Discussion Forum

1)

Consider the lines

$L_{1}=\frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1},$
$L_{2}=\frac{x-4}{1}=\frac{y+3}{1}=\frac{z+3}{2}$
 and the planes p₁:7x+y+2z=3,

 p₂:3x+5y-6z=4, Let  ax+by+cz=d the equation of the line passing through the point of intersection of lines L₁ and L₂ and perpendicular to plane p₁  and p₂.

 Match List I with List II and select the correct answer using the code given below the lists.

 

Answer:

Option A

Explanation:

$L_{1}:\frac{x-1}{2}=\frac{y-0}{-1}=\frac{z-(-3)}{1}$

  Normal of plane p:n=  $\begin{bmatrix}\hat{i} & \hat{j} & \hat{k}\\7 & 1&2\\3&5&-6 \end{bmatrix}$

 = $\hat{i} (-16)-\hat{j}(-42-6)+\hat{k}(32)$

   $= -16\hat{i}+48\hat{j}+32\hat{k}$

 DR's  of normal n=  $\hat{i}-3\hat{j}-2\hat{k}$

 Point of intersection of L₁  and L₂ 

$\Rightarrow $    $ 2K_{1}+1=K_{2}+4 and -k_{1}=k_{2}-3$

 $\therefore$   k₁= 2 and k₂ =1

 $\therefore$  point of intersection (5,-2,-1)

 $\therefore$ Equation of plane ,

  1.(x-5)-3 (y+2)-2(z+1)=0

 $\Rightarrow  $  x-3y-2z-13=0

 $\Rightarrow $   x-3y-2z=13

$\therefore$  a=1, b=-3, c=-2, d=13