Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the normal vector of the required plane be $\vec{n}$. Since the required plane is perpendicular to the given planes, its normal vector will be perpendicular to the normal vectors of the given planes.

The normal vectors of the given planes are ${\vec{n}}_1=2\hat{i}-\hat{j}+2\hat{k}$ and ${\vec{n}}_2=3\hat{i}+6\hat{j}+2\hat{k}$.

The normal vector $\vec{n}$ is given by their cross product:

$\vec{n}={\vec{n}}_1\times {\vec{n}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -1& 2\\ 3& 6& 2\end{array}\right|$

$\vec{n}=\hat{i}(-2-12)-\hat{j}(4-6)+\hat{k}(12-(-3))$

$\vec{n}=-14\hat{i}+2\hat{j}+15\hat{k}$

The equation of the plane passing through $(1,1,1)$ with normal vector $\vec{n}$ is:

$-14(x-1)+2(y-1)+15(z-1)=0$

$-14x+14+2y-2+15z-15=0$

$-14x+2y+15z=3$

Answer: $-14x+2y+15z=3$