Discussion Forum



1)

The equation of the  plane passing through the points with position vectors $ A(2\hat{i}+6\hat{j}-6\hat{k})$, $B(-3\hat{i}+10\hat{j}-9\hat{k})$  and $ C(-5\hat{i}-6\hat{k})$  is 


A) $r.(2\hat{i}-\hat{j}-2\hat{k})=2$

B) $r.(\hat{i}-2\hat{j}-\hat{k})=1$

C) $r. (2\hat{i}+\hat{j}-2\hat{k})=3$

D) $r. (\hat{i}+2\hat{j}-2\hat{k})=3$

Answer:

Option A

Explanation:

Equation of plane passing through any three  given point is 

$\begin{bmatrix}x-x_{1} &y-y_{1}&z-z_{1} \\x_{2}-x_{1} & y_{2}-y_{1}&z_{2}-z_{1}\\x_{3}-x_{1}&y_{3}-y_{1}&z_{3}-z_{1} \end{bmatrix}$

   Points are (-2,6,-6),(-3,10,-9) and (-5,0,-6)

 $\Rightarrow$   $\begin{bmatrix}x+2 &y-6&z+6 \\-1 & 4&-3\\-3&-6&0 \end{bmatrix}=0$

$\Rightarrow $  $(x+2)(0-18)-(y-6)(0-9)+(z+6)(6+12)$=0

$\Rightarrow $    $-18(x+2)+9(y-6)+18(z+6)=0$

$\Rightarrow $     $-18x+9y+18z=36+54-108$

$\Rightarrow $    $-2 \hat{i}+\hat{j}+2\hat{k}=-2$

$\Rightarrow $   $2\hat{i}-\hat{j}-2\hat{k}=2$

$\therefore$    $r.(2\hat{i}-\hat{j}-2\hat{k})=2$