Mathematics - Vector Question with Solution | TestHub

Given,

Equation of plane,

$P:\sqrt{3}x+2y+3z=16$

And $S=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k} : {\alpha }^2+{\beta }^2+{\gamma }^2=1$ which is equation of sphere,

Also distance, $d\left(\alpha , \beta , \gamma \right)$ from $P=\frac{7}{2}$

And relation between distinct vector is given by,$\left|\overrightarrow{u}-\overrightarrow{v}\right|=\left|\overrightarrow{v}-\overrightarrow{w}\right|=\left|\overrightarrow{w}-\overrightarrow{u}\right|$

Now  $V$ : volume of parallelepiped by vectors $\overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w}$

So, using formula of distance $d\left(\alpha , \beta , \gamma \right)$ from $P=\frac{7}{2}$ we get,

$\Rightarrow \frac{\left|\sqrt{3}\alpha +2\beta +3\gamma -16\right|}{\sqrt{3+4+9}}=\frac{7}{2}$

$\Rightarrow \frac{\left|\sqrt{3}\alpha +2\beta +3\gamma -16\right|}{4}=\frac{7}{2}$

$\Rightarrow \left|\sqrt{3}\alpha +2\beta +3\gamma -16\right|=14 ....\left(\mathrm{i}\right)$

Also, ${\alpha }^2+{\beta }^2+{\gamma }^2=1 .......\left(\mathrm{ii}\right)$

Now we know that,

Volume of parallelepiped by vector, $\overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w}$ is given by,

$V=\left[\overrightarrow{u} \overrightarrow{v} \overrightarrow{w}\right]$

$=\overrightarrow{u}·\left(v\times w\right) ......\left(\mathrm{iii}\right)$

$\left|\overrightarrow{u}\right|=\left|\overrightarrow{v}\right|=\left|\overrightarrow{w}\right|=1$ (As they lie on sphere of unit radius)  $......\left(\mathrm{iv}\right)$

$\left|\overrightarrow{u}-\overrightarrow{v}\right|=\left|\overrightarrow{v}-\overrightarrow{w}\right|=\left|\overrightarrow{w}-\overrightarrow{u}\right|$ (Given)

Now squaring, we get,

$\Rightarrow {\left|\overrightarrow{u}-\overrightarrow{v}\right|}^2={\left|\overrightarrow{v}-\overrightarrow{w}\right|}^2={\left|\overrightarrow{w}-\overrightarrow{u}\right|}^2$

$\Rightarrow \underset{\left(\mathrm{A}\right)}{u^2+v^2-2\overrightarrow{u}·\overrightarrow{v}}=\underset{\left(\mathrm{B}\right)}{v^2+w^2-2\overrightarrow{v}·\overrightarrow{w}}=\underset{\left(\mathrm{C}\right)}{\underbrace{w^2+u^2-2\overrightarrow{w}·\overrightarrow{u}}}$

Now from $\left(\mathrm{A}\right)$ and $\left(\mathrm{B}\right)$ we get,

$\Rightarrow u^2+v^2-2\overrightarrow{u}·\overrightarrow{v}=v^2+w^2-2\overrightarrow{v}·\overrightarrow{w}$

$\Rightarrow u^2-w^2=2\overrightarrow{u}·\overrightarrow{v}-2\overrightarrow{v}·\overrightarrow{w}$ $\left[\because \left|\overrightarrow{\mathrm{u}}\right|=\left|\overrightarrow{\mathrm{w}}\right|=1 \left(\mathrm{Given}\right)\right]$

$\Rightarrow \overrightarrow{u}·\overrightarrow{v}=\overrightarrow{v}·\overrightarrow{w}$

Hence, by using $\left(\mathrm{B}\right)$ and $\left(\mathrm{C}\right)$ also, we will get

$\overrightarrow{u}·\overrightarrow{v}=\overrightarrow{v}·\overrightarrow{w}=\overrightarrow{w}·\overrightarrow{u}=m \left(\mathrm{say}\right) .......\left(\mathrm{v}\right)$

$\Rightarrow \overrightarrow{u}, \overrightarrow{v}, \overrightarrow{w}$ are the vectors of an equilateral triangle (say $∆ ABC$)

$d\left(O, P\right)=\frac{16}{\sqrt{3+4+9}}$$=\frac{16}{4}$$=4$ units

$\overrightarrow{OA}=\overrightarrow{u},\overrightarrow{ OB}=\overrightarrow{v}, \overrightarrow{OC}=\overrightarrow{w}$

$\left|\overrightarrow{OA}\right|=\left|\overrightarrow{OB}\right|=\left|\overrightarrow{OC}\right|=1$ (Given)

In an equilateral triangle, circumcentre, orthrocentre and centroid coincide.

