \end{aligned}$ [ By triangle law of addition ]
$B \vec{D}=\frac{1}{2} B \vec{C}$ ..as D is the midpoint of
$\begin{aligned} &=\hat{\imath}-\hat{\jmath}+2 \hat{k} \\ A \vec{D} &=A \vec{B}+B \vec{D} \end{aligned}$ [ By triangle law of addition ]
$\begin{aligned} &=(\hat{\imath}+\hat{\jmath})+(\hat{\imath}-\hat{\jmath}+2 \hat{k}) \\ &=2 \hat{\imath}+2 \hat{k} \\ |A \vec{D}| &=\sqrt{2^{2}+2^{2}}=\sqrt8 \\ &=2 \sqrt{2} \text { units } \end{aligned}$

The RD Sharma class 12 solution of Algebra of vectors exercise 22.6 is an essential chapter to score good marks in the board exam. The RD Sharma class 12th exercise 22.6 consists of 20 questions that cover all the important concepts of the chapter that is mentioned below-

Magnitude of the vector
Unit vector
Addition and subtraction of vectors
Multiplication and division by a scalar quantity
Dot and Cross product
Position vector

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