RD Sharma Class 12 Exercise 23.1 Scalar or Dot Products Solutions Maths

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RD Sharma Class 12 Exercise 23.1 Scaler and dot product Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 27, 2022 05:07 PM IST

The syllabus for the class 12 students is no less when compared to the other grades. As the challenges get bigger, every student spends on preparing for the exams gets extended. Not everyone who practices reaps the fruit; only the ones who practice in the right way can attain it. Students encounter difficulties in solving sums in the chapters like the Scalar and Dot Product. RD Sharma Solutions And as a solution for all their doubts, the RD Sharma Class 12th Exercise 23.1 plays a major role.

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RD Sharma Class 12 Solutions Chapter23 Scaler and dot product - Other Exercise
Scalar or Dot Products Excercise: 23.1
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter23 Scaler and dot product - Other Exercise

Scalar or Dot Products Excercise: 23.1

Scaler or Dot Products Exercise 23.1 Question 1(i)

Answer: 19
Hints: you must the properly of vectors.
Given: $\vec{a}=\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=4\hat{i}-4\hat{j}+7\hat{k}$ Find $\vec{a}.\vec{b}$
Solution:
$\begin{aligned} &\vec{a}=\hat{\imath}-2 \hat{\jmath}+\hat{k} \\\\ &\vec{b}=4 \hat{\imath}-4 \hat{j}+7 \hat{k} \\\\ &\vec{a} \cdot \vec{b}=(\hat{\imath}-2 \hat{\jmath}+\hat{k})(4 \hat{\imath}-4 \hat{\jmath}+7 \hat{k}) \\ \end{aligned}$ $\therefore\left|\begin{array}{l} \hat{\imath} \cdot \hat{i}=1 \\ \hat{\imath} \cdot \hat{\jmath}=0 \end{array}\right|$
$\begin{aligned} &=(4+8+7) \\\\ &=19 \end{aligned}$

Scaler or Dot Products Exercise 23.1 Question 1(ii)

Answer: 2
Hints: you must know the properties of vectors.
Given: $\vec{a}=\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+\hat{k}$
Solution: $\vec{b}=\hat{j}+2\hat{k}$
$\vec{b}=2\hat{j}+\hat{k}$
$\\\vec{a}.\vec{b}=\left ( \hat{j}+2\hat{k} \right ).\left ( 2\hat{i}+\hat{k} \right ) \\\\ = (0\times2)+(1\times0)+(1\times2)$ $\left|\begin{array}{c} \hat{\imath} \cdot \hat{\jmath}=0 \\ \therefore \hat{\jmath} \cdot \hat{k}=0 \\ \hat{k} \cdot \hat{k}=1 \end{array}\right|$
$=+2$

Scaler or Dot Products Exercise 23.1 Question 1(iii)

Answer: 5
Hints: you must know the property of vectors.
Given: $\vec{a}=\hat{j}-\hat{k}$ and $\vec{b}=2\hat{j}+3\hat{j}-2\hat{k}$
Solution:
$\vec{a}=\hat{j}-\hat{k}$
$\vec{b}=2\hat{j}+3\hat{j}-2\hat{k}$ $\left|\begin{array}{l} \hat{\imath} \cdot \hat{\imath}=1 \\ \hat{\imath} \cdot \hat{\jmath}=0 \\ \hat{J} \cdot \hat{\jmath}=1 \\ \hat{\jmath} \cdot \hat{k}=0 \end{array}\right|$
$\vec{a}.\vec{b}=\left ( \hat{j}-\hat{k} \right )\left ( 2\hat{i}+3\hat{j}-2\hat{k} \right )$
$=3+2$
$=5$

Scaler or Dot Products Exercise 23.1 Question 2(i)

Answer: $\lambda =4$
Hints: you must know the rules of solving vectors.
Given: for what value of $\lambda$ , $\vec{a}$ and $\vec{b}$ are $\perp$ to each other,
$\vec{a}=\lambda \hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}= 4\hat{i}-9\hat{j}+2\hat{k}$
Solution: $\vec{a}=\lambda \hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}= 4\hat{i}-9\hat{j}+2\hat{k}$

If $\vec{a}.\vec{b}=0$ them both are perpendicular to each other

$\begin{aligned} &\vec{a} \cdot \vec{b}=(\lambda \hat{\imath}+2 \hat{\jmath}+\widehat{k})(4 \hat{\imath}-9 \hat{\jmath}+2 \hat{k})=0 \\ \end{aligned}$

$\begin{aligned} &=4 \lambda-18+2=0 \\ &=4 \lambda-16=0 \\ \end{aligned}$

$\begin{aligned} &4 \lambda=16 \\ &\Rightarrow \lambda=4 \end{aligned}$

Scaler or Dot Products Exercise 23.1 Question 2(ii)

Answer: $\lambda =\frac{16}{5}$
Hints: if dot products of two vector are equal to zero then vectors are perpendicular to each other
Given: $\vec{a}=\lambda \hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}=5\hat{i}-9\hat{j}+2\hat{k}$
Solution:
$\vec{a}=\lambda \hat{i}+2\hat{j}+\hat{k}$
$\vec{b}=5\hat{i}-9\hat{j}+2\hat{k}$
If $\vec{a}.\vec{b}=0,$ them both $\vec{a}$ and $\vec{b}$ are $\perp$ to each other
$\begin{gathered} \Rightarrow \vec{a} \cdot \vec{b}=(\lambda \hat{\imath}+2 \hat{\jmath}+\widehat{k})(5 \hat{\imath}-9 \hat{\jmath}+2 \hat{k})=0 \\ \end{gathered}$
$\left|\begin{array}{l} \hat{i} \cdot \hat{l}=1 \\ \hat{\imath} \cdot \hat{\jmath}=0 \end{array}\right|$
$5\lambda -18+2=0$
$5\lambda -16=0$
$5\lambda =16$
$\lambda =\frac{16}{5}$

Scaler or Dot Products Exercise 23.1 Question 2(iii)

Answer: $\lambda =3$
Hints: you must know the proper of vectors
Given: $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$ and $\vec{b}=3\hat{i}+2\hat{j}-\lambda \hat{k}$
Solution: $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$
$\vec{b}=3\hat{i}+2\hat{j}-\lambda \hat{k}$
If $\vec{a}.\vec{b}=0$ them both $\vec{a}$ and $\vec{b}$ are perpendicular to each other
$\begin{aligned} &\Rightarrow(2 \hat{\imath}+3 \hat{\jmath}+4 \widehat{k})(3 \hat{\imath}+2 \hat{\jmath}-\lambda \hat{k})=0 \\ &\Rightarrow 6+6-4 \lambda=0 \\ &12-4 \lambda=0 \\ &-4 \lambda=-12 \\ &\lambda=3 \end{aligned}$ $\left|\begin{array}{l} \hat{i} \cdot \hat{l}=1 \\ \hat{\imath} \cdot \hat{\jmath}=0 \end{array}\right|$

Scaler or Dot Products Exercise 23.1 Question 2(iv)

Answer: -3
Hints: you must know the rules of solving vectors.
Given: $\vec{a}=\lambda \hat{i}+3\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+3\hat{k}$
Solution: $\vec{a}=\lambda \hat{i}+3\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+3\hat{k}$
If $\vec{a}.\vec{b}=0$ them both $\vec{a}$ and $\vec{b}$ are perpendicular to each other
$\begin{aligned} &=(\lambda \hat{\imath}+3 \hat{\jmath}+2 \widehat{k})(\hat{\imath}-\hat{\jmath}+3 \hat{k})=0 \\ \end{aligned}$
$\begin{aligned} &\lambda-3+6=0 \\ &\lambda+3=0 \\ &\lambda=-3 \end{aligned}$

Scaler or Dot Products Exercise 23.1 Question 3

Answer: $\frac{\pi }{3}$
Hints: you must know properly of finding angle from vectors.
Given: $\mid \vec{a}\mid =4,$ $\mid \vec{b}\mid =3,$ and $\vec{a}\: .\: \vec{b} =6$
Solution: $\mid \vec{a}\mid =4,$ $\mid \vec{b}\mid =3,$ $\vec{a}\: .\: \vec{b} =6$
Now, we know $\vec{a}.\vec{b}=\mid \vec{a}\mid \mid \vec{b}\mid \cos \theta$
$\begin{aligned} &6=(4)(3)(\cos \theta) \\ &6=(12)(\cos \theta) \\ &\frac{1}{2}=\cos \theta \\ &\theta=\cos ^{-1}\left(\frac{1}{2}\right) \\ &=\frac{\pi}{3} \end{aligned}$

Scaler or Dot Products Exercise 23.1 Question 4

Answer: - 9
Hint: you must know the rules of solving vectors.
Given: $\vec{a}=\hat{i}-\hat{j}$ and $\vec{b}=-\hat{j}+2\hat{k}$
Solution: $\vec{a}=\hat{i}-\hat{j}$ and $\vec{b}=-\hat{j}+2\hat{k}$
Find $\left ( \vec{a}-2\vec{b} \right ).\left ( \vec{a}+\vec{b} \right )$
$\begin{aligned} &(\vec{a}-2 \vec{b})=(\hat{\imath}-\hat{\jmath})-2(-\hat{\jmath}+2 \hat{k}) \\\\ &=(\hat{\imath}-\hat{\jmath}+2 \hat{\jmath}-4 \hat{k}) \\\\ &(\vec{a}+\vec{b})=(\hat{\imath}-\hat{\jmath})+(-\hat{\jmath}+2 \hat{k}) \\ \end{aligned}$
$\begin{aligned} &=\hat{\imath}-\hat{\jmath}-\hat{\jmath}+2 \hat{k} \\\\ &=\hat{\imath}-2 \hat{\jmath}+2 \hat{k} \\\\ &(\vec{a}-2 \vec{b}) \cdot(\vec{a}+\vec{b}) \\\\ &\Rightarrow(\hat{\imath}+\hat{\jmath}-4 \hat{k})(\hat{\imath}-2 \hat{\jmath}+2 \hat{k}) \\\\ &\Rightarrow 1-2-8 \\\\ &\Rightarrow-9 \end{aligned}$

Scaler or Dot Products Exercise 23.1 Question 5(i)

Answer: $\frac{2\pi }{3}$
Hint: you must know the properly of finding angle b/w two vectors.
Given: $\hat{a}=\hat{i}-\hat{j}$ and $\hat{b}=\hat{j}+\hat{k}$
Solution: $\hat{a}=\hat{i}-\hat{j}$ and $\hat{b}=\hat{j}+\hat{k}$
$\begin{aligned} &(\vec{a} \cdot \vec{b})=(\hat{\imath}-\hat{\jmath}) \cdot(\hat{\jmath}+\hat{k}) \\\\ &=(0,-1+0)-1 \end{aligned}$
$|\vec{a}|=\sqrt{(1)^{2}+(-1)^{2}+(0)^{2}} \quad=\sqrt{1+1} \quad \Rightarrow \sqrt{2}$
$|\vec{b}|=\sqrt{(0)^{2}+(1)^{2}+(1)^{2}} \quad=\sqrt{1+1} \quad \Rightarrow \sqrt{2}$

Now we know $\vec{a}.\vec{b}=\mid \vec{a}\mid \mid \vec{b}\mid \cos \theta$

$\begin{aligned} &-1=(\sqrt{2})(\sqrt{2}) \cos \theta \\\\ &\frac{-1}{2}=\cos \theta \\ \end{aligned}$

$\begin{aligned} &\theta=\frac{2 \pi}{3} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 5(ii)

Answer: $\cos ^{-1}\left ( \frac{-34}{63} \right )$
Hint: you must know the properly of findings angle b/w two vectors
Given: $\vec{a}=3\hat{i}-2\hat{j}+6\hat{k}$ and $\vec{b}=4\hat{i}-\hat{j}+8\hat{k}$
Solution: $\vec{a}=3\hat{i}-2\hat{j}+6\hat{k}$ and $\vec{b}=4\hat{i}-\hat{j}+8\hat{k}$
We know $\vec{a}.\vec{b}=\mid \vec{a}\mid \mid \vec{b}\mid \cos \theta$
Now, $\left ( \vec{a}.\vec{b} \right )$ $=\left ( 3\hat{i}-2\hat{j}+6\hat{k} \right )$ $\left ( 4\hat{i}-\hat{j}+8\hat{k} \right )$
$=12+2-48$ $\Rightarrow 14-48=-34$
Magnitude of $\mid \vec{a}\mid =\sqrt{\left ( 3 \right )^{2}+\left ( -2 \right )^{2}+\left ( -6 \right )^{2}}$
$=\sqrt{9+4+36}$ $=\sqrt{49}$
$\Rightarrow 7$
Magnitude of $\mid \vec{b}\mid =\sqrt{\left ( 4 \right )^{2}-\left ( -1 \right )^{2}+\left ( -8 \right )^{2}}$
$=\sqrt{16+1+64}$
$=\sqrt{81}$
$\Rightarrow 9$
Put values in 1
$\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &-34=(7)(9) \cos \theta \\\\ &-34=(63) \cos \theta \\\\ \end{aligned}$
$\begin{aligned} &\frac{-34}{63}=\cos \theta \\\\ &\theta=\cos ^{-1}\left(\frac{-34}{63}\right) \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 5(iii)