Let $D$ be the circumcentre of $∆ABC$, then

$\angle ADB={120}^{\circ }$

$\Rightarrow \cos \angle ADB=\frac{D{A}^2+D{B}^2-A{B}^2}{2\left(DA\right)·\left(DB\right)} ......\left(\mathrm{vi}\right)$

$OE=OD+DE$$=OD+AF$

$\Rightarrow 4=OD+\frac{7}{2}$

$\Rightarrow OD=4-\frac{7}{2}=\frac{1}{2}$

$\Rightarrow DA=\sqrt{O{A}^2-O{D}^2}$$=\sqrt{1-\frac{1}{4}}$

$\Rightarrow DA=\frac{\sqrt{3}}{2}$

$\Rightarrow DA=DB=\frac{\sqrt{3}}{2} .....\left(\mathrm{vii}\right)$

From $\left(\mathrm{vi}\right)$ and $\left(\mathrm{vii}\right)$  we get,

$-\frac{1}{2}=\frac{\frac{3}{4}+\frac{3}{4}-A{B}^2}{2\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}}$

$-\frac{1}{2}=\frac{\frac{3}{2}-A{B}^2}{\frac{3}{2}}$

$\Rightarrow -\frac{1}{2}\times \frac{3}{2}=\frac{3}{2}-A{B}^2$

$\Rightarrow A{B}^2=\frac{3}{2}+\frac{3}{4}$

$\Rightarrow A{B}^2=\frac{9}{4}$

$\Rightarrow AB=\frac{3}{2}=\left|\overrightarrow{u}-\overrightarrow{v}\right|$

$\Rightarrow A{B}^2=\frac{9}{4}=u^2+v^2-2\overrightarrow{u}·\overrightarrow{v}$

$\Rightarrow \frac{9}{4}=1+1-2m$

$\Rightarrow 2m=2-\frac{9}{4}=-\frac{1}{4}$

$\Rightarrow m=-\frac{1}{8} ......\left(\mathrm{viii}\right)$

Volume of parallelepiped,

$V=\left|\left[\overrightarrow{u}\overrightarrow{v}\overrightarrow{w}\right]\right|$

${\left|\overrightarrow{u}\overrightarrow{v}\overrightarrow{w}\right|}^2=\left|\begin{array}{ccc}1& \overrightarrow{u}·\overrightarrow{v}& \overrightarrow{u}·\overrightarrow{w}\\ \overrightarrow{u}·\overrightarrow{v}& 1& \overrightarrow{v}·\overrightarrow{w}\\ \overrightarrow{w}·\overrightarrow{u}& \overrightarrow{w}·\overrightarrow{v}& 1\end{array}\right|$

$=\left|\begin{array}{ccc}1& m& m\\ m& 1& m\\ m& m& 1\end{array}\right|$

$=1\left(1-m^2\right)-m\left(m-m^2\right)+m\left(m^2-m\right)$

$=1-m^2-m^2+m^3+m^3-m^2$

$=1-3m^2+2m^3$

${\left|\begin{array}{lll}\overrightarrow{u}& \overrightarrow{v}& \overrightarrow{w}\end{array}\right|}^2=2m^3-3m^2+1$

$=\left(m-1\right)\left[2m^2-m-1\right]$

$=\left(m-1\right)\left[2m^2-2m+m-1\right]$

$=\left(m-1\right)\left(m-1\right)\left(2m+1\right)$

$={\left(m-1\right)}^2\left(2m+1\right)$

$\Rightarrow \left|\left[\begin{array}{lll}\overrightarrow{u}& \overrightarrow{v}& \overrightarrow{w}\end{array}\right]\right|=\left(m-1\right)\sqrt{\left(2m+1\right)}$

$\Rightarrow V=\left|\left(-\frac{1}{8}-1\right)\sqrt{2\times -\frac{1}{8}+1}\right|$

$\Rightarrow V=\frac{9}{8}\times \frac{\sqrt{3}}{2}$

Hence, $\frac{80}{\sqrt{3}}V=\frac{80}{\sqrt{3}}\times \frac{9}{8}\times \frac{\sqrt{3}}{2}$$=45$