Question 3 & 4, Exercise 3.2

Solutions of Question 3 & 4 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.

If $\vec{r}=\hat{i}-9\hat{j}$, $\vec{a}=\hat{i}+2\hat{j}$ and $\vec{b}=5\hat{i}-\hat{j}$, determine the real number $p$ and $q$ such that $\vec{r}=p\vec{a}+q\vec{b}$.

Given $$\vec{r}=p\vec{a}+q\vec{b}.$$ We put the values of $\vec{r},\vec{a}$ and $\vec{b}$ in the given equation. We get $$\hat{i}-9\hat{j}=p(\hat{i}+2\hat{j})+q(5\hat{i}-\hat{j})$$ $$\implies \hat{i}-9\hat{j}=(p+5q)\hat{i}+(2p-q)\hat{j}.$$ By comparing the coeffients of $\hat{i}$ and $\hat{j}$, we have, $$p+5q=1…(i)$$ $$2p-q=-9 …(ii)$$ Multiply $2$ by (i) and subtract (ii) from (i). We have \[\begin{array}{ccc} 2p&+10q&=2 \\ \mathop+\limits_{-}2p&\mathop-\limits_{+}q&=\mathop-\limits_{+}9 \\ \hline &11q&=11\\ \end{array} \] $$\implies q=1$$
Put the value of $q$ in (i). We have, $$p+5(1)=1 \quad \implies p=-4$$ Hence we have $p=-4$ and $q=1$.

If $\vec{p}=2\hat{i}-\hat{j}$ and $\vec{q}=x\hat{i}+3\hat{j},$ then find the value of $x$ such that $|\vec{p}+\vec{q}|=5.$

We calculate \begin{align}\vec{p}+\vec{q}&=2\hat{i}-\hat{j}+x\hat{i}+3\hat{j}\\ &=(2+x)\hat{i}+2\hat{j}\end{align} Thus \begin{align}|\vec{p}+\vec{q}|&=\sqrt{(2+x)^2+2^2}\\ &=\sqrt{x^2+4x+4+4}\\ &=\sqrt{x^2+4x+8}\end{align} But we are given that \begin{align}&|\vec{p}+\vec{q}|=5 \\ \implies & \sqrt{{{x}^{2}}+4x+8}=5 \\ \implies & x^2+4x+8=25\\ \implies & x^2+4x-17=0\end{align} This is quadratic equation with $a=1$, $b=4$ and $c=-17$, so \begin{align}x&=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\ &=\dfrac{-4\pm \sqrt{4^2-4(1)(-17)}}{2(1)}\\ &=\dfrac{-4\pm \sqrt{84}}{2}\\ &=\dfrac{-4\pm 2\sqrt{21}}{2}\\ &=-2\pm \sqrt{21}\end{align} Thus $$x=-2\pm\sqrt{21}.$$