Vector Algebra

Cross product will represent a vector perpendicular to both the vector.

Dir. Of Motion of P : $=\left(\widehat{i}+\widehat{j}\right)*\left(\widehat{j}*\widehat{k}\right)$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

= $\widehat{k}-\widehat{j}+\widehat{i}$ $\stackrel{}{}\stackrel{}{}$

Dir. Of Motion of Q = $\left(\widehat{i}+\widehat{j}\right)*\left(-\widehat{i}+\widehat{j}\right)$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

$=2\widehat{k}$

$cos\theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{ab}=\frac{2}{2\sqrt[]{3}}=\frac{1}{\sqrt[]{3}}\Rightarrow \theta =co{s}^{-1}\left(\frac{1}{\sqrt[]{3}}\right)$

 $\overrightarrow{a}\cdot \widehat{c}=\frac{\alpha +6+2}{\sqrt[]{1}+4+4}$

$\frac{10}{3}=\frac{8+\alpha }{3}\Rightarrow \alpha =2$

$\overrightarrow{b}*\overrightarrow{c}=\widehat{i}\left(2\beta -8\right)+\widehat{j}\left(10\right)+\widehat{k}\left(6+\beta \right)$

$=-6\widehat{i}+10\widehat{j}+7\widehat{k}$

So = 1

Given : $\overrightarrow{a}=\left(\alpha ,1,-1\right)and\overrightarrow{b}=\left(2,1,-\alpha \right)\overrightarrow{c}=\overrightarrow{a}*\overrightarrow{b}=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill \alpha \hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill -\alpha \hfill \end{array}\right|$

$=\left(-\alpha +1\right)\widehat{i}+\left({\alpha }^2-2\right)\widehat{j}+\left(\alpha -2\right)\widehat{k}$ Projection of $\overrightarrow{c}$ on $\overrightarrow{d}=-\widehat{i}+2\widehat{j}-2\widehat{k}=\left|\overrightarrow{c}.\frac{\overrightarrow{d}}{\left|\overrightarrow{d}\right|}\right|=30\left\{Given\right\}$

$\Rightarrow \left|\frac{\alpha -1-4+2{\alpha }^2-2\alpha +4}{\sqrt[]{1+4+4}}\right|=30$

On solving $\alpha =\frac{-13}{2}$ (Rejected as > 0) and = 7

Any tangent to y² = 24x at (α, β) is βy = 12 (x + α) therefore Slope = $\frac{12}{\beta }$

and perpendicular to 2x + 2y = 5 =>12 =β and α= 6 Hence hyperbola is $\frac{x^2}{6^2}-\frac{y^2}{12^2}$ = 1 and normal is drawn at (10, 16)

therefore equation of normal $\frac{36x}{10}+\frac{144y}{16}=36+144\Rightarrow \frac{x}{50}+\frac{y}{20}=1$ This does not pass through (15, 13) out of given option.

$If{\widehat{n}}_1$ is a vector normal to the plane determined by $\widehat{i}and\widehat{i}+\widehat{j}then{\widehat{n}}_1=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right|=\widehat{k}$

$If{\widehat{n}}_2$ is a vector normal to the pane determined by $\widehat{i}-\widehat{j},$ and $\widehat{i}+\widehat{k}$ then ${\widehat{n}}_2$ = $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=\widehat{i}-\widehat{j}+\widehat{k}$

Vector $\widehat{a}$ is parallel to ${\widehat{n}}_1*{\widehat{n}}_2$ i.e. $\widehat{a}$ is parallel to $\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|=\widehat{i}-\widehat{j}$