Answer: $\frac{\pi }{2}$
Hint: you must know the rules of finding angle b/w two vectors.
Given: $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=4\hat{i}+4\hat{j}-2\hat{k}$
Solution:
$\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$
$\vec{b}=4\hat{i}+4\hat{j}-2\hat{k}$
We know $\vec{a}.\vec{b}=\mid \vec{a}\mid \mid \vec{b}\mid \cos \theta$
Now, $\left ( \vec{a}.\vec{b} \right )=$ $\left ( 2\hat{i}-\hat{j}+2\hat{k} \right )$ $.\left( 4\hat{i}+4\hat{j}-2\hat{k} \right )$ $\left|\begin{array}{l} \hat{l} \cdot \hat{l}=1 \\ \hat{\imath} \cdot \hat{j}=0 \end{array}\right|$
$\left ( 8-4-4 \right )\Rightarrow 0$
Magnitude of $\mid \vec{a}\mid =\sqrt{\left ( 2 \right )^{2}+\left ( -1 \right )^{2}+\left ( 2 \right )^{2}}$
$=\sqrt{4+1+4}$ $=\sqrt{9}=3$
Magnitude of $\mid \vec{b}\mid =\sqrt{\left ( 4 \right )^{2}+\left ( 4 \right )^{2}+\left ( -2 \right )^{2}}$
= $\sqrt{16+16+4}$ $=\sqrt{36}=6$
Put in 1
$\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &0=(3)(6) \cos \theta \\\\ &0=\cos \theta \\\\ &\theta=\frac{\pi}{2} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 5(iv)

Answer: $\cos ^{-1}\left ( \frac{-3}{\sqrt{84}} \right )$

Hint: You must know the rules of finding angle between two vectors.
Given: $\vec{a}=2\hat{i}-3\hat{j}+\vec{k}$ and $\vec{b}=\hat{i}+\hat{j}-2\vec{k}$
Solution:
$\vec{a}=2\hat{i}-3\hat{j}+\vec{k}$
$\vec{b}=\hat{i}+\hat{j}-2\vec{k}$
we Know $\vec{a}.\vec{b}=\mid \vec{a}\mid \mid \vec{b}\mid \cos \theta$ ....(i)
Now, $\left ( \vec{a}.\vec{b} \right )=$ $\left ( 2\hat{i}-3\hat{j}+\vec{k} \right )$ $\left ( \hat{i}+\hat{j}-2\vec{k} \right )$ $\left|\begin{array}{l} \hat{\imath} \cdot \hat{\imath}=1 \\ \hat{\imath} \cdot \hat{\jmath}=0 \end{array}\right|$
$\left ( 2-3-2 \right )\Rightarrow -3$
Magnitude of $\mid \vec{a}\mid =\sqrt{\left ( 2 \right )^{2}+\left ( -3 \right )^{2}+\left ( -2 \right )^{2}}$
= $\sqrt{4+9+4}$ $=\sqrt{8+9}=\sqrt{17}$
Magnitude of $\mid \vec{b}\mid =\sqrt{\left ( 1 \right )^{2}+\left ( 1 \right )^{2}+\left ( -2 \right )^{2}}$
= $\sqrt{1+1+4}$ $=\sqrt{6}$
Put in (1)
$\vec{a}.\vec{b}=\mid \vec{a}\mid \mid \vec{b}\mid \cos \theta$
$-3=\left ( \sqrt{17} \right )\left ( \sqrt{6} \right )\cos \theta$
$-3=\sqrt{84}\cos \theta$
$\cos \theta =\frac{-3}{\sqrt{84}}$
$\theta =\cos ^{-1}\left ( \frac{-3}{\sqrt{84}} \right )$

Scalar or Dot Products Exercise 23.1 Question 5(v)

Answer: $\cos ^{-1}\left ( \frac{-\sqrt{2}}{3} \right )$
Hint: you must know the rules of solving and finding angle b/w two vectors
Given: $\vec{a}=\hat{i}+\hat{2j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$
Solution:
$\vec{a}=\hat{i}+\hat{2j}-\hat{k}$
$\vec{b}=\hat{i}-\hat{j}+\hat{k}$
$We\: \: know \quad \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$
$Now, (\vec{a} \cdot \vec{b})=(\hat{\imath}+2 \hat{\jmath}-\hat{k})$ $(\hat{\imath}-\hat{\jmath}+\hat{k})$
= $=(1-2-1)$
$=-2$
$Magnitude of |\vec{a}|=\sqrt{(1)^{2}+(2)^{2}+(-1)^{2}}$
$=\sqrt{1+4+1} \quad=\sqrt{6}$
$Magnitude of |\vec{b}|=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}}$
$=\sqrt{1+1+1} \quad=\sqrt{3}$
Put in 1
$\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &(-2)=(\sqrt{6})(\sqrt{3}) \cos \theta \\\\ &-2=\sqrt{18} \cos \theta \\\\ &\frac{-2}{\sqrt{3 \times 3 \times 2}}=\cos \theta \end{aligned}$
$\begin{aligned} &\frac{-2}{3 \sqrt{2}}=\cos \theta \\\\ &\frac{-\sqrt{2} \times \sqrt{2}}{3 \sqrt{2}}=\cos \theta \\\\ &\frac{-\sqrt{2}}{3}=\cos \theta \\\\ &\theta=\cos ^{-1}\left(\frac{-\sqrt{2}}{3}\right) \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 6

Answer: $\frac{\pi }{3},\frac{2\pi }{3},\frac{\pi }{4}$
Hint: you must know the rules of finding angle
Given: $\vec{a}=\hat{i}-\hat{j}+\sqrt{2}\hat{k}$ makes with co-ordinate ones.
Solution: Let $\theta$ , be the angle between $\vec{a}$ and x-axis
$\begin{aligned} &\therefore|\vec{a}|=\sqrt{(1)^{2}+(-1)^{2}+(\sqrt{2})^{2}} \\ &=\sqrt{1+1+2} \\ &=\sqrt{2}=2 \end{aligned}$
$\vec{b}=\hat{i}$ (Because it is a unit vector along x-axis)
$\begin{aligned} &|\vec{b}|=\sqrt{(1)^{2}+0+0}=1 \\ &\vec{a} \cdot \vec{b}=(\hat{\imath}-\hat{\jmath}+\sqrt{2} \hat{k})(\hat{\imath}) \quad\left|\begin{array}{l} \hat{\imath} \cdot \hat{\imath}=1 \\ \hat{\imath} \cdot \hat{\jmath}=0 \end{array}\right| \\ &\Rightarrow 1 \\ &\text { Now, }(\vec{a} \cdot \vec{b})=|\vec{a}||\vec{b}| \cos \theta \end{aligned}$
$\begin{aligned} &1=(2)(1) \cos \theta \\ &\frac{1}{2}=\cos \theta \\ &\theta=\frac{\pi}{3} \\ \end{aligned}$

Now, Let $\theta _{2}$ be the angle between $\vec{a}$ and y-axis.
$\begin{aligned} &|\vec{a}|=\sqrt{(1)^{2}+(-1)^{2}+(\sqrt{2})^{2}}=\sqrt{4} \quad=2 \\ &\vec{b}=\hat{j} \\ \end{aligned}$ (become it is a unit vector along y-axis)
$\mid \vec{b}|=\sqrt{(1)^{2}+0+0} \quad=\sqrt{1} \quad \Rightarrow 1$
$\begin{aligned} &\vec{a} \cdot \vec{b}=(\hat{\imath}-\hat{\jmath}+\sqrt{2} \hat{k})(\hat{j}) \\ &\Rightarrow(-1) \quad\left|\begin{array}{l} \hat{i} \cdot \hat{\imath}=1 \\ \hat{\imath} \cdot \hat{\jmath}=0 \end{array}\right| \end{aligned}$
$Now, (\vec{a} \cdot \vec{b})=|\vec{a}||\vec{b}| \cos \theta$
$\begin{aligned} &-1=(2)(1) \cos \theta \\ &-1=2 \cos \theta_{2} \\ &\cos \theta_{2}=\frac{-1}{2} \\ &\theta_{2}=\frac{2 \pi}{3} \end{aligned}$

Now, Let $\theta _{3}$ be the angle between $\vec{a}$ and y-axis.

$\begin{aligned} &|\vec{a}|=\sqrt{(1)^{2}+(-1)^{2}+(\sqrt{2})^{2}}=\sqrt{4} \quad=2 \\ &\vec{b}=\hat{k} \\ \end{aligned}$ $(become\: \: \hat{k} is \: \: unit\: \: vector \: \: along \: \: z-axis)$

$\begin{aligned} &|\vec{b}|=\sqrt{0+0+(1)^{2}}=1 \\ &\vec{a} \cdot \vec{b}=(\hat{\imath}-\hat{\jmath}+\sqrt{2} \hat{k})(\hat{k}) \quad\left|\begin{array}{l} \hat{k} \cdot \hat{k}=1 \\ \hat{k} \cdot \hat{\imath}=0 \end{array}\right| \\ &=(\sqrt{2}) \end{aligned}$

$Now, (\vec{a} \cdot \vec{b})=|\vec{a}||\vec{b}| \cos \theta$

$\begin{aligned} &\sqrt{2}=(2)(1) \cos \theta \\ &\frac{\sqrt{2}}{2}=\cos \theta \\ &\frac{2}{\sqrt{2}}=\cos \theta \\ &\theta=\frac{\pi}{4} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 7(i)

Answer: $\hat{i}+2\hat{j}+\hat{k}$
Hint: you must know the rules of finding vector from given dot product values
Given: dot product of $\hat{i}+\hat{j}-3\hat{k},\hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}+4\hat{k}$ are 0,5 and 8 respectively,
Solution: Let $a\hat{i}+b\hat{j}+c\hat{k}$ be the required vector
Give that,
$\begin{aligned} &\left(a \hat{\imath}+\widehat{b}_{j}+c \hat{k}\right) \cdot(\hat{\imath}+\hat{\jmath}-3 \hat{k})=0 \\ &a+b-3 c=0 \quad-----(1) \\ &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}-3 \hat{k})=5 \\ &a+3 b-2 c=5 \quad-----(2) \\ &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(2 \hat{\imath}+\hat{\jmath}+4 \hat{k})=8 \\ &2 a+b+4 c=8 \quad-----(3) \end{aligned}$

From 1 and 2, subtract

$\begin{aligned} &(a+b-3 c)-(a+3 b-2 c)=0-5\\ \end{aligned}$

$\begin{aligned} &a+b-3 c-a-3 b-2 c=-5\\ &-2 b-c=-5\\ &c=5-2 b\\ \end{aligned}$

From 2 and 3, subtract

$\begin{aligned} &(a+3 b-2 c)-(2 a+3 b+4 c)=5-8\\ \end{aligned}$

$\begin{aligned} &a+3 b-2 c-2 a-b-4 c=-3\\ &-a+2 b-6 c=-3\\ &a-2 b+6 c=3\\ &a=3+2 b-6 c \end{aligned}$

Where $c=5-2b$

$\begin{aligned} &a=3+2 b-6 c(5-2 b) \\ &=3+2 b-3 c+12 b \\ &a=27+14 b \end{aligned}$

Put value of (a) and (c) in 1

$\begin{aligned} &A+b-3 c=0 \\ \end{aligned}$

$\begin{aligned} &(-27+14 b)+b-3(5-2 b)=0 \\\\ &-27+14 b+b-15+6 b=0 \\ \end{aligned}$

$\begin{aligned} &21 b-42=0 \end{aligned}$

$21b=42$

$B=\frac{42}{21}$

$B=2$

Now put value of b in (c) and (a)

$c=5-2b$

$c=5-2\times 2$

$=5-4$

$c=1$

and $a=-27+14b$

$=-27+14\left ( 2 \right )$

$=-27+28$

$a=1$

$a=1,b=2,c=1$

So, $a \hat{\imath}+b \hat{\jmath}+c \hat{k}=\hat{\imath}+2 \hat{\jmath}+\hat{k}$

Scalar or Dot Products Exercise 23.1 Question 7(ii)

Answer: $2\hat{i}-\hat{j}+\hat{k}$
Hint: you must know the rules of finding vector from gives dot product values.
Given: $\hat{i}-\hat{j}+\hat{k}$ , $2\hat{i}+\hat{j}-3\hat{k},$ $\hat{i}+\hat{j}+\hat{k},$ are 4,0 and 2 respectively
Solution: $\hat{i}-\hat{j}+\hat{k}$ , $2\hat{i}+\hat{j}-3\hat{k},$ $\hat{i}+\hat{j}+\hat{k},$ are 4,0 and 2 respectively
Let $a\hat{i}+b\hat{j}+c\hat{k},$ be the required vector.
Given that
$\begin{aligned} &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=4 \\ &\Rightarrow a-b+c=4 \\ &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(2 \hat{\imath}+\hat{\jmath}-3 \hat{k})=0 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow 2 a+b-3 c=0 \quad-1 \\ &\Rightarrow 2 a \end{aligned}$
$\begin{aligned} &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k}) \\\\ &\Rightarrow a+b+c=2 \\ \end{aligned}$

From 1 and 2 subtract
$\begin{aligned} &(a-b+c)-(a+b+c)=4-2 \\\\ &a-b+c-a-b-c=2 \\\\ &2 b=2 \\\\ &B=-1 \\ \end{aligned}$

