====== Question 4 and 5 Exercise 3.3 ======
Solutions of Question 4 and 5 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.

=====Question 4=====
Show that the vector $\hat{i}+7 \hat{j} + 3 \hat{k}$ is perpendicular to both $\hat{i}-\hat{j}+2 \hat{k}$ and $2 \hat{i}-$ $\hat{j}+3 \hat{k}$.
====Solution====
Let $\vec{a}=\hat{i}+7 \hat{j}+3 \hat{k}$, $\vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c} = 2 \hat{i}-\hat{j}-3 \hat{k}$.
Then
\begin{align}\vec{a} \cdot \vec{b}&=(\hat{i}+7 \hat{j}+3 \hat{k}) \cdot(\hat{i}-\hat{j}+2 \hat{k}) . \\
\Rightarrow \vec{a} \cdot \vec{b}&=1(1)+7(-1)+3(2) \\
\Rightarrow \vec{a} \cdot \vec{b}&=1-7+6=0 \\
\Rightarrow \vec{a} \perp \vec{b} \cdot \text { Now } \\
\vec{a} \cdot \vec{c}&=(\hat{i}+7 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+\hat{j}-3 \hat{k}) \\
\Rightarrow \vec{a} \cdot \vec{c}&=1(2)+7(1)+3(-3) \\
\Rightarrow \vec{a} \cdot \vec{c}&=2+7-9=0 . \\
\Rightarrow \vec{a} \perp \vec{c}\end{align}

=====Question 5=====
Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$.
Find a vector that is orthogonal to both $\vec{a}$ and $\vec{b}$.
====Solution====
We know that $\vec{c}=\vec{a} \times \vec{b}$ is a vector that is orthogonal to both $\vec{a}$ and $\vec{b}$. Therefore,
\begin{align}\vec{c}&=\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 1 \\
2 & 1 & -1
\end{array}\right| \end{align}
expanding by $R_1$ we have
\begin{align}
& \vec{c}=(-2-1) \hat{i}+(-1-2) \hat{j}+(1-4) \hat{k} \\
& \Rightarrow \vec{c}=-3 \hat{i}-3 \hat{j}-3 \hat{k}\end{align}
is the desired vector perpendicular to both $\vec{a}$ and $\vec{b}$.



====Go To====
<text align="left"><btn type="primary">[[math-11-kpk:sol:unit03:ex3-3-p2 |< Question 2 & 3]]</btn></text>
<text align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3-3-p4|Question 6 >]]</btn></text>