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- Question 12, 13 & 14, Exercise 3.2
- e, \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
- Question 3 & 4, Exercise 3.2
- t{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}.$$ B... mits_{+}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 $q=1$. =====Ques... given that \begin{align}&|\vec{p}+\vec{q}|=5 \\ \implies & \sqrt{{{x}^{2}}+4x+8}=5 \\ \implies & x^2+4x+8=
- Question 9 & 10, Exercise 3.2
- }+2\hat{k})]\\ &=2(\hat{i}-2\hat{j}+2\hat{k})\\ \implies 6\hat{a}&=2\hat{i}-4\hat{j}+4\hat{k}\end{align} T... )}{3}\\ &=\dfrac{-\hat{i}+4\hat{j}+\hat{k}}{3}\\ \implies \overrightarrow{OR}&=-\dfrac{1}{3}\hat{i}+\dfrac{... i}-2\hat{j}+\hat{k}-2\hat{i}+2\hat{j}+2\hat{k}\\ \implies\quad \text{externally}\quad \overrightarrow{OR}&=
- Question 5 & 6, Exercise 3.2
- ow{OA}=\overrightarrow{OC}-\overrightarrow{OD}\\ \implies &(\hat{i}+4\hat{j})-(-2\hat{i}-3\hat{j})=5\hat{j}-x\hat{i}-y\hat{j}\\ \implies &3\hat{i}+7\hat{j}=-x\hat{i}+(5-y)\hat{j}\end{ali
- Question 11, Exercise 3.2
- t{j})}{7}\\ &=\dfrac{1}{7}(8\hat{i}+27\hat{j})\\ \implies \overrightarrow{OH}&=\dfrac{8}{7}\hat{i}+\dfrac{2... i}+2\hat{j})]\\ &=-(6-12)\hat{i}-(-9-8)\hat{j}\\ \implies \overrightarrow{OK}&=6\hat{i}+17\hat{j}\end{align
- Question 5(i) & 5(ii) Exercise 3.5
- cdot \vec{a} \times \vec{b}&=0\end{align}\\ Which implies that $\vec{a} \times \vec{b} \perp \vec{a}$. For... cdot(\vec{a} \times \vec{b})&=0.\end{align} Which implies that $\vec{a} \times \vec{b}\perp \vec{b}$. ====