Mathematics - Vector Question with Solution | TestHub

(A) Equation of the line passing through origin is $$\begin{array}{rl}& \frac{x}{a}=\frac{y}{b}=\frac{z}{c}\\ & \therefore \left|\begin{array}{ccc}2& 1& -1\\ 1& -2& 1\\ a& b& c\end{array}\right|=0\\ & \Rightarrow a(-1)-b(3)+c(-5)=0\\ & \Rightarrow -a-3b-5c=0\\ & \Rightarrow a+3b+5c=0\\ & \text{ Also, }\left|\begin{array}{ccc}\frac{8}{3}& -3& 1\\ 2& -1& 1\\ a& b& c\end{array}\right|=0\end{array}$$ $$\begin{array}{rl}& \therefore a(-2)-b\left(\frac{2}{3}\right)+c\left(\frac{10}{3}\right)=0\\ & \Rightarrow 2a+\frac{2b}{3}-\frac{10c}{3}=0\\ & 3a+b-5c=0\\ & \end{array}$$ From Eqs. (i) and (ii), $$\begin{array}{rl}\frac{a}{-20}& =\frac{b}{20}=\frac{c}{-8}\\ \frac{a}{5}& =\frac{b}{-5}=\frac{c}{4}\end{array}$$ Equation of line is $$\frac{x}{5}=\frac{y}{-5}=\frac{z}{4}=\lambda \text{ (say) }$$ Also, $\frac{x-2}{1}=\frac{y-1}{2}=\frac{z+1}{1}=k_1$ (say) Now, $\frac{x-\frac{8}{3}}{2}=\frac{y+3}{-1}=\frac{z-1}{1}=k_2$ (say) Point on (iii) is $(5\lambda ,-5\lambda ,+4\lambda )$ Point on (iv) is $$\left(2+k_1,1-2k_1,-1+k_1\right)$$ Point on $(\mathrm{v})$ is $$\left(\frac{8}{3}+2k_2,-3-k_2,1+k_2\right)$$ On solving, $2+k_1+1-2k_1=0$ $$-k_1+3=0$$ $$\begin{array}{rl}{k}_1& =3\\ P& \equiv (5,-5,2)\end{array}$$ Again, for $Q$ $$\begin{array}{r}\frac{8}{3}+2k_2-3-k_2=0\\ k_2-\frac{1}{3}=0\\ k_2=\frac{1}{3}\\ Q\equiv \left(\frac{10}{3},\frac{-10}{3},\frac{4}{3}\right)\end{array}$$ $$\text{ Now, }\begin{array}{rl}PQ& =\sqrt{{\left(\frac{5}{3}\right)}^2+{\left(\frac{5}{3}\right)}^2+{\left(\frac{2}{3}\right)}^2}\\ & =\frac{\sqrt{54}}{3}\end{array}$$ $$P{Q}^2=d^2=\frac{54}{9}=6$$ (B) ${\tan }^{-1}\left(\frac{x+3-x+3}{1+\left(x^2-9\right)}\right)={\tan }^{-1}\left(\frac{3}{4}\right)$ $$\begin{array}{rlrl}\Rightarrow & & \frac{6}{x^2-8}& =\frac{3}{4}\\ \Rightarrow & & 3x^2& =48\\ \Rightarrow & x& =\pm 4& \end{array}$$ (C) $$\begin{array}{rl}& (\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}})\cdot \left(\overrightarrow{\mathbf{b}}+\frac{\overrightarrow{\mathbf{a}}-\mu \overrightarrow{\mathbf{b}}}{4}\right)=0\\ & \Rightarrow (\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}})\cdot (4\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}}-\mu \overrightarrow{\mathbf{b}})=0\\ & \text{ Also, }2\left|\overrightarrow{\mathbf{b}}+\frac{\overrightarrow{\mathbf{a}}-\mu \overrightarrow{\mathbf{b}}}{4}\right|=|\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}|\\ & \Rightarrow 2\left|\frac{(4-\mu ){\overrightarrow{\mathbf{b}}}^2-{\overrightarrow{\mathbf{a}}}^2=0}{4}\right|=|\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}|\end{array}$$ $$\begin{array}{rl}& \Rightarrow \frac{(4-\mu {)}^2{\overrightarrow{\mathbf{b}}}^2}{4}+\frac{{\overrightarrow{\mathbf{a}}}^2}{4}={\overrightarrow{\mathbf{b}}}^2+{\overrightarrow{\mathbf{a}}}^2\\ & \Rightarrow \frac{3{\overrightarrow{\mathbf{a}}}^2}{4}=\frac{(4-\mu {)}^2-4}{4}\cdot {\overrightarrow{\mathbf{a}}}^2\\ & 3{\overrightarrow{\mathbf{a}}}^2=C(4-\mu {)}^2-4{\overrightarrow{\mathbf{b}}}^2\end{array}$$ From Eqs. (i) and (ii) $$\begin{array}{rl}& 3(4-\mu )=(4-\mu {)}^2-4\\ & (4-\mu {)}^2-3(4-\mu )-4=0\\ & \Rightarrow \mu =0,5\\ & \mu =5\text{ is not admissible. }\end{array}$$ (D) $f(0)=9$, $$\begin{array}{rl}& f(x)=\frac{\sin \left(\frac{9x}{2}\right)}{\sin \frac{x}{2}}\\ & =\left(3-4{\sin }^2\frac{x}{2}\right)\left(3-4{\sin }^2\frac{3x}{2}\right)\end{array}$$ $$\begin{array}{rl}& =9-12{\sin }^2\frac{x}{2}-12{\sin }^2\frac{3x}{2}\\ & +16{\sin }^2\frac{x}{2}\cdot {\sin }^2\frac{3x}{2}\\ & =9-6(1-\cos x)-6(1-\cos 3x)\\ & +4(1-\cos x)(1-\cos 3x)\\ & =1+6\cos x+6\cos 3x-4\cos x\\ & -4\cos 3x+4\cos x\cos 3x\\ & \text{ Let }I=\frac{2}{\pi }{\int }_{-\pi }^{\pi }\frac{\sin \frac{9x}{2}}{2}dx\\ & =\frac{4}{\pi }{\int }_0^{\pi }1+2\cos x+2\cos 3x\\ & =\frac{4}{\pi }\times \pi +2(\cos 4x+\cos x)dx\\ & =4\end{array}$$