====== 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.
=====Question 2=====
Find $\lambda$ and $\mu$ if\\
$$(\hat{i}+3 \hat{j}+9 \hat{k}) \times(3 \hat{i}-\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0} \text {. }$$\\
====Solution====
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}
=====Question 3=====
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 $\vec{a}+\vec{b}$.\\
====Solution====
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}
====Go To====
[[math-11-kpk:sol:unit03:Review-ex3-p1 |< Question 1]]
[[math-11-kpk:sol:unit03:Review-ex3-p3|Question 4 & 5 >]]