⇒Vectors

Correct answer option is (D)

$\begin{array}{rl}& (\overline{a}-\overline{d})\cdot (\overline{b}-\overline{c})=(\overline{b}-\overline{d})(\overline{c}-\overline{a})=0\\ & \overline{AD}\cdot \overline{BC}=\overline{BD}\cdot \overline{CA}=0\\ & \Rightarrow \overline{AD}\perp \overline{BC}\text{ and }\overline{BD}\perp \overline{CA}\end{array}$

$\Rightarrow D\text{ is the orthocentre of }\mathrm{△}ABC\text{. }$

Correct answer option is (C)

Let $\theta$ be the angle between $\overline{\mathrm{AB}}$ and $\overline{\mathrm{AD}}$

$\begin{array}{rl}\therefore \cos \theta & =\frac{\overline{AB}\cdot \overline{AD}}{|\overline{AB}|\overline{AD}\mid }\\ & =\frac{(2\hat{i}+10\hat{j}+11\hat{k})\cdot (-\hat{i}+2\hat{j}+2\hat{k})}{\sqrt{4+100+121}\sqrt{1+4+4}}\\ & =\frac{-2+20+22}{\sqrt{225}\sqrt{9}}\\ & =\frac{40}{45}\\ & =\frac{8}{9}\\ \therefore \sin \theta & =\sqrt{1-{\left(\frac{8}{9}\right)}^2}=\frac{\sqrt{17}}{9}\end{array}$

$\alpha$ is the angle of rotation of AD The angle between side $AB$ and $AD$

$\begin{array}{rl}& =\alpha +\theta \\ & ={90}^{\circ }\text{...[Given]}\end{array}$

$\therefore \cos (\alpha +\theta )=\cos \left({90}^{\circ }\right)$

$\therefore \cos \alpha \cos \theta -\sin \alpha \sin \theta =0$

$\therefore 8\cos \alpha =\sqrt{17}\sin \alpha$

$\therefore 64{\cos }^2\alpha =17(1-{\cos }^2\alpha )$

$\therefore 81{\cos }^2\alpha =17$

$\therefore \cos \alpha =\frac{\sqrt{17}}{9}$

Correct answer option is (C)

Consider option (C)

$(3\hat{i}+2\hat{j}+6\hat{k})\cdot \left(\frac{-8\hat{i}+3\hat{j}+3\hat{k}}{\sqrt{82}}\right)=0$

This is valid for only option (C)

$\therefore$ Option (c) is correct.

Correct answer option is (B)

Let $\theta$ be the angle between $\overline{\mathrm{AB}}$ and $\overline{\mathrm{AD}}$

$\begin{array}{rl}\therefore \cos \theta & =\frac{\overline{\mathrm{AB}}\cdot \overline{\mathrm{AD}}}{|\overline{\mathrm{AB}}||\overline{\mathrm{AD}}|}\\ & =\frac{(2\hat{\mathrm{i}}+10\hat{\mathrm{j}}+11\hat{\mathrm{k}})\cdot (-\hat{\mathrm{i}}+2\hat{\mathrm{j}}+2\hat{\mathrm{k}})}{\sqrt{4+100+121}\sqrt{1+4+4}}\\ & =\frac{-2+20+22}{\sqrt{225}\sqrt{9}}\\ & =\frac{40}{45}\\ & =\frac{8}{9}\end{array}$

$\therefore \sin \theta =\sqrt{1-{\left(\frac{8}{9}\right)}^2}=\frac{\sqrt{17}}{9}$

$\alpha$ is the angle of rotation of AD.

$\therefore$ The angle between side $\mathrm{AB}$ and $\mathrm{AD}$

$\begin{array}{rl}& =\alpha +\theta \\ & ={90}^{\circ }....[\text{Given}]\end{array}$

$\begin{array}{rl}& \therefore \cos (\alpha +\theta )=\cos \left({90}^{\circ }\right)\\ & \therefore \cos \alpha \cos \theta -\sin \alpha \sin \theta =0\\ & \therefore 8\cos \alpha =\sqrt{17}\sin \alpha \\ & \therefore 64{\cos }^2\alpha =17(1-{\cos }^2\alpha )\\ & \therefore 81{\cos }^2\alpha =17\\ & \therefore \cos \alpha =\frac{\sqrt{17}}{9}\end{array}$

Correct answer option is (C)

$|\bar{\mathrm{u}}|=1,|\overrightarrow{\mathrm{v}}|=2,|\bar{\mathrm{w}}|=3$

According to the given condition, (Projection of $\bar{\mathrm{v}}$ along $\bar{\mathrm{u}}$ ) $=(\text{ Projection of }\overline{w}\text{ along }\overline{u})$

$\begin{array}{rl}& \therefore \frac{\bar{\mathrm{v}}\cdot \bar{\mathrm{u}}}{|\bar{\mathrm{u}}|}=\frac{\bar{\mathrm{w}}\cdot \bar{\mathrm{u}}}{|\bar{\mathrm{u}}|}\\ & \therefore \bar{\mathrm{v}}\cdot \bar{\mathrm{u}}=\bar{\mathrm{w}}\cdot \bar{\mathrm{u}}\\ & \therefore (\bar{\mathrm{w}}-\bar{\mathrm{v}})\cdot \bar{\mathrm{u}}=0\text{.... (i)}\end{array}$

Now consider, $|\bar{\mathrm{u}}-\bar{\mathrm{v}}+\bar{\mathrm{w}}|=\sqrt{|\bar{\mathrm{u}}+\bar{\mathrm{w}}-\bar{\mathrm{v}}{|}^2}$

$\begin{array}{rl}& =\sqrt{|\bar{\mathrm{u}}{|}^2+|\bar{\mathrm{w}}-\bar{\mathrm{v}}{|}^2+2\bar{\mathrm{u}}\cdot (\bar{\mathrm{w}}-\bar{\mathrm{v}})}\\ & =\sqrt{(1{)}^2+|\bar{\mathrm{w}}-\bar{\mathrm{v}}{|}^2+0}\text{.... [From (i)]}\\ & =\sqrt{1+|\bar{\mathrm{w}}{|}^2+|\bar{\mathrm{v}}{|}^2-2(\bar{\mathrm{w}}\cdot \bar{\mathrm{v}})}\\ & =\sqrt{1+9+4+0}...[\because \bar{w}\text{and}\bar{v}\text{are perpendicular}]\\ & =\sqrt{14}\end{array}$