# Question 12, 13 & 14, Exercise 3.2

Solutions of Question 12, 13 & 14 of Exercise 3.2 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 $\alpha ,$ so that $|\alpha \hat{i}+(\alpha +1)\hat{j}+2\hat{k}|=3$.

We are given \begin{align}|\alpha \hat{i}+(\alpha +1)\hat{j}+2\hat{k}|&=3.\end{align} This gives \begin{align}\sqrt{(\alpha )^2+(\alpha +1)^2+(2)^2}&=3.\end{align} Taking square on both sides, we have, \begin{align}&{\alpha ^2+(\alpha +1)^2}+4=9\\ \implies & {\alpha ^2+\alpha ^2}+2\alpha +1+4=9\\ \implies & 2{\alpha ^2}+2\alpha +5-9=0\\ \implies & 2{\alpha ^2}+2\alpha -4=0\\ \implies & {\alpha ^2}+\alpha -2=0.\end{align} This is quadratic equation in $\alpha$. $$a=1, \quad b=1,\quad c=-2$$ \begin{align}\alpha &=\dfrac{-1\pm \sqrt{(1)^2-4(1)(-2)}}{2(1)}\\ & =\dfrac{-1\pm \sqrt{9}}{2}\\ & =\dfrac{-1\pm 3}{2}\\ \implies \alpha & =\dfrac{-1+3}{2}=1\quad \text{or} \quad \alpha =\dfrac{-1-3}{2}=-2\end{align} $\alpha=1$ or $\alpha =-2$, no real value of $\alpha$

If $\overrightarrow{u}=2\hat{i}+3\hat{j}+4\hat{k}$,$\overrightarrow{v}=-\hat{i}+3\hat{j}-\hat{k}$ and $\overrightarrow{w}=\hat{i}+6\hat{j}-z\hat{k}$ represents the sides of a triangle. Find the value of $z.$

\begin{align}\vec{u}+\vec{v}&=\vec{w}\\ (2\hat{i}+3\hat{j}+4\hat{k})+(-\hat{i}+3\hat{j}-\hat{k})&=\hat{i}+6\hat{j}-z\hat{k}\\ \hat{i}+6\hat{j}+3\hat{k}&=\hat{i}+6\hat{j}-z\hat{k}\end{align} By comparison $\hat{i},\hat{j}$ and $\hat{k}.$ we have,
$$3=-z$$ $$-z=3$$ $$\Rightarrow \,\,\,z=-3$$

The position vectors of the points $A,B,C$ and $D$ are $2\hat{i}-\hat{j}+\hat{k}$,$3\hat{i}+\hat{j},$ $2\hat{i}+4\hat{j}-2\hat{k}$ and $-\hat{i}-2\hat{j}+\hat{k}$ respectively. Show that $\overrightarrow{AB}$ is parallel to $\overrightarrow{CD}.$

Position vector of $A$ is $\overrightarrow{OA}=2\hat{i}-\hat{j}+\hat{k}$.

Position vector of $B$ is $\overrightarrow{OB}=3\hat{i}+\hat{j}$.

Position vector of $C$ is $\overrightarrow{OC}=2\hat{i}+4\hat{j}-2\hat{k}$.

Position vector of $D$ is $\overrightarrow{OD}=-\hat{i}-2\hat{j}+\hat{k}$.

Now, we have \begin{align}\overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\ &=(3\hat{i}+\hat{j})-(2\hat{i}-\hat{j}+\hat{k})\\ \implies \overrightarrow{AB}&=\hat{i}+2\hat{j}-\hat{k}…. (i)\end{align} Also \begin{align}\overrightarrow{CD}&=\overrightarrow{OD}-\overrightarrow{OC}\\ &=(-\hat{i}-2\hat{j}+\hat{k})-(2\hat{i}+4\hat{j}-2\hat{k})\\ &=-3\hat{i}-6\hat{j}+3\hat{k}\\ &=-3(\hat{i}+2\hat{j}-\hat{k})\\ \implies \overrightarrow{CD}&=-3\overrightarrow{AB} \text{ by using (i)}\end{align} This gives $$\overrightarrow{AB}\,\,\parallel \,\,\overrightarrow{CD}.$$