Question

A plane passes through the points $A(1,2,3),B(2,3,1)$ and $C(2,4,2)$. If $O$ is the origin and $P$ is $(2,–1,1)$, then the projection of vector$\left(OP\right)$ on this plane is of length:

• A. $\sqrt{\frac{2}{5}}$
• B. $\sqrt{\frac{2}{3}}$
• C. $\sqrt{\frac{2}{11}}$
• D. $\sqrt{\frac{2}{7}}$

Answer 1

The correct option is C

$\sqrt{\frac{2}{11}}$

Explanation for the correct option:

Step 1: Use the determinant form of the plane to get the equation:

Put the points in the determinant form of the plane and simplify.

$\left|\begin{array}{ccc}x-1& y-2& z-3\\ 1& 1& -2\\ 1& 2& -1\end{array}\right|=0$

To get the planar equation, solve the determinant.

$\begin{array}{rcl}\left(x-1\right)\left(-1+4\right)-\left(y-2\right)\left(-1+2\right)+\left(z-3\right)\left(2-1\right)& =& 0\\ 3x-3-y+2+z-3& =& 0\\ 3x-y+z-4& =& 0\end{array}$

Step 2: Find the equation of plane's normal:

To find the plane's normal, use the determinant form.

$$\begin{gathered} \overrightarrow{n}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1& 1& -2\\ 0& 1& 1\end{array}\right| \\ \Rightarrow \overrightarrow{n}=3\hat{i}-\hat{j}+\hat{k} \end{gathered}$$

Also, $\overrightarrow{l}=2\hat{i}-\hat{j}+\hat{k}$

Step 3: Find the projection of the line on the plane:

Find the sine of the projection angle of the line on the plane.

$\begin{array}{rcl}\cos \theta & =& \frac{6+1+1}{\sqrt{6}\sqrt{11}}\\ & =& \frac{8}{\sqrt{66}}\\ & \Rightarrow & \sin \theta =\sqrt{\frac{2}{66}}\end{array}$

Find the projection of the line on the plane.

Projection$$\begin{gathered} =\left|\overrightarrow{l}\right|\sin \theta \\ =\sqrt{\frac{2}{11}} \end{gathered}$$

Hence, option (C) is the correct solution.