From 1 and 2 subtract
$\begin{aligned} &(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k}) \\ \end{aligned}$
$\begin{aligned} &\Rightarrow a+b+c=2 \\\\ &(a-b+c)-(a+b+c)=4-2 \\\\ &a-b+c-a-b-c=2 \\ \end{aligned}$
$2b=2$
$B=-1$

From 1 and 2 subtract
$\begin{aligned} \\\\ &(a-b+c)-(2 a+b-3 c)=4-0 \\\\ &a-b+c-2 a-b+3 c=4 \\\\ &-a-2 b+4 c=4 \\\\ &a+2 b-4 c=-4 \\\\ &a=-4-2 b+4 c \end{aligned}$

Put b = -1

$=-4-2\left ( -1 \right )+4c$

$=-4+2+4c$

$A=4c-2$

Now from 3

$\begin{aligned} &\mathrm{A}+\mathrm{b}+\mathrm{c}=2 \\ &(4 \mathrm{c}-2)+(-1)+\mathrm{c}=2 \\ &4 \mathrm{c}-2-1+\mathrm{c}=2 \\ \end{aligned}$

$\begin{aligned} &5 \mathrm{c}-3=2 \\ &5 \mathrm{c}=3+2 \\ &5 \mathrm{c}=5 \\ &\mathrm{C}=1 \end{aligned}$

Put value of b and c in 1

$a-b+c=4$

$\begin{aligned} &a-(-1)+1=4 \\ &a+1+1=4 \\ &a+2=4 \\ &a=2 \\ &a=2, \quad b=-1, \quad c=1 \\ &\therefore a \hat{\imath}+b \hat{\jmath}+c \hat{k} \cdot 2 \hat{\imath}-\hat{\jmath}+\hat{k} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 8(i)

Answer: proved
Hint: you must know the rules of solving vectors.
Given: $\hat{a}$ and $\hat{b}$ are unit vector in dined at angle $\theta$ then prove $\cos \frac{\theta }{2}=\frac{1}{2}\mid \hat{a}+\hat{b}\mid$
Solution: given that $\hat{a}$ and $\hat{b}$ are unit vectors
$\therefore \mid \hat{a}\mid =1,\mid \hat{b}\mid =1$
We have,
$\begin{aligned} &|\hat{a}+\hat{b}|^{2}=\mid \widehat{a \mid}^{2}+\mid \widehat{b \mid}^{2}+2 \hat{a} \cdot \hat{b}\\ \end{aligned}$
$\begin{aligned} &=1+1+2|\hat{a}||\hat{b}| \cos \theta\\ &=2+2(1)(1) \cos \theta\\ \end{aligned}$
$\begin{aligned} &|\hat{a}+\hat{b}|^{2}=2+2 \cos \theta\\ \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{|\hat{a}+\hat{b}|^{2}-2}{2}\\ \end{aligned}$
$\begin{aligned} &|\hat{a}+\hat{b}|^{2}=|\widehat{a}|^{2}+|\hat{b}|^{2}-2 \hat{a} \cdot \hat{b}\\ \end{aligned}$
$\begin{aligned} &=(1)^{2}+(1)^{2}-2|\hat{a}||\hat{b}| \cos \theta\\ \end{aligned}$
$\begin{aligned} &|\hat{a}+\hat{b}|^{2}=2-2 \cos \theta\\ \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{2-|\hat{a}+\hat{b}|^{2}}{2} \end{aligned}$

Now,

$\begin{aligned} &\cos \theta=\sqrt{\frac{1+\cos \theta }{2}} \end{aligned}$

$\begin{aligned} &=\sqrt{\frac{1+\frac{|\hat{a}+\hat{b}|^{2}-2}{2}}{2}} \\ &=\sqrt{\frac{2+|\hat{a}+\hat{b}|^{2}-2}{4}} \\ &=\sqrt{\frac{|\hat{a}+\hat{b}|^{2}}{4}} \\ &\cos \frac{\vartheta}{2} \Rightarrow \frac{1}{2}|\hat{a}+\hat{b}| \quad=\text { Proved } \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 8(ii)

Answer: Proved
Hint: you must know the property of solving vectors.
Given: if $\hat{a}$ and $\hat{b}$ are unit vectors inclined at angle $\theta$ , prove that $\tan \frac{\theta }{2}=\mid \frac{\hat{a}-\hat{b}}{\hat{a}+\hat{b}}\mid$
Solution: given that $\hat{a}$ and $\hat{b}$ are unit vectors
So $\mid \hat{a}\mid =1,\mid \hat{b}\mid =1$
We have,
$\begin{aligned} &|\hat{a}-\hat{b}|^{2}=|\widehat{a}|^{2}+|\hat{b}|^{2}-2 \hat{a} \cdot \hat{b} \\ &=1+1-2|\hat{a}||\hat{b}| \cos \theta \\ &|\hat{a}-\hat{b}|^{2}=2-2 \cos \theta \\ &\cos \theta=\frac{2-|\hat{a}-\hat{b}|^{2}}{2} \end{aligned}$
And $\begin{aligned} &\sin \frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{2}} \\ \end{aligned}$
$\begin{aligned} &\sqrt{\frac{1-\frac{2-|\hat{a}-\hat{b}|^{2}}{2}}{2}} \\ &\Rightarrow \sqrt{\frac{2-2 + |\hat{a}-\hat{b}|^{2}}{4}} \end{aligned}$
$\begin{aligned} &\Rightarrow \sqrt{\frac{|\hat{a}-\hat{b}|^{2}}{4}} \\ &\sin \frac{\theta}{2}=\frac{1}{2}|\hat{a}-\hat{b}| \end{aligned}$
And similarly $\begin{aligned} &\cos \frac{\theta}{2}=\frac{1}{2}|\hat{a}+\hat{b}| \end{aligned}$
$\begin{aligned} &\therefore \tan \frac{\theta}{2}=\frac{\sin \theta / 2}{\cos \theta / 2}=\frac{\frac{1}{2}|\hat{a}-\hat{b}|}{\frac{1}{2}|\hat{a}+\hat{b}|} \\ &\Rightarrow \frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 9

Answer: Proved
Hint: you must know rule of proving vector properties
Given: if the sum of two unit vectors is a unit vector prove that magnitude of their difference is $\sqrt{3}$
Solution: Let their unit vectors are a, b, c. given, sum of unit vectors is a unit vector
$\begin{aligned} &\therefore a+b=c \\\\ &|\hat{c}|^{2}=|\hat{a}+\hat{b}|^{2} \\\\ &|\hat{c}|^{2}=|a|^{2}+|b|^{2}+2|a||b| \cos \theta \\\\ &=1=1+1+2 \cos \theta \end{aligned}$ $[\therefore|a|-|b|=|c|=1 \text { (unit vector) }]$
$\cos \theta =\frac{-1}{2}$
Now $\begin{aligned} &|\hat{a}-\hat{b}|^{2}=|a|^{2}+|b|^{2}-2|a||b| \cos \theta \\ \end{aligned}$
$\begin{aligned} &|\hat{a}-\hat{b}|^{2}=[1+1+1] \\ &|\hat{a}-\hat{b}|^{2}=[3] \\ &|a-b|=\sqrt{3} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 10

Answer: proved
Hint: you must know the rules of proving vector
Given: if $\vec{a},\vec{b},\vec{c}$ are these mutually perpendicular unit vectors, prove that $\mid \vec{a}+\vec{b}+\vec{c}\mid =\sqrt{3}$
Solution: given that $\vec{a},\vec{b}$ and $\vec{c}$ are unit vectors
So, $\mid \vec{a}\mid =1,\mid \vec{b}\mid =1,\mid \vec{c}\mid =1$
Since they are mutually perpendicular
$\begin{aligned} &\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0 \\ \end{aligned}$
Now, $\begin{aligned} &|\vec{a}+\vec{b}+\vec{c}|^{2}=|\hat{a}|^{2}+|\hat{b}|^{2}+|\hat{c}|^{2}+2 \hat{a} \cdot \hat{b}+2 \hat{b} \cdot \hat{c}+2 \hat{c} \cdot \hat{a} \\ \end{aligned}$
$\begin{aligned} &=1+1+1+0+0+0 \\\\ &|\vec{a}+\vec{b}+\vec{c}|^{2} \Rightarrow 3 \\\\ &|\vec{a}+\vec{b}+\vec{c}| \Rightarrow \sqrt{3} \end{aligned}$
=Proved

Scalar or Dot Products Exercise 23.1 Question 12

Answer: proved
Hint: you must know the rules of solving vectors.
Given: show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to coordinate
Solution: Let $\theta$ , be the angle $b/w \ \ \ \vec{a}$ and x-axis,
$\begin{aligned} &|\vec{a}|=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}}=\sqrt{1+1+1}=\sqrt{3} \\\\ &\vec{b}=\hat{i}(\text { Because } \hat{i} \text { is the unit vector along } \mathrm{x}-\mathrm{axis}) \\\\ &|\vec{b}|=\sqrt{(1)^{2}+0+0}=\sqrt{1} \quad \Rightarrow 1 \\\\ &\vec{a} \cdot \vec{b}=1+0+0=1 \end{aligned}$
$\begin{aligned} &|\vec{b}|=\sqrt{(1)^{2}+0+0}=\sqrt{1} \quad \Rightarrow 1 \\\\ &\vec{a} \cdot \vec{b}=1+0+0=1 \\\\ &\cos \theta_{1}=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{1}{(\sqrt{3})(1)} \quad \Rightarrow \frac{1}{\sqrt{3}} \\\\ &\therefore \theta_{1}=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{aligned}$ 1
$\\\text{Let} \ \theta_{2} \text{ be the angle between } \vec{a} \: and \: \mathrm{y}-axis$
$\begin{aligned} &|\vec{a}|=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}=\sqrt{3}} \\\\ &\vec{b}=\hat{J} \\\\ &|\vec{b}|=\sqrt{0+(1)^{2}+0} \quad=\sqrt{1} \Rightarrow 1 \\\\ &(\vec{a} \cdot \vec{b})=0+1+0=1 \\\\ &\Rightarrow 1 \end{aligned}$
we Know $\vec{a} \cdot \vec{b}=\cos \theta|\vec{a}||\vec{b}|$
$\begin{aligned} &1=\sqrt{3} \cos \theta_{2} \\ &\frac{1}{\sqrt{3}}=\cos \theta_{2} \\ &\therefore \theta_{2}=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{aligned}$

Again , Let $\theta _{3}$ be the angle between $\vec{a}$ and z-axis

$\begin{aligned} &|\vec{a}|=\sqrt{(1)^{2}+(1)^{2}+(1)^{2}=\sqrt{3}} \\\\ &\vec{b}=\hat{k} \quad \text { (Because } \hat{k} \text { is unit vector along } z-\text { axis }) \\\\ &|\vec{b}|=\sqrt{0+0+(1)^{2} \quad=\sqrt{1}}=1 \\\\ &\vec{a} \cdot \vec{b}=(\hat{\imath}-\hat{\jmath}+\hat{k}) \cdot(\hat{k}) \\\\ &\Rightarrow(0+0+1) \quad \Rightarrow 1 \end{aligned}$

We Know $\vec{a} \cdot \vec{b}=\cos \theta|\vec{a}||\vec{b}|$

$1=\sqrt{3} \cos \theta$

$\frac{1}{\sqrt{3}}=\cos \theta$

$\therefore \theta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

From 1, 2, 3 the given vector is equally inclined to the co-ordinate ones.

Scalar or Dot Products Exercise 23.1 Question 13

Answer: proved
Hint: you must know the rules of solving vectors
Given: show that the vectors
$\begin{aligned} &\vec{a}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}), \quad \vec{b}=\frac{1}{7}(3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}) \\ &\vec{c}=\frac{1}{7}(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \end{aligned}$
Solution: we Have
$\begin{aligned} \vec{a}=\frac{1}{7}((2)^{2}+(3)^{2}+(6)^{2}) \\ \end{aligned}$
$\begin{aligned} =\frac{1}{7} \sqrt{4+9+36} \\ \end{aligned}$
$\begin{aligned} =\frac{1}{7} \sqrt{49} \quad \Rightarrow \frac{7}{7} \end{aligned}$
$\begin{aligned} &=1 \\ \end{aligned}$
$\begin{aligned} &\vec{b}=\frac{1}{7} \sqrt{(3)^{2}+(-6)^{2}+(2)^{2}} \\\\ &=\frac{1}{7} \sqrt{9+36+4} \\ \end{aligned}$
$\begin{aligned} &=\frac{1}{7} \sqrt{49} \\\\ &=1 \\\\ &\vec{c}=\frac{1}{7} \sqrt{(6)^{2}+(2)^{2}+(-3)^{2}} \\\\ &=\frac{1}{7} \sqrt{36+4+9} \end{aligned}$
$=\frac{1}{7}\Rightarrow 1$
And
$\begin{aligned} &\vec{a} \cdot \vec{b}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}) \cdot \frac{1}{7}(3 \hat{\imath}-6 \hat{j}+2 \hat{k}) \\ &=\frac{1}{49}(6-18+12) \\ &\Rightarrow 0 \end{aligned}$
$\begin{aligned} &\vec{b} \cdot \vec{c}=\frac{1}{7}(3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}),(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \\ &=\frac{1}{49}(18-12-6) \\ &=0 \end{aligned}$
$\begin{aligned} &\vec{c} \cdot \vec{a}=\frac{1}{7}(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \cdot \frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}) \\ &=\frac{1}{7}(12+6+18) \\ &\Rightarrow 0 \end{aligned}$
So, $\mid \vec{a}\mid =\mid \vec{b}\mid =\mid \vec{c}\mid =1$
And $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$

Therefore, the given vectors are mutually perpendicular unit vectors.

