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.

Show that the vector $\hat{i}+7 \hat{j} + 3 \hat{k}$ is perpendicular to both $\hat{i}-\hat{j}+2 \hat{k}$ and $2 \hat{i}-$ $\hat{j}+3 \hat{k}$.

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 $$\begin{align}\vec{a} \cdot \vec{b}&=(\hat{i}+7 \hat{j}+3 \hat{k}) \cdot(\hat{i}-\hat{j}+2 \hat{k}) . \\ \Rightarrow \vec{a} \cdot \vec{b}&=1(1)+7(-1)+3(2) \\ \Rightarrow \vec{a} \cdot \vec{b}&=1-7+6=0 \\ \Rightarrow \vec{a} \perp \vec{b} \cdot \text { Now } \\ \vec{a} \cdot \vec{c}&=(\hat{i}+7 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+\hat{j}-3 \hat{k}) \\ \Rightarrow \vec{a} \cdot \vec{c}&=1(2)+7(1)+3(-3) \\ \Rightarrow \vec{a} \cdot \vec{c}&=2+7-9=0 . \\ \Rightarrow \vec{a} \perp \vec{c}\end{align}$$

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 $\vec{b}$.

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 $\vec{b}$.

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