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The distance of line $3 y-2 z-1=0=3 x-z+4$ from the point $(2,-1,6)$ is :
Option: 1 $2 \sqrt{5}$
Option: 2 $2 \sqrt{6}$
Option: 3 $\sqrt{26}$
Option: 4 $4 \sqrt{2}$

Answers (1)

best_answer

Let PQ be perpendicular to the given line

Now Q lies on both the planes,

So $3a-2c-1= 0---\left ( i \right )$ and
$3a-c+4= 0---\left ( ii \right )$

$Also\: PQ\perp line\Rightarrow PQ\perp \left ( \vec{n_{1} }\times\vec{n_{2}} \right )$
$\Rightarrow \vec{PQ}\cdot \left ( \vec{n_{1}} \times \vec{n_{2}}\right )= 0$
$\Rightarrow\left [ \vec{PQ}\; \; \vec{n_{1}} \; \; \vec{n_{2}} \right ] = 0$
$\Rightarrow \begin{vmatrix} a-2&b+1 &c-6 \\ 3& 0 & -1\\ 0&3 &-2 \end{vmatrix}= 0$
$\Rightarrow 3\left ( a-2 \right )-\left ( b+1 \right )\left ( -6 \right )+\left ( c-6 \right )9= 0$
$\Rightarrow a-2+2b+2+3c-18= 0$
$\Rightarrow a+2b+3c= 18---\left ( iii \right )$

Solving (i),(ii) & (iii)

$a= 0,b= 3,c= 4\Rightarrow Q\left ( 0,3,4 \right )$
$PQ= 2\sqrt{6}$

Posted by

Kuldeep Maurya

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