Scalar or Dot Products Exercise 23.1 Question 14

Answer: proved
Hint: you must know the rules of solving vectors
Given: for any two vectors $\vec{a}$ and $\vec{b}$
Show: $\begin{aligned} &(\vec{a}+\vec{b}) \cdot(\vec{a} - \vec{b})=0 \: \: \: \: \: \: \: \quad \Leftrightarrow|\vec{a}|=|\vec{b}| \\ \end{aligned}$
Solution: we have
$\begin{aligned} &(\vec{a}+\vec{b}) \cdot(\vec{a} - \vec{b})=0 \: \: \: \: \: \: \: \: \: \quad\left[(a+b)(a-b)=\left(a^{2}-b^{2}\right)\right] \\ &|\vec{a}|^{2}-|\vec{b}|^{2}=0 \\ &|\vec{a}|=|\vec{b}| \end{aligned}$
$=$ Proved

Scalar or Dot Products Exercise 23.1 Question 15

Answer: $\lambda =-2$
Hint: you must know the rules of solving vectors
Given: $\begin{aligned} &\vec{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k} \\ \end{aligned}$ $Find \lambda$
$\begin{aligned} &\vec{b}=\hat{\imath}+\hat{\jmath}-2 \hat{k} \\ \end{aligned}$ $Such\: \: \vec{a} \: \: is \perp to \: \: \lambda \: \: \vec{b}+c$
$\begin{aligned} &\vec{c}=\hat{\imath}+3 \hat{\jmath}-\hat{k} \end{aligned}$
Solution: The given vectors are $\begin{aligned} &\vec{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k} \\ \end{aligned}$
$\begin{aligned} &\vec{b}=\hat{\imath}+\hat{\jmath}-2 \hat{k} \\ \end{aligned}$ and $\begin{aligned} &\vec{c}=\hat{\imath}+3 \hat{\jmath}-\hat{k} \end{aligned}$
Now,
$\begin{aligned} &\lambda \vec{b}+\vec{c}=\lambda(\hat{\imath}+\hat{\jmath}-2 \hat{k})+\hat{\imath}+3 \hat{\jmath}-\hat{k} \\ &=(\lambda+1) \hat{\imath}+(\lambda+3) \hat{\jmath}-(2 \lambda+1) \hat{k} \end{aligned}$

It is given that

$\begin{aligned} &\vec{a} \perp(\lambda \vec{b}+\vec{c}) \\\\ &\Rightarrow \vec{a}.(\lambda \vec{b}+\vec{c})=0 \\ \end{aligned}$

$\begin{aligned} &\Rightarrow 2(\lambda+1)-(\lambda+3)-(2 \lambda+1)=0 \\\\ &2 \lambda+2-\lambda-3-2 \lambda-1=0 \\\\ &\lambda=-2 \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 16

Answer: $\lambda =\pm 1$
Hint: you must know the rules of solving vectors
Given: if $\vec{p}=5\hat{i}+\lambda \hat{j}-3\hat{k}$ and $\vec{q}=\hat{i}+3\hat{j}-5\hat{k}$
Then find the value of $\lambda$ , so $\vec{p}+\vec{q}$ and $\vec{p}-\vec{q}$ are perpendicular vectors
Solution: given that
$\begin{aligned} &\vec{p}=5 \hat{\imath}+\lambda \hat{\jmath}-3\hat{k}\text { and } \\ &\vec{q}=\hat{\imath}+3 \hat{\jmath}-5 \hat{k} \\ &\vec{p}+\vec{q}=(5 \hat{\imath}+\lambda \hat{\jmath}-3 \hat{k})+(\hat{\imath}+3 \hat{j}-5 \hat{k}) \end{aligned}$
$\begin{aligned} &=6 \hat{\imath}+(\lambda+3) \hat{\jmath}-8 \hat{k} \\ \end{aligned}$
$\begin{aligned} &\vec{p}-\vec{q}=(5 \hat{\imath}+\lambda \hat{\jmath}-3 \hat{k})-(\hat{\imath}+3 \hat{\jmath}-5 \hat{k}) \\ &=4 \hat{\imath}+(\lambda-3) \hat{\jmath}+2 \hat{k} \end{aligned}$

Given that is $\vec{p}+\vec{q}$ orthogonal to $\vec{p}-\vec{q}$

$\begin{aligned} &\Rightarrow(\vec{p}+\vec{q})(\vec{p}-\vec{q})=0 \\\\&\Rightarrow[6 \hat{\imath}+(\lambda+3) \hat{\jmath}-8 \hat{k}] \cdot[4 \hat{\imath}+(\lambda-3) \hat{\jmath}+2 \hat{k}]=0 \\ \end{aligned}$

$\begin{aligned} &\Rightarrow 24+\lambda^{2}-9-16=0 \\\\ &\lambda^{2}=1 \\\\ &\therefore \lambda=\pm 1 \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 17

Answer: $\overrightarrow{B}_{1}= \frac{1}{5}\left ( 3\hat{i}+4\hat{j}+5\hat{k} \right )$
$\overrightarrow{B}_{2}=\frac{1}{5}\left ( 13\hat{i}+9\hat{j}-15\hat{k} \right )$
Hint: you must know the rule of solving vectors
Given: if $\vec{a}=3\hat{i}+4\hat{j}+5\hat{k} \: \: \: \vec{B}=2\hat{i}+\hat{j}-4\hat{k}$
Then express $\vec{B}$ in the form of $\vec{B}=\overrightarrow{B}_{1}+\overrightarrow{B}_{2}$ where $\overrightarrow{B}_{1}$ is parallel to $\vec{a}$ and $\overrightarrow{B_{2}}$ is perpendicular to $\vec{a}$
Solution: given that $\begin{aligned} &\vec a=3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k}, \ \ B=2 \hat{\imath}+\hat{\jmath}-4 \hat{k} \\ \end{aligned}$
Also,
$\begin{aligned} &\vec{B}=\overrightarrow{B_{1}}+\overrightarrow{B_{2}} \\ &\overrightarrow{B_{2}}=\vec{B}-\overrightarrow{B_{1}} \quad 1 \end{aligned}$

Since $\overrightarrow{B_{1}}$ parallel to $\overrightarrow{a}$

$\begin{aligned} &\overrightarrow{B_{1}}=t \vec{a} \\ &\overrightarrow{B_{1}}=t(3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k}) \\ &=3 t \hat{\imath}+4 t \hat{\jmath}+5 t \hat{k} \end{aligned}$

Substituting the value of $\overrightarrow{B_{1}}$ and $\vec{a}$

$\begin{aligned} &\overrightarrow{B_{2}}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k}-(3 t \hat{\imath}+4 t \hat{\jmath}+5 t \hat{k}) \\\\ &=(2-3 t) \hat{\imath}+(1-4 t) \hat{\jmath}+(-4-5 t) \hat{k} \end{aligned}$

Since $\overrightarrow{B}$ is $\perp$ to $\vec{a}$

$\begin{aligned} &\overrightarrow{B_{2}} \cdot \vec{a}=0 \\ &\Rightarrow[(2-3 t) \hat{\imath}+(1-4 t) \hat{\jmath}+(-4-5 t) \hat{k}] \cdot[3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k}]=0 \\ &\Rightarrow 3(2-3 t)+4(1-4 t)+5(-4-5 t)=0 \\ &\Rightarrow 6-9 t+4-16 t-20-25 t=0 \\ &\Rightarrow-50 t=10 \\ &t=\frac{-1}{5} \end{aligned}$

$\therefore$ we get

$\begin{aligned} &\overrightarrow{B_{1}}=-\frac{1}{5}(3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k}) \\ &\overrightarrow{B_{2}}=\frac{1}{5}(13 \hat{\imath}+9 \hat{j}-15 \hat{k}) \\ &=\frac{1}{5}(13 \hat{\imath}+9 \hat{\jmath}-15 \hat{k}) \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 18

Answer: proved
Hint: you must know the rules of solving vectors
Given: if either $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$ then $\vec{a}.\vec{b}=0$
But the inverse need not be true, justify with example
Solution: let us assume that either.
$\mid \vec{a}\mid =0$ or $\mid \vec{b}\mid =0$
Then, $\vec{a}.\vec{b}=\mid \vec{a}\mid \mid \vec{b}\mid\cos \theta =0$
Now let us assume that $\vec{a}.\vec{b}=0$
$\Rightarrow \mid \vec{a}\mid \mid \vec{b}\mid\cos \theta =0$
But here we cannot say that either $\mid \vec{a}\mid =0$ or $\mid \vec{b}\mid =0$
For example, Let,
$\begin{aligned} &\vec{a}=2 \hat{\imath}+\hat{\jmath}+3 \hat{k} \text { and } \vec{b}=-3 \hat{\imath}+2 \hat{k} \\\\ &\text { Here, }|\vec{a}|=\sqrt{4+1+9}=\sqrt{14} \quad \neq 0 \\\\ &|\vec{b}|=\sqrt{9+0+4}=\sqrt{13} \quad \neq 0 \\\\ &\text { But }(\vec{a} \cdot \vec{b})=(2 \hat{\imath}+\hat{\jmath}+3 \hat{k}) \cdot(-3 \hat{\imath}+2 \hat{k}) \\\\ &=-6+0+6 \\\\ &\Rightarrow 0 \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 19

Answer: proved
Hint: you must know the rules of solving vectors
Given: $\vec{b}=\hat{i}-3\hat{j}+5\hat{k},$
$\vec{c}=2\hat{i}+\hat{j}-4\hat{k},$ form a right angle
Solution: Let ABC be the given triangle
$\begin{aligned} &\overrightarrow{A C}=\vec{b}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k} \\ &\overrightarrow{C B}=\vec{a}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \\ &\overrightarrow{A B}=\vec{c}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k} \\ &\vec{a} \cdot \vec{b}=3+6+5=14 \\ &\vec{b} \cdot \vec{c}=2-3-20=-21 \\ &\vec{a} \cdot \vec{c}=6-2-4=0\\&\operatorname{So} \overrightarrow{A B} \text { is } \perp \text { to } \overrightarrow{CB} \end{aligned}$ $\mid \therefore \Delta ABC \text{is a right angle triangle} \mid$

Scalar or Dot Products Exercise 23.1 Question 20

Answer: $\lambda =8$
Hint: You must know the rules of solving vector.
Given:
$\begin{aligned} &\vec{a}=2 \hat{\imath}+2 \hat{\jmath}+3 \hat{k} \\ &\vec{b}=-\hat{\imath}+2 \hat{\jmath}+\hat{k} \\ &\vec{c}=3 \hat{\imath}+\hat{\jmath} \end{aligned}$
$\vec{a}+\lambda \vec{b} \text{ is perpendicular to }\vec{c} \ then \ find \: \lambda$
Solution: We have,
$\begin{aligned} &\vec{a}=2 \hat{\imath}+2 \hat{\jmath}+3 \hat{k}, \quad \vec{b}=-\hat{\imath}+2 \hat{\jmath}+\hat{k}, \quad \vec{c}=3 \hat{\imath}+\hat{\jmath} \\ &\therefore \vec{a}+\lambda \vec{b}=2 \hat{\imath}+2 \hat{\jmath}+3 \hat{k}+\lambda(-\hat{\imath}+2 \hat{\jmath}+\hat{k}) \\ &\vec{a}+\lambda \vec{b}=(2-\lambda) \hat{\imath}+(2+2 \lambda) \hat{\jmath}+(3+\lambda) \hat{k} \end{aligned}$
$Given \: \: that \: \: \vec{a}+\lambda \vec{b} \: \: is \perp to \ \vec{c}$
$\therefore(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0 \\$
$\Rightarrow[(2-\lambda) \hat{\imath}+(2+2 \lambda) \hat{\jmath}+(3+\lambda) \hat{k}] \cdot(3 \hat{\imath}+\hat{\jmath}) \\$
$\Rightarrow 3(2-\lambda)+1(2+2 \lambda)=0 \\$
$\Rightarrow 6-3 \lambda+2+2 \lambda=0 \\$
$\lambda=8$

