Mathematics - Vector Question with Solution | TestHub

If two vectors $a_1\hat{i}+b_1\hat{j}+c_1\hat{k}$ and $a_2\hat{i}+b_2\hat{j}+c_2\hat{k}$ are perpendicular, then $a_1{a}_2+b_1{b}_2+c_1{c}_2=0.$

Given $\overrightarrow{OP}\perp \overrightarrow{OQ}$

$\Rightarrow -x+2y-3x=0$

$\Rightarrow y=2x ...\left(i\right)$

Also, $\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$

$\Rightarrow \overrightarrow{PQ}=\left(-1-x\right)\hat{i}+\left(2-y\right)\hat{j}+\left(3x+1\right)\hat{k}$

And $\left|\overrightarrow{PQ}\right|=\sqrt{20}$

$\Rightarrow \sqrt{{\left(-1-x\right)}^2+{\left(2-y\right)}^2+{\left(3x+1\right)}^2}=\sqrt{20}$

$\Rightarrow (x+1{)}^2+(y-2{)}^2+(1+3x{)}^2=20$

Put the value from equation $\left(i\right),$ to get

$(x+1{)}^2+(2x-2{)}^2+(1+3x{)}^2=20$

$\Rightarrow x^2+2x+1+4x^2-8x+4+1+6x+9x^2=20$

$\Rightarrow 14x^2+6=20$

$\Rightarrow 14x^2=14$

$\Rightarrow x=\pm 1,$ but given $x>0$

$\Rightarrow x=1$ and $y=2.$

Now $\overrightarrow{OP}, \overrightarrow{OQ},\overrightarrow{OR}$ are coplanar and we know that the three vectors $a_1\hat{i}+b_1\hat{j}+c_1\hat{k}, a_2\hat{i}+b_2\hat{j}+c_2\hat{k}$ and $a_3\hat{i}+b_3\hat{j}+c_3\hat{k}$ are coplanar, then $\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x& y& -1\\ -1& 2& 3x\\ 3& z& -7\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}1& 2& -1\\ -1& 2& 3\\ 3& z& -7\end{array}\right|=0$

$\Rightarrow 1(-14-3z)-2(7-9)-1(-z-6)=0$

$\Rightarrow \mathrm{z}=-2$

$\therefore x^2+y^2+z^2=1+4+4=9.$