# Question 2 & 3 Review Exercise 3

Solutions of Question 2 & 3 of Review Exercise 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.

Find $\lambda$ and $\mu$ if
$$(\hat{i}+3 \hat{j}+9 \hat{k}) \times(3 \hat{i}-\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0} \text {. }$$

We are given
\begin{align}(\hat{i}+3 \hat{j}+9 \hat{k}) \times(3 \hat{i}-\lambda \hat{j}+\mu \hat{k})&=\vec{O} \\ \Rightarrow\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 9 \\ 3 & -\lambda & \mu \end{array}\right|&=\vec{O} \\ \Rightarrow(3 \mu+9 \lambda)\hat{i}-(\mu-27) \hat{j}+(-\lambda-9) \hat{k}&=\overrightarrow{0} \\ \Rightarrow \mu-27=0 \text { and }-\lambda-9&=0 \\ \Rightarrow \mu=27 \text { and } \lambda&=-9 .\end{align}

If $\vec{a}=9 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}-2 \hat{j}-\hat{k}$, then find a unit vector parallel to $\vec{a}+\vec{b}$.

Let $\hat{n}$ be unit normal in direction of $\vec{a}+\vec{b}$, then
\begin{align} \hat{n}&=\dfrac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|} \\ \vec{a}+\vec{b}&=(9 \hat{i}-\hat{j}+\hat{k})+(2 \hat{i}-2 \hat{j}-\hat{k}) \\ \Rightarrow \quad \vec{a}+\vec{b}&=11 \hat{i}-3 \hat{j} \\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{(11)^2+(9)^2} \\ \Rightarrow|\vec{a}+\vec{b}|&=\sqrt{202}\\ &=\dfrac{11 \hat{i}-3 \hat{j}}{\sqrt{202}}\\ \text { Now } \hat{n}&=\dfrac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}\\ &=\dfrac{11 \hat{i}-3 \hat{j}}{\sqrt{202}}\\ &=\dfrac{1}{\sqrt{202}}(11 \hat{i}-3 \hat{j})\end{align}