Scalar or Dot Products Exercise 23.1 Question 21

Answer: $< A=\frac{\pi }{4},< B=\frac{\pi }{2},< C=\frac{\pi }{4}$
Hint: You must know the rules of finding angle of triangle
Given: Find angle of triangle $A\left ( 0,-1,-2 \right ),B\left ( 3,1,4 \right )$ and $C\left ( 5,7,1 \right )$
Solution: Given that,
$\overrightarrow{O A}=0 \vec{\imath}-1 \hat{\jmath}-2 \hat{k}, \overrightarrow{O B}=3 \vec{\imath}+1 \hat{\jmath}+4 \hat{k}, \overrightarrow{O C}=5 \vec{\imath}+7 \hat{\jmath}+1 \hat{k}$
$\begin{aligned} &\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\ &\overrightarrow{A B}=3 \vec{\imath}+2 \hat{\jmath}+6 \hat{k} \\ &\overrightarrow{A B}=\sqrt{9+4+36} \\ &\overrightarrow{A B}=\sqrt{49} \\ &\overrightarrow{A B}=7 \end{aligned}$
$\begin{aligned} &\overrightarrow{B A}=\overrightarrow{O A}-\overrightarrow{O B} \\ &\overrightarrow{B A}=-3 \vec{\imath}+2 \hat{\jmath}-6 \hat{k} \\ &\overrightarrow{B A}=\sqrt{9+4+36} \\ &\overrightarrow{B A}=\sqrt{49} \\ &\overrightarrow{B A}=7 \\ &\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \end{aligned}$
$\begin{aligned} &\overrightarrow{B C}=2 \vec{\imath}+6 \hat{\jmath}-3 \hat{k} \\ &\overrightarrow{B C}=\sqrt{4+36+9} \\ &\overrightarrow{B C}=\sqrt{49} \\ &\overrightarrow{B C}=7 \end{aligned}$
$\begin{aligned} &\overrightarrow{C B}=\overrightarrow{O B}-\overrightarrow{O C} \\ &\overrightarrow{C B}=-2 \vec{\imath}-6 \hat{\jmath}+3 \hat{k} \\ &\overrightarrow{C B}=\sqrt{4+36+9} \\ &\overrightarrow{C B}=\sqrt{49} \\ &\overrightarrow{C B}=7 \end{aligned}$
$\begin{aligned} &\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C} \\ &\overrightarrow{C A}=-5 \vec{l}-8 \hat{j}-3 \hat{k} \\ &\overrightarrow{C A}=\sqrt{25+64+9} \\ &\overrightarrow{C A}=\sqrt{98} \\ &\overrightarrow{C A}=7 \sqrt{2} \end{aligned}$
$\begin{aligned} &\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A} \\ &\overrightarrow{A C}=5 \vec{\imath}+8 \hat{\jmath}+3 \hat{k} \\ &\overrightarrow{A C}=\sqrt{25+64+9} \\ &\overrightarrow{A C}=\sqrt{98} \\ &\overrightarrow{A C}=7 \sqrt{2} \end{aligned}$
$\begin{aligned} &\operatorname{Cos} A=\frac{\overrightarrow{A B} \cdot \overrightarrow{A C}}{|\overrightarrow{A B}||\overrightarrow{A C}|} \\\\ &\operatorname{Cos} A=\frac{15+16+18}{(7)(7 \sqrt{2})} \\\\ &\operatorname{Cos} A=\frac{49}{49 \sqrt{2}} \\\\ &\operatorname{Cos} A=\frac{1}{\sqrt{2}} \end{aligned}$
$\begin{aligned} &A=\operatorname{Cos}^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ &A=\frac{\pi}{4} \\ &\operatorname{Cos} B=\frac{\overrightarrow{B A} \cdot \overrightarrow{B C}}{|\overrightarrow{B A}||\overrightarrow{B C}|} \end{aligned}$
$\begin{aligned} &\operatorname{Cos} B=\frac{-6-12+18}{(7)(7)} \\ &\operatorname{Cos} B=\frac{0}{49} \\ &\operatorname{Cos} B=0 \\ &B=\operatorname{Cos}^{-1}(0) \\ &B=\frac{\pi}{2} \end{aligned}$
$\begin{aligned} &\operatorname{Cos} C=\frac{\overrightarrow{C B} \cdot \overrightarrow{C A}}{|\overrightarrow{C B}||\overrightarrow{C A}|} \\ &\operatorname{Cos} C=\frac{10+48-9}{(7)(7 \sqrt{2})} \\ &\operatorname{Cos} C=\frac{49}{49 \sqrt{2}} \end{aligned}$
$\begin{aligned} &\operatorname{Cos} C=\frac{1}{\sqrt{2}} \\ &C=\operatorname{Cos}^{-1}\left(\frac{1}{\sqrt{2}}\right) \\ &C=\frac{\pi}{4} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 22

Answer: $\mid \vec{a\mid }=\mid \vec{b}\mid =1$
Hint: You must know the angle of solving vectors.
Given: Find the magnitude of two vectors $\vec{a }$ and $\vec{b}$ that are the same magnitude, are inclined at 60°, whose scalar product is $\frac{1}{2}$ .
Solution: Given that angle between $\vec{a }$ and $\vec{b}$ is 30°
Also,
$\mid \vec{a\mid }=\mid \vec{b}\mid$
$\begin{aligned} &\qquad \vec{a} \cdot \vec{b}=\frac{1}{2}, \ \ \ \ \theta=60^{\circ} \\ \end{aligned}$
we Know
$\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\ &\frac{1}{2}=|\vec{a}||\vec{b}| \cos 60^{\circ} \\ &\frac{1}{2}=|\vec{a}|^{2} \frac{1}{2} \\ &|\vec{a}|^{2}=1 \\ &|\vec{a}|=1 \\ &\therefore|\vec{a}|=|\vec{b}|=1 \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 23

Answer: Proved
Hint: You must know the rules of solving vectors
Given: Show that the point whose position vectors are,
$\vec{a}=4\hat{i}-3\hat{j}+\hat{k}, \ \ \ \vec{b}=2\hat{i}-4\hat{j}+5\hat{k}, \ \ \ \ \ \vec{c}=\hat{i}-\hat{j}$ from a right triangle.
Solution: Given that,
$\begin{aligned} &\vec{a}=\overrightarrow{O A} \Rightarrow 4 \hat{\imath}-3 \hat{\jmath}+\hat{k} \\ &\vec{b}=\overrightarrow{O B} \Rightarrow 2 \hat{\imath}-4 \hat{\jmath}+5 \hat{k} \\ \end{aligned}$
$\begin{aligned} &\qquad \vec{c}=\overrightarrow{O C} \Rightarrow \hat{\imath}-\hat{\jmath} \\ \end{aligned}$
$\begin{aligned} &\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \Rightarrow-2 \hat{\imath}-\hat{\jmath}+4 \hat{k} \\ \end{aligned}$
$\begin{aligned} &\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \Rightarrow-\hat{\imath}+3 \hat{\jmath}-5 \hat{k} \\\\ &\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C} \Rightarrow 3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \\ \end{aligned}$
$\begin{aligned} &\overrightarrow{A B} \cdot \overrightarrow{B C}=2-3-20 \Rightarrow-21 \neq 0 \\ &\overrightarrow{B C} \cdot \overrightarrow{C A}=-3-6-5 \Rightarrow-14 \neq 0 \\ &\overrightarrow{A B} \cdot \overrightarrow{C A}=-6+2+4 \Rightarrow 0 \end{aligned}$
So, $\overrightarrow{AB}$ is perpendicular to $\overrightarrow{CA}$
So, $\Delta ABC$ is a right angle triangle.

Scalar or Dot Products Exercise 23.1 Question 24

Answer: $\cos ^{-1}\left ( \frac{10}{\sqrt{102}} \right )$
Hint: You must know the rules of solving vectors.
Given: If vertices A,Band C of $\Delta ABC$ have position vectors $\left ( 1,2,3, \right )$ $\left ( -1,0,0 \right )$ and $\left ( 0,1,2 \right )$ respectively. What is the magnitude of $< ABC$ ?
Solution: Given that,
$\begin{aligned} &\overrightarrow{O A}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k} \\ &\overrightarrow{O B}=-\hat{\imath}+0 \hat{\jmath}+0 \hat{k} \\ &\overrightarrow{O C}=0 \hat{\imath}+1 \hat{\jmath}+2 \hat{k} \\ &\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \end{aligned}$
$\begin{aligned} &\overrightarrow{A B}=-2 \hat{\imath}-2 \hat{\jmath}-3 \hat{k} \\ & \mid \overrightarrow{A B} \mid=\sqrt{4+4+9} \\ & \mid \overrightarrow{A B} \mid=\sqrt{17} \\ &\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \end{aligned}$
$\begin{aligned} &\overrightarrow{B C}=\hat{\imath}+\hat{\jmath}+2 \hat{k} \\ &\mid \overrightarrow{B C} \mid=\sqrt{1+1+4} \\ & \mid \overrightarrow{B C} \mid=\sqrt{6} \\ &\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C} \\ \end{aligned}$
$\begin{aligned} &\overrightarrow{C A}=\hat{\imath}+\hat{\jmath}+\hat{k} \\ & \mid \overrightarrow{C A} \mid=\sqrt{1+1+1} \\ & \mid \overrightarrow{C A} \mid=\sqrt{3} \end{aligned}$
$\begin{aligned} &\operatorname{Cos}<A B C=\frac{|\overrightarrow{A B} \cdot \overrightarrow{B C}|}{|\overrightarrow{A B}||\overrightarrow{B C}|} \\ &\operatorname{Cos}<A B C=\frac{|-2-2-6|}{|\sqrt{17}||\sqrt{6}|} \\ &\operatorname{Cos}<A B C=\frac{10}{\sqrt{102}} \end{aligned}$
$< ABC=\cos ^{-1}\left ( \frac{10}{\sqrt{102}} \right )$

Scalar or Dot Products Exercise 23 .1 Question 25

Answer: Proved
Hint: You must know the rules of solving vectors.
Given: If A, B and C have position vectors $\left ( 0,1,1, \right )$ $\left ( 3,1,5 \right )$ and $\left ( 0,3,3 \right )$ respectively. Show $\Delta ABC$ right angled at C.
Solution: Given that,
$\begin{aligned} &\overrightarrow{O A}=0 \hat{\imath}+\hat{\jmath}+\hat{k} \\ &\overrightarrow{O B}=3 \hat{\imath}+1 \hat{\jmath}+5 \hat{k} \\ &\overrightarrow{O C}=0 \hat{\imath}+3 \hat{\jmath}+3 \hat{k} \\ \end{aligned}$
$\begin{aligned} &\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B} \\ &\overrightarrow{B C}=-3 \hat{\imath}+2 \hat{\jmath}-2 \hat{k} \\ &\overrightarrow{C A}=\overrightarrow{O A}-\overrightarrow{O C} \\ &\overrightarrow{C A}=0 \hat{\imath}-2 \hat{\jmath}-2 \hat{k} \end{aligned}$

Now,

$\begin{aligned} &\overrightarrow{B C} \cdot \overrightarrow{C A}=0-4+4 \\ &\overrightarrow{B C} \cdot \overrightarrow{C A}=0 \end{aligned}$

So, $\overrightarrow{BC}$ is perpendicular to $.\overrightarrow{CA}$

So, $\Delta ABC$ is right angled at C.

Scalar or Dot Products Exercise 23.1 Question 26

Answer: 2
Hint: You must know the rules of solving vectors.
Given: Find the projection of $\vec{b}+\vec{c}$ on $\vec{a}$ , where $\vec{a}=2\hat{i}-2\hat{j}+\hat{k}$ , $\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$
$\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$
Solution: Given that,
$\vec{a}=2\hat{i}-2\hat{j}+\hat{k}$
$\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$ and
$\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$
$\begin{aligned} &\vec{b}+\vec{c}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k}+2 \hat{i}-\hat{\jmath}+4 \hat{k} \\ &\vec{b}+\vec{c}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k} \end{aligned}$

Projection of $\vec{b}+\vec{c}$ on $\vec{a}$ is

$\begin{aligned} &\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac{(3 \hat{\imath}+\hat{\jmath}+2 \hat{k}) \cdot(2 \hat{\imath}-2 \hat{\jmath}+\hat{k})}{\left|\sqrt{(2)^{2}+(-2)^{2}+(1)^{2}}\right|} \\ &\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac{6-2+2}{|\sqrt{4+4+1}|} \\ &\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac{6}{|\sqrt{9}|} \\ &\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac{6}{3} \\ &\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=2 \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 27

Answer: Proved
Hint: You must know the rules of solving vectors.
Given: 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 orthogonal.
Solution: $\vec{a}=5\hat{i}-\hat{j}-3\hat{k}$ and $\vec{b}=\hat{i}+3\hat{j}-5\hat{k}$
$\begin{aligned} \therefore \vec{a}+\vec{b}=5 \hat{\imath}-\hat{\jmath}-3 \hat{k}+\hat{\imath}+3 \hat{\jmath}-5 \hat{k} \\ \end{aligned}$
$\begin{aligned} \vec{a}+\vec{b}=6 \hat{i}+2 \hat{\jmath}-8 \hat{k} \end{aligned}$
And
$\begin{aligned} &\vec{a}-\vec{b}=5 \hat{\imath}-\hat{\jmath}-3 \hat{k}-\hat{\imath}+3 \hat{j}-5 \hat{k} \\\\ &\vec{a}-\vec{b}=4 \hat{\imath}-4 \hat{\jmath}+2 \hat{k} \\ \end{aligned}$
Now,
$\begin{aligned} &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(6 \hat{\imath}+2 \hat{\jmath}-8 \hat{k}) \cdot(4 \hat{\imath}-4 \hat{\jmath}+2 \hat{k}) \\\\ &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=24-8-16 \end{aligned}$
$\left ( \vec{a}.\vec{b} \right ) . \left ( \vec{a}-\vec{b} \right )=0$

