Question

Find the equation of the plane perpendicular to vector $2\hat{i}+3\hat{j}+6\hat{k}$ and passing through the point $\hat{i}+5\hat{j}+3\hat{k}$. Also find the distance of the plane from the origin.

Answer 1

Since plane is $\perp$ to$(2\hat{i}+3\hat{j}+6\hat{k})$ & pass through $\hat{i}+5\hat{j}+3\hat{k}$ then equation is

$\left[\right(x-1)\hat{i}+(y-5)\hat{j}+(z-3\left)\hat{k}\right]\cdot [2\hat{i}+3\hat{j}+6\hat{k}]=0$

$\Rightarrow 2(x-1)+3(y-5)+6(z-3)=0$

$\Rightarrow 2x+3y+6z-35=0$

Dist from origin $d=\frac{|0+0+0-35|}{\sqrt{4+9+36}}$

$d=\left(\frac{35}{7}\right)$

$d=5$.