Given $\overrightarrow{b}=\widehat{i}-2\widehat{j}+2\widehat{k}$

consine of acute angle between $\widehat{a}and\widehat{b}$ = $\left|\frac{\widehat{a}.\widehat{b}}{\left|\widehat{a}\right|.\left|\widehat{b}\right|}\right|=\frac{1}{\sqrt[]{2}}$

obtuse angle between $\widehat{a}and\widehat{b}=\frac{3\pi }{4}$

For angle to the acute $\overrightarrow{U}.\overrightarrow{V}>0$

$\Rightarrow a{\left(lo{g}_{e}b\right)}^2-12+6a\left(lo{g}_{e}b\right)>0$

$\forall b>1$

Let log_e b = t > 0 as b > 1

Y = at² + 6at – 12 & y > 0 $\forall$ t > 0

$\Rightarrow a\in ?$

 $x=2\sqrt[]{2}cost\sqrt[]{sin2t}$

$\frac{dx}{dt}=\frac{2\sqrt[]{2}cos3t}{\sqrt[]{sin2t}}$

$y\left(t\right)=2\sqrt[]{2}sint\sqrt[]{sin2t}$

$\frac{dy}{dx}=\frac{2\sqrt[]{2}sin3t}{\sqrt[]{sin2t}}$

$\therefore \frac{1+{\left(\frac{dy}{dx}\right)}^2}{\frac{d^{2}y}{d{x}^2}}=\frac{1+1}{-3}=\frac{-2}{3}$

$\left(\widehat{a}+\widehat{b}\right).\left(\widehat{a}+2\widehat{b}+2\left(\widehat{a}*\widehat{b}\right)\right),\widehat{a}.\widehat{b}=\frac{1}{\sqrt[]{2}}$

= 1 + $\frac{1}{\sqrt[]{2}}+\sqrt[]{2}$ + 2 + 0 + 0

$=3+\frac{3}{\sqrt[]{2}}$

$\left|\widehat{a}+2\widehat{b}+2{\left(\widehat{a}*\widehat{b}\right)}^2\right|=1+4+4\left(\frac{1}{2}\right)+2\sqrt[]{2}+0=7+2\sqrt[]{2}$

$\Rightarrow \left(2+\sqrt[]{2}\right)\left(7+2\sqrt[]{2}\right)co{s}^2\theta =9+\frac{9}{2}+9\sqrt[]{2}=\frac{27+18\sqrt[]{2}}{2}$

$\Rightarrow co{s}^2\theta =\frac{27+18\sqrt[]{2}}{2}*\frac{1}{18+11\sqrt[]{2}}=\frac{\left(27+18\sqrt[]{2}\right)\left(18-11\sqrt[]{2}\right)}{2\left(324-242\right)}$

$=\frac{486+324\sqrt[]{2}-297\sqrt[]{2}-396}{2*82}$

164 cos² (θ)= 90 + $2$$7$

 $\left(\stackrel{?}{a}.\stackrel{?}{c}\right)\stackrel{?}{b}?\left(\stackrel{?}{a}.\stackrel{?}{b}\right)\stackrel{?}{c}$

$=\stackrel{?}{b}+?\stackrel{?}{c}\left(given\right)$

$\stackrel{?}{a}.\stackrel{?}{c}=1,?=?\stackrel{?}{a}.\stackrel{?}{b}$

$?=?\left(3,1,0\right).\left(1,2,1\right)$

= (3 + 2)= 5

Be the vectors along the diagonals of a parallelogram having are 2.

$\therefore \frac{1}{2}\left|\overrightarrow{a}*\overrightarrow{b}\right|=2\sqrt[]{2}$

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|sin\theta =4\sqrt[]{2}$

$\Rightarrow \left|\overrightarrow{b}\right|sin\theta =4\sqrt[]{2}$ ……. (i)

And

$\Rightarrow \overrightarrow{c}.\overrightarrow{b}=-2{\left|\overrightarrow{b}\right|}^2=-128......(ii)$

$\Rightarrow \left|\overrightarrow{c}\right|=16\sqrt[]{2}.......(iii)$

From (ii) and (iii)

$\left|\overrightarrow{c}\right|\left|\overrightarrow{b}\right|cos\alpha =-128$

$\Rightarrow cos\alpha =\frac{1}{\sqrt[]{2}}$

$\alpha =\frac{3\pi }{4}$