So, $\vec{a}+\vec{b}$ is orthogonal of $\vec{a}-\vec{b}$

Scalar or Dot Products Exercise 23.1 Question 28

Answer: $\frac{\pi }{3},\frac{1}{\sqrt2}\hat{i},\frac{1}{2}\hat{j},\frac{1}{2}\hat{k}$
Hint: You must know the rules of solving vectors.
Given: A unit vector $\vec{a}$ makes angle $\frac{\pi }{4}$ and $\frac{\pi }{3}$ with $\hat{i}$ and $\hat{j}$ respectively and an acute angle $\theta$ and $\hat{k}$ . Find the angle $\theta$ and components of $\vec{a}$
Solution: Let
$\vec{a}=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=1 \Rightarrow(1)$ $[because\: \: \vec{a}\: \: is \: \: a \: \: unit\: \: vector ]$
Now,
$\begin{aligned} &\vec{a} \cdot \hat{\imath}=a_{1} \\ &\Rightarrow|\vec{a}||\hat{\imath}| \cos \frac{\pi}{4}=a_{1}\left[\text { because the angle between } \vec{a} \text { and } \hat{\imath} \text { is } \frac{\pi}{4}\right] \\ &\Rightarrow a_{1}=(1)(1)\left(\frac{1}{\sqrt{2}}\right) \\ &\Rightarrow a_{1}=\frac{1}{\sqrt{2}} \end{aligned}$

Again,

$\begin{aligned} &\Rightarrow|\vec{a}||\hat{\imath}| \cos \frac{\pi}{3}=a_{2} \\ &\Rightarrow(1)(1)\left(\frac{1}{2}\right)=a_{2} \\ &\Rightarrow \frac{1}{2}=a_{2} \end{aligned}$

Now from (1)

$\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)^{2}+a_{2}^{2}=1\left[\right. because the angle between \vec{a} and \hat{\jmath} is \left.\frac{\pi}{3}\right]$

$\begin{aligned} &\frac{1}{2}+\frac{1}{4}+a_{3}^{2}=1 \\ &\frac{3}{4}+a_{3}^{2}=1 \\ &a_{3}^{2}=1-\frac{3}{4} \end{aligned}$

$a_{3}=\frac{1}{2}$

NOw,

$\begin{aligned} &\vec{a} \cdot \hat{k}=a_{3} \\ &\Rightarrow|\vec{a}||\hat{k}| \cos \theta=\frac{1}{2}[\text { because the angle between } \vec{a} \text { and } \hat{k} \text { is } \theta] \\ \end{aligned}$

$\begin{aligned} &\Rightarrow(1)(1) \cos \theta=\frac{1}{2} \\\\ &\theta=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3} \end{aligned}$

And,

$\vec{a}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{2}\hat{j}+\frac{1}{2}\hat{k}$

Scalar or Dot Products Exercise 23.1 Question 29

Answer: $0$
Hint: You must know the rules of solving vectors.
Given: If two vectors $\vec{a}$ and $\vec{b}$ are such that $\mid \vec{a}\mid =2,\mid \vec{b}\mid =1,$ and $\vec{a}.\vec{b}=1$ then find the value of $\left ( 3\vec{a}-5\vec{b} \right ).\left ( 2\vec{a}+7\vec{b} \right )$
Solution: Given that,
$\mid \vec{a}\mid =2,\mid \vec{b}\mid =1,$ and $\vec{a}.\vec{b}=1$
Now,
$\begin{aligned} &(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b}) \\\\ &=6|\vec{a}|^{2}+21 \vec{a} \cdot \vec{b}-10 \vec{b} \cdot \vec{a}-35|\vec{b}|^{2} \\ \end{aligned}$
$\begin{aligned} &=6|\vec{a}|^{2}+21 \vec{a} \cdot \vec{b}-10 \vec{a} \cdot \vec{b}-35|\vec{b}|^{2} \\\\ &=6|\vec{a}|^{2}+11 \vec{a} \cdot \vec{b}-35|\vec{b}|^{2} \\\\ &=6(2)^{2}+11(1)-35(1)^{2} \\\\ &=24+11-35 \\\\ &=0 \end{aligned}$

Scaler or Dot Products Exercise 23.1 Question 30(i)

Answer: 3
Hint: You must know the rules of solving vectors.
Given: $\left ( \vec{x}-\vec{a} \right ).\left ( \vec{x}+\vec{a} \right )=8$
Solution: Given that $\vec{a}$ is a unit vector.
$\begin{aligned} &\Rightarrow|\vec{a}|=1 \rightarrow(1) \\ &(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8 \\ &|\vec{x}|^{2}-|\vec{a}|^{2}=8 \\ &|\vec{x}|^{2}-1^{2}=8 \\ &|\vec{x}|^{2}=9 \\ &|\vec{x}|=3 \end{aligned}$

Scaler or Dot Products Exercise 23.1 Question 30(ii)

Answer: $\sqrt{13}$
Hint: You must know the rules of solving vectors.
Given: $\left ( \vec{x}-\vec{a} \right ).\left ( \vec{x}+\vec{a} \right )=12$
Solution: Given that $\mid \vec{a}\mid =1$ unit vector
$\mid \vec{a}\mid =1$
$\begin{aligned} &(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12 \\ &|\vec{x}|^{2}-|\vec{a}|^{2}=12 \\ &|\vec{x}|^{2}-1^{2}=12 \\ &|\vec{x}|^{2}=13 \\ &|\vec{x}|=\sqrt{13} \end{aligned}$

Scaler or Dot Products Exercise 23.1 Question 31(i)

Answer: $\mid \vec{a}\mid =4 ,\mid \vec{b}\mid =2$
Hint: You must know the rules of solving vectors.
Given: Find $\mid \vec{a}\mid$ and $\mid \vec{b}\mid$ b if $\left ( \vec{a}+\vec{b} \right ).\left ( \vec{a}-\vec{b} \right )=12$ and
Solution: Given,
$\begin{aligned} &|\vec{a}|=2|\vec{b}| \text { and }(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=12 \\\\ &|\vec{a}|^{2}-|\vec{b}|^{2}=12 \\ \end{aligned}$
$\begin{aligned} (2|\vec{b}|)^{2}-|\vec{b}|^{2}=12 \\ \end{aligned}$
$\qquad 3|\vec{b}|^{2}=12 \\$
$|\vec{b}|^{2}=4 \\$
$|\vec{b}|=2 \\$
$|\vec{a}|=2|\vec{b}| \Rightarrow 2(2)=4 \\$
$\therefore|\vec{a}|=4 \text { and }|\vec{b}|=2$

Scaler or Dot Products Exercise 23.1 Question 31(ii)

Answer: $\mid \vec{a}\mid =\frac{8\sqrt{8}}{\sqrt{63}}$ , $\mid \vec{b}\mid =\sqrt{\frac{8}{63}}$
Hint: You must know the rules of solving vectors.
Given: $\left ( \vec{a}+\vec{b} \right ).\left ( \vec{a}-\vec{b} \right )=8$ and $\mid \vec{a}\mid =8\mid \vec{b}\mid$
Solution: Given that,
$\begin{aligned} &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8 \text { and }|\vec{a}|=8|\vec{b}| \\\\ &|\vec{a}|^{2}-|\vec{b}|^{2}=8 \\\\ &(8|\vec{b}|)^{2}-|\vec{b}|^{2}=8 \end{aligned}$
$\begin{aligned} &64|\vec{b}|^{2}-|\vec{b}|^{2}=8 \\\\ &63|\vec{b}|^{2}=8 \\\\ &|\vec{b}|^{2}=\frac{8}{63} \\ \end{aligned}$
$\begin{aligned} &|\vec{b}|=\sqrt{\frac{8}{63}} \\\\ &\therefore|\vec{a}|=\frac{8 \sqrt{8}}{\sqrt{63}} \text { and }|\vec{b}|=\sqrt{\frac{8}{63}} \end{aligned}$

Scaler or Dot Products Exercise 23.1 Question 31(iii)

Answer: $\mid \vec{a}\mid =2,\mid \vec{b}\mid =1$
Hint: You must know the rules of solving vectors.
Given: $\left ( \vec{a}+\vec{b} \right ).\left ( \vec{a}-\vec{b} \right )=3$ and $\mid \vec{a}\mid =2\mid \vec{b}\mid$ and
Solution: Given that,
$\begin{aligned} &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=3 \text { and }|\vec{a}|=2|\vec{b}| \\\\ &|\vec{a}|^{2}-|\vec{b}|^{2}=3 \\\\ &(2|\vec{b}|)^{2}-|\vec{b}|^{2}=3 \end{aligned}$
$\begin{aligned} &4|\vec{b}|^{2}-|\vec{b}|^{2}=3 \\\\ &3|\vec{b}|^{2}=3 \\\\ &|\vec{b}|^{2}=1 \\ \end{aligned}$
$\begin{aligned} &|\vec{b}|=1 \\\\ &|\vec{a}|=2|\vec{b}| \\\\ &|\vec{a}|=2(1) \\\\ &|\vec{a}|=2 \\\\ &\therefore|\vec{a}|=2 \text { and }|\vec{b}|=1 \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 32(i)

Answer: $\sqrt{13}$
Hint: You must know the rules of solving vectors.
Given: Find $\mid \vec{a}-\vec{b}\mid$ if $\mid \vec{a}\mid =2$ , $\mid \vec{b}\mid =5$ , & $\vec{a} .\vec{b}=8$
Solution: $\mid \vec{a}\mid =2$ , $\mid \vec{b}\mid =5$ & $\vec{a} .\vec{b}=8$
We know that,
$\begin{aligned} &|\vec{a}-\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}-2(\vec{a} \cdot \vec{b}) \\ &|\vec{a}-\vec{b}|^{2}=(2)^{2}+(5)^{2}-2(8) \\ &|\vec{a}-\vec{b}|^{2}=4+25-16 \\ &\therefore|\vec{a}-\vec{b}|=\sqrt{13} \end{aligned}$

Scaler or Dot Products Exercise 23.1 Question 32(ii)

Answer: $\sqrt{23}$
Hint: You must know the rules of solving vectors.
Given: Find $\mid \vec{a}-\vec{b}\mid$ if $\mid \vec{a}\mid =3$ , $\mid \vec{b}\mid =4$ & $\vec{a}.\vec{b}=1$
Solution:
$\begin{aligned} &|\vec{a}-\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}-2(\vec{a} \cdot \vec{b}) \\ &|\vec{a}-\vec{b}|^{2}=(3)^{2}+(4)^{2}-2(1) \\ &|\vec{a}-\vec{b}|^{2}=9+16-2 \\ &|\vec{a}-\vec{b}|^{2}=23 \\ &\therefore|\vec{a}-\vec{b}|=\sqrt{23} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 32(iii)

Answer: $\sqrt{5}$
Hint: You must know the rules of solving vectors.
Given: Find $\mid \vec{a}-\vec{b}$ if $\mid \vec{a}\mid =2$ , $\mid \vec{b}\mid =3$ & $\vec{a}.\vec{b}=4$
Solution:
$\mid \vec{a}\mid =2$ , $\mid \vec{b}\mid =3$ & $\vec{a}.\vec{b}=4$

We know,

$\begin{aligned} &|\vec{a}-\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}-2(\vec{a} \cdot \vec{b}) \\ &|\vec{a}-\vec{b}|^{2}=(2)^{2}+(3)^{2}-2(4) \\ &|\vec{a}-\vec{b}|^{2}=4+9-8 \\ \end{aligned}$

$\begin{aligned} &\qquad|\vec{a}-\vec{b}|^{2}=5 \\ \end{aligned}$

$\begin{aligned} &\therefore|\vec{a}-\vec{b}|=\sqrt{5} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 33(i)

Answer: $\frac{\pi }{4}$
Hint: You must know the rules of solving vectors.
Given: Find the angle between two vectors $\vec{a}$ and $\vec{b}$ if $\mid \vec{a}\mid =\sqrt{3}$ , $\mid \vec{b}\mid =2$ & $\vec{a} . \vec{b} =\sqrt{6}$
Solution: Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$
$\mid \vec{a}\mid =\sqrt{3}$ , $\mid \vec{b}\mid =\sqrt{2}$ & $\vec{a} . \vec{b} =\sqrt{6}$
We know that,
$\begin{aligned} &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\\\ &\sqrt{6}=(\sqrt{3})(2) \cos \theta \\\\ &\cos \theta=\frac{\sqrt{6}}{2 \sqrt{3}} \\\\ &\cos \theta=\frac{1}{\sqrt{2}} \\\\ &\theta=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \\\\ &\theta=\frac{\pi}{4} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 34

