Question

The angle between the line $r=\left(i+2j+3k\right)+\lambda \left(2i+3j+4k\right)$and the plane $r$ . $\left(i+2j-2k\right)=0$ is

• A. $0°$
• B. $60°$
• C. $30°$
• D. $90°$
• E. $45°$

Answer 1

The correct option is A

$0°$

Find the angle between the line and plane.

Consider the given equation as,

$r=\left(i+2j+3k\right)+\lambda \left(2i+3j+4k\right)$

$\left(i+2j-2k\right)=0$

We know that the general form of vector equation

$$\begin{gathered} \Rightarrow \overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \\ \Rightarrow \overrightarrow{r}.\overrightarrow{n}=\overrightarrow{d} \\ \Rightarrow \sin \theta =\frac{\overrightarrow{b}.\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|} \end{gathered}$$

Let,

$$\begin{gathered} \overrightarrow{b}=2\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k} \\ \overrightarrow{n}=\overrightarrow{i}+2\overrightarrow{j}-2\overrightarrow{k} \end{gathered}$$

Then,

$$\begin{gathered} \overrightarrow{b}.\overrightarrow{n}=\left(2\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k}\right)\left(\overrightarrow{i}+2\overrightarrow{j}-2\overrightarrow{k}\right) \\ \overrightarrow{b}.\overrightarrow{n}=\left(2\times 1+3\times 2+4\times -2\right) \\ \overrightarrow{b}.\overrightarrow{n}=\left(2+6-8\right) \\ \overrightarrow{b}.\overrightarrow{n}=0 \end{gathered}$$

Therefore,

$$\begin{gathered} \sin \theta =\frac{0}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|} \\ \sin \theta =0 \\ \theta ={\sin }^{-1}0 \\ \theta =0° \end{gathered}$$

Hence, the correct answer is Option A.