====== Question 11, Exercise 3.3 ======
Solutions of Question 11 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 11 (i)=====
Show that the vectors $3 \hat{i}-2 \hat{j}+$ $\hat{k} . \quad \hat{i}-3 \hat{j}-5 \hat{k}$ and $2 \hat{i}+\hat{j}-4 \hat{k}$ form a right angle triangle.
====Solution====
Let $\vec{a}=3 \hat{i}-2 \hat{j}+\hat{k}$. $\vec{b}=\hat{i}-3 \hat{j}+5 \hat{k}$ and $\vec{c}=2 \hat{i}+\hat{j}-4 \hat{k}$.
Then 
\begin{align}|\vec{a}|&=\sqrt{(3)^2+(-2)^2+(1)^2}\\
\Rightarrow \quad |\vec{a}|&=\sqrt{14},\\
|\vec{b}|&=\sqrt{(1)^2+(-3)^2+(5)^2} \\
|\vec{b}|&=\sqrt{35},and\\
|\vec{c}|&= \sqrt{(2)^2+(1)^2+(-4)^2}&=\sqrt{21}\\
|\vec{a}|^2+|\vec{c}|^2&=|\vec{b}|^2\\
14+21&=35\\
35&=35\end{align}
Thus by Pytagorous theorem, the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ represent the sides of
triangle and they form right angle triangle. Also if we see
\begin{align}\vec{a} \cdot \vec{c}&=-(3 \hat{i}-2 \hat{j}+\hat{k}) \cdot(2 \hat{i}+\hat{j}-4 \hat{k}) \\
\Rightarrow \vec{a} \cdot \vec{c}&=6 - 2-4=0 .\end{align}
$\therefore \quad \vec{a} \cdot \vec{c}$. or sides represented by $\vec{a}$ and $\vec{c}$ form right angle with each other.

=====Question 11 (ii)=====
Show that $P(1,0,1), Q(1,1,1)$ and $R(1,1.0)$ forms a right isosceles triangle.
====Solution====
We find vectors representing the sides of triangle from the ver-tices given.
\begin{align}\overrightarrow{P Q}&=\overrightarrow{O Q} - \overrightarrow{O P} \\
\Rightarrow \overrightarrow{P Q}&=(\hat{i}+\hat{j}+\hat{k})-(\hat{i}+\hat{k})=\hat{j} \\
\overrightarrow{Q R}&=\overrightarrow{O R}-\overrightarrow{O Q} \\
\Rightarrow \quad \overrightarrow{Q R}&=(\hat{i}+\hat{j})-(\hat{i}+\hat{j}+\hat{k})=-\hat{k} \\
\overrightarrow{P R}&=\overrightarrow{O R}-\overrightarrow{O P} \\
\Rightarrow \overrightarrow{P R}&=(\hat{i}+\hat{j})-(\hat{i}-\hat{k})=\hat{j}-\hat{k}\\
\text { Now }|\overrightarrow{P Q}|&=\sqrt{(1)^2}=1 \\
|\overrightarrow{Q R}|&=\sqrt{(-1)^2}=1, \quad \text { and } \\
|\overrightarrow{P R}|&=\sqrt{(1)^2+(-1)^2}=\sqrt{2} .\end{align}
We observe that\\
\begin{align}|\overrightarrow{P Q}|^2+|\overrightarrow{Q R}|^2&=|\overrightarrow{P R}|^2 .\end{align}\\
$\Rightarrow \quad P(1.0 .1) \quad Q(1,1,1) \text { and } R(1.1,0)$ forms a right angle triangle.\\
Also $|\overrightarrow{P Q}|=|\overrightarrow{Q R}|$, so the the right angle triangle is also isosceles.







====Go To====
<text align="left"><btn type="primary">[[math-11-kpk:sol:unit03:ex3-3-p6 |< Question 9 & 10]]</btn></text>
<text align="right"><btn type="success">[[math-11-kpk:sol:unit03:ex3-3-p8|Question 12 & 13 >]]</btn></text>