Answer: $6\hat{i}+2\hat{k},-\hat{i}-2\hat{j}+3\hat{k}$
Hint: You must know the rules of solving vectors.
Given: Express the vector $\vec{a}=5\hat{i}-2\hat{j}+5\hat{k}$ as the sum of the two vectors such that one is parallel to vector $\vec{b}=3\hat{i}+\hat{k}$ and other is perpendicular to $\vec{b}.$
Solution: Given that,
$\vec{a}=5\hat{i}-2\hat{j}+5\hat{k}$ and $\vec{b}=3\hat{i}+\hat{k}$
Let $\vec{x}.$ and $\vec{y}.$ be such that
$\begin{aligned} &\vec{a}=\vec{x}+\vec{y} \\\\ &\vec{y}=\vec{a}-\vec{x} \rightarrow(1) \\ \end{aligned}$
Since, $\vec{x}.$ is parallel to $\vec{b}.$
$\begin{aligned} &\vec{x}=t(3 \hat{\imath}+\hat{k}) \\\\ &\vec{x}=3 t \hat{\imath}+t \hat{k} \\ \end{aligned}$
Substituting the values, $\vec{x}.$ and $\vec{a}.$
$\begin{aligned} &\vec{y}=5 \hat{\imath}-2 \hat{\jmath}+5 \hat{k}-3 t \hat{\imath}+t \hat{k} \\\\ &\vec{y}=(5-3 t) \hat{\imath}-2 \hat{\jmath}+(5-t) \hat{k} \\ \end{aligned}$
Since $\vec{y}.$ is perpendicular to $\vec{b}.$
$\begin{aligned} &\vec{y} \cdot \vec{b}=0 \\ &{[(5-3 t) \hat{\imath}-2 \hat{\jmath}+(5-t) \hat{k}] \cdot 3 \hat{\imath}+\hat{k}=0} \\ &3(5-3 t)+0+(5-t)=0 \\ &15-9 t+5 t=0 \\ &20-10 t=0 \\ &t=2 \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 35

Answer: $\sqrt{2\sqrt{3}}=\mid \vec{a}\mid =\mid \vec{b}\mid$
Hints: You must know the value of same magnitude incident at an angle of 30°, $\vec{a} .\vec{b}=3$ . Find $\mid \vec{a} \mid .\mid \vec{b}\mid$
Solution: Given that angle between $\vec{a}$ and $\vec{b}$ is 30°
Now,
$\begin{aligned} &|\vec{a}| \cdot|\vec{b}| \text { and } \vec{a} \cdot \vec{b}=3 \\\\ &\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\ \end{aligned}$
$\begin{aligned} &3=|\vec{a}||\vec{a}| \cos 30^{\circ} \\\\ &3=|\vec{a}|^{2}\left(\frac{\sqrt{3}}{2}\right) \\ \end{aligned}$
$\begin{aligned} &|\vec{a}|^{2}=\frac{6}{\sqrt{3}} \\\\ &|\vec{a}|^{2}=2 \sqrt{3} \\\\ &|\vec{a}|=\sqrt{2 \sqrt{3}} \\\\ &|\vec{b}|=\sqrt{2 \sqrt{3}} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 36

Answer: $\left ( -\frac{1}{2}\hat{i}-\hat{j}+\frac{1}{2} \hat{k}\right )+\frac{5}{2}\left ( \hat{i}+\hat{k} \right )$
Hints: You must know the rules of solving vectors.
Given: Express $2\hat{i}-\hat{j}+3\hat{k}$ as the sum of a vector parallel and a vector perpendicular to $2\hat{i}+4\hat{k}-2\hat{k}.$
Solution:
Let, $\vec{a}=2\hat{i}-\hat{j}+3\hat{k}$
$\vec{b}=2\hat{i}+4\hat{j}-2\hat{k}$
And $\vec{x}$ and $\vec{y}$ be such that,
$\Rightarrow \vec{a}=\vec{x}+\vec{y}$
$\Rightarrow \vec{y}=\vec{a}-\vec{x}$
Since $\vec{x}$ is parallel to $\vec{b}$
$\begin{aligned} &\Rightarrow \vec{x}=t \vec{b} \\ &\Rightarrow \vec{x}=t(2 \hat{\imath}+4 \hat{\jmath}-2 \hat{k}) \\ &\Rightarrow \vec{x}=2 t \hat{\imath}+4 t \hat{\jmath}-2 t \hat{k} \end{aligned}$

Substituting the values of $\vec{x}$ and $\vec{a}$

$\begin{aligned} &\Rightarrow(2 \hat{\imath}-\hat{\jmath}+3 \hat{k})-(2 t \hat{\imath}+4 t \hat{\jmath}-2 t \hat{k}) \\\\ &\Rightarrow(2-2 t) \hat{\imath}+(-1-4 t) \hat{\jmath}+(3+2 t) \hat{k} \\ \end{aligned}$
Since y is perpendicular to $\vec{b}$

$\begin{aligned} &\Rightarrow \vec{y} \cdot \vec{b}=0 \\ &\Rightarrow[(2-2 t) \hat{\imath}+(-1-4 t) \hat{\jmath}+(3+2 t) \hat{k}] \cdot(2 \hat{\imath}+4 \hat{\jmath}-2 \hat{k})=0 \\ &\Rightarrow 2(2-2 t)+4(-1-4 t)-2(3+2 t)=0 \\ &\Rightarrow 4-4 t-4-10 t-6-4 t=0 \\ &\Rightarrow-24 t=6 \end{aligned}$

$\begin{aligned} &\Rightarrow t=\left(\frac{-1}{4}\right) \\ &\therefore \vec{x}=2\left(\frac{-1}{4}\right) \hat{\imath}+4\left(\frac{-1}{4}\right) \hat{\jmath}-2\left(\frac{-1}{4}\right) \hat{k} \\ &\vec{x}=\frac{-1}{2} \hat{\imath}-\hat{\jmath}+\frac{1}{2} \hat{k} \\ &\vec{y}=\left[2-2\left(\frac{-1}{4}\right)\right] \hat{\imath}+\left[-1-4\left(\frac{-1}{4}\right)\right] \hat{\jmath}+\left[3+2\left(\frac{-1}{4}\right)\right] \hat{k} \\ &\vec{y}=\frac{5}{2} \hat{\imath}+\frac{5}{2} \hat{k}=\frac{5}{2}(\hat{\imath}+\hat{k}) \end{aligned}$
So,

$\vec{a}=\vec{x}+\vec{y}$

$\vec{a}=\left ( -\frac{1}{2}\hat{i}-\hat{j}+\frac{1}{2} \hat{k}\right )+\frac{5}{2}\left ( \hat{i}+\hat{k} \right )$

Scalar or Dot Products Exercise 23.1 Question 37

Answer: $-\hat{i}-\hat{j}-\hat{k},7\hat{i}-2\hat{j}-5\hat{k}$
Hint: You must know the rules of solving vectors.
Given: Decompose the vector $6\hat{i}-3\hat{j}-6\hat{k}$ into vector which are parallel and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$
Solution:
Let $\vec{a}=6\hat{i}-3\hat{j}-6\hat{k}$
$\vec{b}=\hat{i}+\hat{j}+\hat{k}$
And $\vec{x}$ and $\vec{y}$ be such that,
$\Rightarrow \vec{a}=\vec{x}+\vec{y}$
$\Rightarrow \vec{y}=\vec{a}-\vec{x}$
Since $\vec{x}$ is parallel to $\vec{b}$
$\begin{aligned} &\Rightarrow \vec{x}=t \vec{b} \\ &\Rightarrow \vec{x}=t(\hat{\imath}+\hat{\jmath}+\hat{k}) \\ &\Rightarrow \vec{x}=t \hat{\imath}+t \hat{\jmath}+t \hat{k} \end{aligned}$
Substituting the values of $\vec{x}$ and $\vec{a}$
$\begin{aligned} &\vec{y}=(6 \hat{i}-3 \hat{\jmath}-6 \hat{k})-(t \hat{\imath}+t \hat{\jmath}+t \hat{k}) \\ &\vec{y}=(6-t) \hat{\imath}+(-3-t) \hat{\jmath}+(-6-t) \hat{k} \end{aligned}$
Since y is perpendicular to $\vec{b}$
$\begin{aligned} &\Rightarrow \vec{y} \cdot \vec{b}=0 \\ &\Rightarrow[(6-t) \hat{\imath}+(-3-t) \hat{\jmath}+(-6-t) \hat{k}] \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=0 \\ &\Rightarrow 1(6-t)+1(-3-t)+1(-6-t)=0 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow-3-3 t=0 \\ &\Rightarrow t=-1 \\ &\vec{x}=-\hat{\imath}-\hat{\jmath}-\hat{k} \\ &\vec{y}=7 \hat{\imath}-2 \hat{\jmath}-5 \hat{k} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 38

Answer: $\sqrt{73}$
Hint: You must know the rules of solving vectors.
Given: $\begin{aligned} &\vec{a}=5 \hat{\imath}-\hat{\jmath}+7 \hat{k} \quad ; \vec{b}=\hat{\imath}-\hat{\jmath}+\lambda \hat{k} \\ \end{aligned}$ , Find $\lambda \vec{a}+\vec{b}$ is orthogonal to $\vec{a}-\vec{b}$
Solution:
Given that,
$\begin{aligned} &\vec{a}=5 \hat{\imath}-\hat{\jmath}+7 \hat{k} \quad ; \vec{b}=\hat{\imath}-\hat{\jmath}+\lambda \hat{k} \\ &\therefore \vec{a}+\vec{b}=5 \hat{i}-\hat{\jmath}+7 \hat{k}+\hat{\imath}-\hat{\jmath}+\lambda \hat{k} \\ &\vec{a}+\vec{b}=6 \hat{\imath}-2 \hat{\jmath}+(7+\lambda) \hat{k} \end{aligned}$
And
$\begin{aligned} &\vec{a}-\vec{b}=5 \hat{\imath}-\hat{\jmath}+7 \hat{k}-(\hat{\imath}-\hat{\jmath}+\lambda \hat{k}) \\ &\vec{a}-\vec{b}=4 \hat{\imath}+0 \hat{\jmath}+(7-\lambda) \hat{k} \\ \end{aligned}$
Give that $\vec{a}+\vec{b}$ borthogonal $\vec{a}-\vec{b}$

$\begin{aligned} &(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0 \\\\ &{[6 \hat{\imath}-2 \hat{\jmath}+(7+\lambda) \hat{k}] \cdot(4 \hat{\imath}+0 \hat{\jmath}+(7-\lambda) \hat{k})=0} \\\\ &24+0+49-\lambda^{2}=0 \\\\ &\lambda^{2}=73 \\\\ &\lambda=\sqrt{73} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 39

Answer: $\vec{b}$ is any vector.
Hint: You must know the rules of solving vectors.
Given: If $\vec{a}.\vec{a}=0$ and $\vec{a}.\vec{b}=0$ . What can you conclude about the vector $\vec{b}$ ?
Solution:
Given that $\vec{a}.\vec{a}=0$
$\Rightarrow \mid \vec{a}\mid^{2} =0$
$\Rightarrow \mid \vec{a}\mid =0$
Also given that,
$\vec{a}.\vec{b}=0$
$\begin{aligned} &\Rightarrow|\vec{a}||\vec{b}| \cos \theta=0[\text { where } \theta \text { is the angle between } \vec{a} \text { and } \hat{b}] \\ &\Rightarrow 0|\vec{b}| \cos \theta=0 \\ &\Rightarrow 0=0 \end{aligned}$

So, it means that for any vector $\vec{b},$ the given equation $\vec{a}.\vec{b}=0$ is satisfied

Scalar or Dot Products Exercise 23.1 Question 40

Answer: Proved
Hints: You must know the rules of solving vectors.
Given: If $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$ then prove that it is perpendicular to both $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$
Solution: Given that $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$
$\Rightarrow \vec{c}.\vec{a}=0$ and $\Rightarrow \vec{c}.\vec{b}=0$
Now,
$\begin{aligned} &\Rightarrow \vec{c}(\vec{a}+\vec{b})=\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b} \\\\ &\Rightarrow 0+0=0 \end{aligned}$
So, $\vec{c}$ is perpendicular to $\vec{a}+\vec{b}$
Again,
$\begin{aligned} &\vec{c} \cdot(\vec{a}-\vec{b})=\vec{c} \cdot \vec{a}-\vec{c} \cdot \vec{b} \\ &\vec{c} \cdot(\vec{a}-\vec{b})=0-0 \\ &\vec{c} \cdot(\vec{a}-\vec{b})=0 \end{aligned}$

So, $\vec{c}$ is perpendicular to $\vec{a}-\vec{b}$

Scalar or Dot Products Exercise 23.1 Question 41

Answer: $\begin{aligned} &\left(\frac{\vec{a}-\vec{b}}{a b}\right)^{2} \\ \end{aligned}$

