Question 7 & 8 Exercise 3.3

Solutions of Question 7 & 8 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.

Given the vectors $\vec{a}$ and $\vec{b}$ as $\vec{a}=-\dfrac{3}{2} \hat{j}+\dfrac{4}{5} \hat{k} \cdot \vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}$. Find in each case the projection of $\vec{a}$ on $\vec{b}$ and $\vec{b}$ on $\vec{a}$.

$\vec{a}=-\dfrac{3}{2} \hat{j}+\dfrac{4}{5} \hat{k}\quad$ $\vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}$ We compute the dot product $$\begin{align}\vec{a} \cdot \vec{b}&=\left(-\dfrac{3}{2} \hat{j}+\dfrac{4}{5} \hat{k}\right) \cdot(\hat{i}-2 \hat{j}-2 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{b}&=0(1)+\left(-\dfrac{3}{2}\right)(-2)+\dfrac{4}{5}(-2) \\ \Rightarrow \quad \vec{a} \cdot \vec{b}&=3-\dfrac{8}{5}=\dfrac{7}{5} \ldots \ldots \ldots . .(1) \\ \text { Also }|\vec{a}|&=\sqrt{\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{4}{5}\right)^2} \\ \Rightarrow|\vec{a}|&=\sqrt{\dfrac{9}{4}+\dfrac{16}{25}} \\ \Rightarrow|\vec{a}|&=\sqrt{\dfrac{225+64}{100}}=\dfrac{\sqrt{287}}{10}=\dfrac{17}{10} \ldots . .(2) \\ \text { and }|\vec{b}|&=\sqrt{(1)^2+(-2)^2+(-2)^2} \\ \Rightarrow|\vec{b}|&=\sqrt{9}=3 \ldots \ldots \ldots \ldots(3)\end{align}$$ Projection of $\vec{a}$ on $\vec{b}=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

Projection of $\vec{a}$ on $\vec{b}=\dfrac{7}{15}$.

Projection of $\vec{b}$ on $\vec{a}=\dfrac{\vec{a} \cdot \vec{b}}{| \vec{a}|}$

Projection of $\vec{b}$ on $\vec{a}=\dfrac{14}{17}$

Given the vectors $\vec{a}$ and $\vec{b}$ as $\vec{a}=-3 \hat{i}+\hat{j}+2 \hat{k} \vec{b}=-\hat{i}-\hat{j}+5 \hat{k}$. Find in each case the projection of $\vec{a}$ on $\vec{b}$ and $\vec{b}$ on $\vec{a}$.

We compute the dot product
$$\begin{align}\vec{a} \cdot \vec{b}&=(-3 \hat{i}+\hat{j}+2 \hat{k}) \cdot(-\hat{i}-\hat{j}+5 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{b}&=-3(-1)+(1)(-1)+2(5) \\ \Rightarrow \vec{a} \cdot \vec{b}&=2+10=12 \ldots \ldots \ldots \ldots(1)\\ \mid \bar{a} i&=\sqrt{(-3)^2+(1)^2+(2)^2}\\ \Rightarrow|\bar{a}|&=\sqrt{14}.\\ |\vec{b}|&=\sqrt{(-1)^2+(-1)^2+(5)^2}\\ \Rightarrow|\vec{b}|&=\sqrt{27}.\end{align}$$ Projection of $\vec{a}$ on $\vec{b}=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
Projection of $\vec{a}$ on $\vec{b}=\dfrac{12}{\sqrt{27}}=\dfrac{4}{\sqrt{3}}$.
Projection of $\vec{b}$ on $\vec{a}=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$
Projection of $\vec{b}$ on $\vec{a}=\dfrac{12}{\sqrt{14}}$

What is the cosine of the angle which the vector $\sqrt{2} \hat{i}+\hat{j}+\hat{k}$ makes with $y$ axis.

Let $\vec{a}=\sqrt{2} \hat{i}+\hat{j}+\hat{k}$ and $\bar{h}=\hat{j} \quad$ unil vector along $y-a x i s$.
The cosine of angie between the given vector and $y-a x i s$ is now actually cosine of angle between $\vec{d}$ and $\vec{b}$.
Now $\vec{a} \cdot \vec{b}=(\sqrt{2} \hat{i}+\hat{j}+\hat{k}) \cdot(\hat{j})$\ $$\begin{align}\Rightarrow \vec{a} \cdot \vec{b}&=1 \text { and } \\ \vec{a} \mid &=\sqrt{(\sqrt{2})^2+(1)^2+(1)^2} \\ \Rightarrow|\vec{a}|&=\sqrt{4}=2 . \text { and } \\ \vec{b} \mid&=\sqrt{(1)^2}=1.\end{align}$$ Now from dot product, we have
$$\cos \theta=\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\dfrac{1}{2.1}=\dfrac{1}{2} \text {. }$$

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