Question

The equation of the plane perpendicular to the $yz-$ plane and passing through the point $(1,-2,4)$ and $(3,-4,5)$ is

• A. $y+2z=5$
• B. $2y+z=5$
• C. $y+2z=6$
• D. $2y+z=6$

Answer 1

The correct option is C $y+2z=6$

Plane is perpendicular to yz plane

So it have x-axis as parallel to it

Find a vector lying in the plane using two given point as-

$\overrightarrow{5}=(3-1)\hat{i}+(-4-(-2)\hat{j}+(5-A)\hat{k}$

$\overrightarrow{5}=2\hat{i}-2\hat{j}+\hat{k}$

parallel vector, $\overrightarrow{p}=1.\hat{i}+0.\hat{j}+0.\hat{k}$

So normal to the plane,

$\overrightarrow{N}=\overrightarrow{S}\times \overrightarrow{P}$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 1\\ 1& 0& 0\end{array}\right|$

$=\hat{j}+2\hat{k}$

equation of plane using a point and normal,

$\Rightarrow 0.(x-1)+1.(y-(-2)+2.(z-4)=0$

$\Rightarrow y+2+2z-8=0$

$\Rightarrow y+2z=6$