Question

The angle between a normal to the plane $2x-y+2z-1=0$ and the $z-\text{axis}$ is

• A. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• B. ${\sin }^{-1}\left(\frac{2}{3}\right)$
• C. ${\cos }^{-1}\left(\frac{2}{3}\right)$
• D. ${\sin }^{-1}\left(\frac{1}{3}\right)$
• E. ${\sin }^{-1}\left(\frac{3}{5}\right)$

Answer 1

The correct option is C

${\cos }^{-1}\left(\frac{2}{3}\right)$

Explanation for the correct option(s)

Find the angle between a normal to the plane.

Consider the given Equation in vector form,

$$\begin{gathered} \overrightarrow{a}=2\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k} \\ \overrightarrow{b}=\overrightarrow{k} \end{gathered}$$

We know that,

$\cos \theta =\frac{\left|\overrightarrow{a}.\overrightarrow{b}\right|}{\left|\stackrel{\rightharpoonup }{a}\right|\left|\overrightarrow{b}\right|}$

Then,

$$\begin{gathered} \cos \theta =\frac{\left|\left(2\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k}\right).\left(\overrightarrow{k}\right)\right|}{\left|2\overrightarrow{i}-\overrightarrow{j}+2\overrightarrow{k}\right|\left|\overrightarrow{k}\right|} \\ \cos \theta =\frac{\left|0+0+2\right|}{\sqrt{2^2-1^2+2^2}\sqrt{1^2}}\left[\because i.k=0,i.j=0,k.k=1\right] \\ \cos \theta =\frac{2}{2-1+2} \\ \cos \theta =\frac{2}{3} \\ \theta ={\cos }^{-1}\frac{2}{3} \end{gathered}$$

Hence, the correct answer is Option C.