Question 2 and 3 Exercise 3.3

Solutions of Question 2 and 3 of Exercise 3.3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Write a unit vector in the direction of the sum of the vectors: $$\vec{a}=2 \hat{i} + 2 \hat{j}-5 \hat{k}, \quad \vec{b}=2 \hat{i}+\hat{j}-7 \hat{k}$$ .

We first find the sum $$\begin{align}\vec{a}+\vec{b}&=(2 \hat{i}+2 \hat{j}-5 \hat{k})+(2 \hat{i}+\hat{j}-7 \hat{k}) \\ \Rightarrow &=4 \hat{i}+3 \hat{j}-12 \hat{k}\\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{(4)^2+(3)^2+(-12)^2} \\ \Rightarrow &=\sqrt{16+9+144} \\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{169}=13\end{align}$$ Now let say $\hat{c}$ be the unit vector $x$ the sum of $\vec{a}$ and $\vec{b}$ then $$\begin{align}\hat{c}&=\dfrac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\dfrac{4 \hat{i}+3 \hat{j}-12 k}{13} \\ \Rightarrow &=\dfrac{1}{13}(4 \hat{i}+3 \hat{j}-12 \hat{k})\\ \Rightarrow &=\dfrac{4}{13}\hat{i}+\dfrac{3}{13}\hat{j}-\dfrac{12}{13}\hat{k}\\ \Rightarrow &=\dfrac{1}{13}(4\hat{i}+3\hat{j}-12\hat{k})\end{align}$$

Find the angles between the pairs of vectors: $\hat{i}-\hat{j}+\hat{k}, \quad-\hat{i}+\hat{j}+2 \hat{k}$

Let $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}+2 \hat{k}$. Let $\theta$ be the angle hetween $\vec{a}$ and $\vec{b}$ $$\begin{align}\text { then } \cos \theta&=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \ldots \ldots \ldots(1) \\ \vec{a} \cdot \vec{b}&=(\hat{i}-\hat{j}+\hat{k}) \cdot(-\hat{i}+\hat{j}+2 \hat{k}) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=1(\cdot 1)+(-1)(1)+1(2) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-1-1+2=0\end{align}$$ $\vec{a}$ and $\vec{b}$ are orthogonal.
$$\Rightarrow \theta=90^{\prime \prime}$$ .

Find the angles between the pairs of vectors: $3 \hat{i}+4 \hat{j}, \quad 2 \hat{j}-5 \hat{k}$

Let $\vec{a}=3 \hat{i}+4 \hat{j}$ and $\vec{b}=2 \hat{j}-5 \hat{k}$. Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$ $$\begin{align}\text { then } \quad \cos \theta &=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \ldots \ldots \ldots \ldots(1) \\ \vec{a} \cdot \vec{b}&=(3 \hat{i}+4 \hat{j}) \cdot(2 \hat{j}-5 \hat{k}: \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=3(0)+(4)(2)+0(-5) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=8 \\ \text { Also }|\vec{a}|&=\sqrt{(3)^2+(4)^2}=\sqrt{9+19} \\ \Rightarrow \quad|\vec{a}|&=\sqrt{25}=5\\ \text { and }|\vec{b}|&=\sqrt{(2)^2+(-5)^2}\\ \Rightarrow|\vec{b}|&=\sqrt{29}\end{align}$$ Now putting all these in (1) $$\begin{align}\cos \theta&=\dfrac{8}{5 \cdot \sqrt{29}} \\ \Rightarrow \theta&=\cos ^{-1}\left(\dfrac{8}{5 \cdot \sqrt{29}}\right) \\ \Rightarrow \theta&=\cos ^1(0.2971)=72.72^{\circ}=73^{\circ}\text{(approximately)}\end{align}$$

Find the angles between the pairs of vectors: $2 \hat{i}-3 \hat{k}, \quad \hat{i}+\hat{j}+\hat{k}$

Let $\vec{a}=2 \hat{i}-3 \hat{k}$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$.

Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$ then $$\cos \theta=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$ . $$\begin{align}\vec{a} \cdot \vec{b}&=(2 \hat{i}-3 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k}) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=2(1)+(0)(1)+(-3)(1) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=-1\end{align}$$ $$\begin{align}\text { Also }|\vec{a}|&=\sqrt{(2)^2+(-3)^2} \\ \Rightarrow|\vec{a}|&=\sqrt{4+9}=\sqrt{13} . \\ \text { and }|\vec{b}|&=\sqrt{(1)^2+\left(1^{12}+(1)^2\right.} \\ \Rightarrow|\vec{b}|&=\sqrt{3}\end{align}$$ Now putting all these in (1) $$\begin{align}\cos \theta&=\dfrac{-1}{\sqrt{3} \cdot \sqrt{13}} \\ \Rightarrow \theta&=\cos ^{-1}\left(\dfrac{-1}{\sqrt{3} \cdot \sqrt{13}}\right) \\ \Rightarrow \theta&=\cos ^1(-0.1601)=99^{\circ}\text{(approximately)}\end{align}$$

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