Maths 1 Marks

$\overrightarrow{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}$

We know projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ $= \frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}|}$ 

$= \frac{8}{\sqrt{4 + 36 + 9}} = \frac{8}{7}$

Unit vector in the direction $\overrightarrow{a} = \frac{\overrightarrow{a}}{|\overrightarrow{b}|} = \hat{a}$

$\Rightarrow \hat{a} = \frac{2\hat{i}- 6\hat{j} + 3\hat{k}}{\sqrt{4 + 36 + 9}}$

$\Rightarrow \hat{a} = \frac{2}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{3}{7} \hat{k}$

$\Rightarrow (3\hat{i} + 2\hat{j} +9\hat{k} )=\lambda(\hat{i} + {p\hat{j} }+ 3\hat{k})$

$\Rightarrow \lambda = 3,\ 2 = \lambda p,\ 9 =3\lambda$

$\text{or} \lambda = 3, p = \frac{2}{3}$

$\overrightarrow{a}.\overrightarrow{b} = 0$

$\Rightarrow(2\hat{i} + \lambda\hat{j} + \hat{k}) .( \hat{i} - 2\hat{j} + 3\hat{k}) = 0$

$\Rightarrow 2 - 2\lambda + 3 = 0 \Rightarrow \lambda= \frac{5}{2}.$

Unit vector in the direction of $\overrightarrow{a} = \frac{\overrightarrow{a}}{\overrightarrow{|a|}}$

$= \frac{3\hat{i} - 2\hat{j} + 6\hat{k}}{\sqrt{3^{2} +(-2)^{2}+ 6^{2}}} = \frac{1}{7}(3\hat{i} - 2\hat{j} + 6\hat{k})$

$\overrightarrow{b} = \hat{i} - \hat{j} +\hat{k} \Rightarrow \overrightarrow{|b|} = \sqrt{1^{2} + (1)^{2} + (-1)^{2}} = \sqrt{3}$

$\overrightarrow{a}.\overrightarrow{b} = \overrightarrow{|a|} \overrightarrow{|b|} \cos\Theta$

$\Rightarrow 1-1-1 = \sqrt{3}.\sqrt{3} \cos \theta \Rightarrow -1 = 3 \cos\theta$

$\Rightarrow \cos\theta = -\frac{1}{3} \Rightarrow \theta = \cos^{-1} \bigg(-\frac{1}{3}\bigg)$

$\Rightarrow\ \text{x}=0$$$ and $\text{y}=0$ $[\because\ \vec{\text{a}}$ and $\vec{\text{b}}$ are non-collinear vectors$]$

$=\overrightarrow{\text{PR}}+\overrightarrow{\text{RP}}$ $\Big[\therefore\ \overrightarrow{\text{PQ}}+\overrightarrow{\text{QR}}=\overrightarrow{\text{PR}}\Big]$

$=\vec0$