Question

The angle between the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2}$ and the plane $x+y+4=0$ is

• A. $0°$
• B. $30°$
• C. $45°$
• D. $90°$

Answer 1

The correct option is C

$45°$

Explanation for correct option:

Given the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2}$and the plane $x+y+4=0$

We know that the general equation of plane is , $ax+by+cz+d=0$.

Then the normal vector $n$ to the plane can be defined as $\overrightarrow{n}=\hat{i}+\hat{j}$ using the equation of the plane $x+y+4=0$.

And the vector $b$ is the direction ratios along the line is defined as $\overrightarrow{b}=2\hat{i}+\hat{j}-2\hat{k}$.

Now we know that ,

$\begin{array}{l}\cos (90-\theta )=\frac{|\overrightarrow{n}.\overrightarrow{b}|}{\left|\overrightarrow{n}\right|\left|\overrightarrow{b}\right|}\\ =\frac{2·1+1·1}{\sqrt{1+1}\sqrt{4+1+4}}\\ =\frac{2+1}{\sqrt{1+1}\sqrt{4+1+4}}\\ =\frac{3}{\sqrt{2}\sqrt{9}}\\ =\frac{3}{\sqrt{2}\times 3}\\ =\frac{1}{\sqrt{2}}\end{array}$

We know that $\cos \left(90-\theta \right)=\sin \theta$, using the same we have,

$\begin{array}{l}⟹\cos (90-\theta )=\frac{1}{\sqrt{2}}\\ ⟹\sin \left(\theta \right)=\frac{1}{\sqrt{2}}\\ ⟹\theta ={\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)\\ ⟹\theta =45°\end{array}$

Therefore the angle between the line and the plane is $45°$

Hence, the correct answer is option (C).