Hint: You must know the rules of solving vectors.
Given: If $\begin{aligned} &|\vec{a}|=a \quad \text { and }|\vec{b}|=b \\ \end{aligned}$
Prove $\begin{aligned} &\left(\frac{\vec{a}}{a^{2}}-\frac{\vec{b}}{b^{2}}\right)^{2}=\left(\frac{\vec{a}-\vec{b}}{a b}\right)^{2} \\ \end{aligned}$
Solution
$\begin{aligned} &\left(\frac{\vec{a}}{a^{2}}-\frac{\vec{b}}{b^{2}}\right)^{2}\end{aligned}$
$\begin{aligned} &=\left|\frac{\vec{a}}{a^{2}}\right|^{2}+\left|\frac{\vec{b}}{b^{2}}\right|^{2}-\frac{2 \vec{a} \cdot \vec{b}}{a^{2} b^{2}} \\ \end{aligned}$
$\begin{aligned} &=\frac{|\vec{a}|^{2}}{a^{4}}+\frac{|\vec{b}|^{2}}{b^{4}}-\frac{2 \vec{a} \cdot \vec{b}}{a^{2} b^{2}} \\\\ &=\frac{a^{2}}{a^{4}}+\frac{b^{2}}{b^{4}}-\frac{2 \vec{a} \cdot \vec{b}}{a^{2} b^{2}} \end{aligned}$
$\begin{aligned} &=\frac{1}{a^{2}}+\frac{1}{b^{2}}-\frac{2 \vec{a} \cdot \vec{b}}{a^{2} b^{2}} \\ \end{aligned}$
$\begin{aligned} &=\frac{b^{2}+a^{2}-2 \vec{a} \cdot \vec{b}}{a^{2} b^{2}} \\\\ &=\left(\frac{\vec{a}}{a^{2}}-\frac{\vec{b}}{b^{2}}\right)^{2} \\\\ &=\left(\frac{\vec{a}-\vec{b}}{a b}\right)^{2} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 42

Answer: Proved
Hint: You must know the rules of solving vectors.
Given: If $\vec{a},\vec{b},\vec{c}$ are three non-coplanar vectors, such that $\vec{a}.\vec{a}=\vec{d}.\vec{b}=\vec{d}.\vec{c}=0$ then show that $\vec{d}$ is the null vector.
Solution:
Given that $\vec{a}\cdot \vec{a}=0$
So, either $\vec{a}=0$ or $\vec{d}$ $\perp$ $\vec{a}$
Similarly, $\vec{d}\cdot \vec{b}=0$
So, $\vec{d}=0$ or $\vec{d}$ $\perp$ $\vec{b}$
Also, $\vec{d}\cdot \vec{c}=0$
So, $\vec{d}=0$ or $\vec{d}\perp \vec{c}$
But $\vec{d}$ cannot be perpendicular to $\vec{a},\vec{b},\vec{c}$ as $\vec{a},\vec{b},\vec{c}$ are non-coplanar
So, $\vec{d}=0$ , $\vec{d}$ is null vector.

Scalar or Dot Products Exercise 23.1 Question 43

Answer: Proved
Hint: You must know the rules of solving vectors.
Given: If $\vec{a}$ vector $\vec{a}$ is perpendicular to two non-collinear vector $\vec{b}$ and $\vec{c}$ then show that $\vec{a}$ is $\perp$ to every vector in the plane of $\vec{b}$ and $\vec{c}$ .
Solution: Given that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$
$\vec{a}\cdot \vec{b}=0$ and $\vec{a}\cdot \vec{c}=0$
Now, let $\vec{x}$ be any vector in plane of $\vec{b}$ and $\vec{c}$
Then, $\vec{x}$ is the linear combination of $\vec{b}$ and $\vec{c}$
$\vec{x}=x \vec{b}+y \vec{c} \text { for some } \mathrm{r} \text { and } \mathrm{y}$
$Now,$
$\begin{aligned} &\vec{a} \cdot \vec{x}=\vec{a} \cdot(x \vec{b}+y \vec{c}) \\ &\vec{a} \cdot \vec{x}=x(\vec{a} \cdot \vec{b})+y(\vec{a} \cdot \vec{c}) \\ &\vec{a} \cdot \vec{x}=x(0)+y(0) \\ &\vec{a} \cdot \vec{x}=0 \end{aligned}$
$Thus, \vec{a} \: is \: perpendicular\: to\: \vec{x}$
$That \: is, \vec{a} \: is \: perpendicular \: to\: every \: vector \: in \: plane \: \vec{b} \: and \: \vec{c}$

Scalar or Dot Products Exercise 23.1 Question 44

Answer: $\begin{aligned} &\cos \theta=\frac{|\vec{a}|^{2}-|\vec{b}|^{2}-\mid \vec{a}^{2}}{2|\vec{b}||\vec{c}|} \\ \end{aligned}$
Hint: You must know the rules of solving vectors.
Given: If $\begin{aligned} &\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}, \end{aligned}$ show angle $\begin{aligned} \theta \end{aligned}$ between vector $\begin{aligned} \vec{b} \text { and } \vec{c} \end{aligned}$ is given $\begin{aligned} \frac{|\vec{a}|^{2}-|\vec{b}|^{2}-|\vec{c}|^{2}}{2|\vec{b}||\overrightarrow{\mid}|} \\ \end{aligned}$ $\begin{aligned} \frac{|\vec{a}|^{2}-|\vec{b}|^{2}-|\vec{c}|^{2}}{2|\vec{b}||\overrightarrow{c\mid}} \\ \end{aligned}$
Solution: Given,
$\begin{aligned} &\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \\ &\vec{b}+\vec{c}=-\vec{a} \\ \end{aligned}$
$\begin{aligned} &|\vec{b}+\vec{c}|^{2}=|-\vec{a}|^{2} \\ &|\vec{b}|^{2}+|\vec{c}|^{2}+2 \vec{b} \vec{c}=|\vec{a}|^{2} \\ \end{aligned}$
$\begin{aligned} &2 \vec{b} \vec{c}=|\vec{a}|^{2}-|\vec{b}|^{2}-|\vec{c}|^{2} \\\\ &2|\vec{b}||\vec{c}| \cos \theta=|\vec{a}|^{2}-|\vec{b}|^{2}-|\vec{c}|^{2} \\\\ &\therefore \cos \theta=\frac{|\vec{a}|^{2}-|\vec{b}|^{2}-|\vec{c}|^{2}}{2|\vec{b}||\vec{c}|} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 45

Answer: $-25$
Hint: You must know the rules of solving vectors.
Given: $\begin{aligned} &\vec{u}+\vec{v}+\vec{w}=0 \text { if }|\vec{u}|=3,|\vec{v}|=4,|\vec{w}|=5 \cdot Find\: \vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u} \\ \end{aligned}$
Solution: $\begin{aligned} &\qquad \vec{u}+\vec{v}+\vec{w}=0 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow|\vec{u}+\vec{v}+\vec{w}|=0 \\ &\Rightarrow|\vec{u}+\vec{v}+\vec{w}|^{2}=0 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow|\vec{u}|^{2}+|\vec{v}|^{2}+|\vec{w}|^{2}+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u})=0 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow(3)^{2}+(4)^{2}+(5)^{2}+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u})=0 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow 9+16+25+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u})=0 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow 50+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u})=0 \\\\ &\Rightarrow 2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u})=-50 \\\\ &\therefore \vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}=-25 \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 46

Answer: $xt\left ( -3,-2 \right )\: \: U\left ( 2,3 \right )$
Hints: You must know the rules of solving vectors.
Given: $\begin{aligned} &\vec{a}=x^{2} \hat{\imath}+2 \hat{\jmath}-2 \hat{k} \\ \end{aligned}$ , $\begin{aligned} &\vec{b}=\hat{\imath}-\hat{\jmath}+\hat{k} \\ \end{aligned}$ , $\begin{aligned} &\vec{c}=x^{2} \hat{\imath}+5 \hat{\jmath}-4 \hat{k} \end{aligned}$
Solution: We have,
$\begin{aligned} &\vec{a}=x^{2} \hat{\imath}+2 \hat{\jmath}-2 \hat{k} \\ &\vec{b}=\hat{\imath}-\hat{\jmath}+\hat{k} \\ &\vec{c}=x^{2} \hat{\imath}+5 \hat{\jmath}-4 \hat{k} \end{aligned}$
Let $\theta _{1}$ be the angle between $\vec{a}$ and $\vec{b}$
$\theta _{2}$ be the angle between $\vec{b}$ and $\vec{c}$
Given that $\theta _{1}$ is acute and $\theta _{2}$ is obtuse
$\begin{aligned} &\Rightarrow \cos \theta_{1}>0 \quad \text { and } \theta_{2}<0 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}>0 \text { and } \frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\cdot \vec{c}|}<0 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{x^{2}-4}{\sqrt{x^{4}+4+4} \sqrt{1+1+1}}>0 \quad \text { and } \frac{x^{2}-9}{\sqrt{x^{4}+25+16} \sqrt{1+1+1}}<0 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow x^{2}-4>0 \quad \text { and } x^{2}-9<0 \end{aligned}$
$\begin{aligned} &\Rightarrow x t(-\infty,-2) U(2, \infty) \text { and } x t(-3,3) \\ \end{aligned}$
$\begin{aligned} &\Rightarrow x t(-3,-2) U(2,3) \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 48

Answer: $\frac{-11}{2}$
Hint: You must know the rules of solving vectors.
Given: If $\vec{a}$ and $\vec{b}$ are two non-collinear unit vectors such that $\mid \vec{a}+\vec{b}\mid =\sqrt{3}$ .Find $\left ( 2\vec{a}-5\vec{b} \right )\cdot \left ( 3\vec{a}+\vec{b} \right )$
Solution: We have,
$\mid \vec{a}+\vec{b}\mid =\sqrt{3}$
Squaring both sides
$\begin{aligned} &\Rightarrow|\vec{a}+\vec{b}|^{2}=3 \\ &\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=3 \\ &\Rightarrow 1+1+2 \vec{a} \cdot \vec{b}=3 \end{aligned}$ $[because\: \vec{a} \: and\: \vec{b} \: are\: unit \: vector ]$
$\begin{aligned} &\Rightarrow 2+2 \vec{a} \cdot \vec{b}=3 \\ &\Rightarrow 2 \vec{a} \cdot \vec{b}=3-2 \\ &\Rightarrow \vec{a} \cdot \vec{b}=\frac{1}{2} \end{aligned}$
Now,
$\left ( 2\vec{a}-5\vec{b} \right )\cdot \left ( 3\vec{a}+\vec{b} \right )$
$\begin{aligned} &\Rightarrow 6|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}-15 \vec{b} \cdot \vec{a}-5|\vec{b}|^{2} \\ &\Rightarrow 6|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}-15 \vec{b} \cdot \vec{a}-5|\vec{b}|^{2} \quad[\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}] \\ &\Rightarrow 6|\vec{a}|^{2}-13 \vec{a} \cdot \vec{b}-5|\vec{b}|^{2} \\ &\Rightarrow 6(1)-13\left(\frac{1}{2}\right)-5(1) \\ &\Rightarrow 1-\frac{13}{2} \\ &\Rightarrow \frac{-11}{2} \end{aligned}$

Scalar or Dot Products Exercise 23.1 Question 49

Answer: Proved
Hints: You must know the rules of solving vectors.
Given: If $\vec{a}\vec{b}$ are two vector such that, $\mid \vec{a}+\vec{b}\mid=\mid \vec{b}\mid$ , prove $\vec{a}+2\vec{b}$ is to $\vec{a}$
Solution: Given that,
$\mid \vec{a}+\vec{b}\mid=\mid \vec{b}\mid$
Squaring both sides,
$\begin{aligned} &\Rightarrow|\vec{a}+\vec{b}|^{2}=|\vec{b}|^{2} \\ &\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}+2 \vec{a} \cdot \vec{b}=|\vec{b}|^{2} \\ &\Rightarrow|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b}=0 \end{aligned}$

Now,

$\begin{aligned} &\Rightarrow(\vec{a}+2 \vec{b}) \cdot \vec{a} \\ &\Rightarrow \vec{a} \cdot \vec{a}+2 \vec{b} \vec{a} \\ &\Rightarrow|\vec{a}|^{2}+2 \vec{a} \cdot \vec{b} \quad \therefore[\vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b}] \end{aligned}$

$\Rightarrow 0$
So, $\vec{a}+2\vec{b}$ is perpendicular to $\vec{a}$ .

Scalar or Dot Products Exercise 23.1 Question 50

Answer: $\frac{\pi }{3}$
Hints: You must know the rules of solving vectors.
Given: Let $\vec{a}$ and $\vec{b}$ be unit vector. If the vectors $\vec{c}=\vec{a}+2\vec{b},\vec{d}=5\vec{a}-4\vec{b}$ , are perpendicular to each other. Find the angle between vector $\vec{a}$ and $\vec{b}$
Solution: a and b are unit vectors, ie, $\mid a\mid =\mid b\mid =1$
$\vec{c}=\vec{a}+2\vec{b}\: \: and\: \: \vec{d}=5\vec{a}-4\vec{b}$
c and d are perpendicular to each other,
$c\cdot d=0$
Angle between a and b
$\begin{aligned} &c \cdot d=0 \\ &(a+2 b)(5 a-4 b)=0 \\ &5 a \cdot a-4 a \cdot b+10 b \cdot a-8 b \cdot b=0 \\ &6(a \cdot b)=3 \\ &a \cdot b=\frac{1}{2} \quad \therefore\left[\text { because } \cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\right] \end{aligned}$
So, angle between a and b is $\frac{\pi }{3}$

Chapter 23 of class 12 mathematics consists of only two exercises, ex 23.1 and ex 23.2. The first exercise, ex 23.1, has around 67 questions, including its subparts in the textbook. This exercise revolves around the projection of vectors, unit and position vector, the angle between two vectors, dot product, etc. The best reference book to find the answers to all these questions is the RD Sharma Class 12 Chapter 23 Exercise 23.1. Students and the teachers can use this book to clarify their doubts and get to know the various methods in which each of these sums can be solved.

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