Question

The angle between the line \(\overrightarrow{r}=\hat{i}+2\hat{j}+t(3\hat{i}+2\hat{j}-\hat{k})\) and the plane 2x-3y-z=1 is

• A. \(\sin^{-1}(\frac{1}{196})\)
• B. \(\sin^{-1}(\frac{1}{14})\)
• C. \(\cos^{-1}(\frac{1}{14})\)
• D. \(\cos^{-1}(\frac{13}{14})\)
• E. \(\sin^{-1}(\frac{13}{14})\)

Answer 1

The Correct Option is B

The given equation of the line is:

\[ \overrightarrow{r} = \hat{i} + 2\hat{j} + t(3\hat{i} + 2\hat{j} - \hat{k}) \] 

The direction vector of the line is:

\[ \overrightarrow{d} = 3\hat{i} + 2\hat{j} - \hat{k} \]

The given plane equation is:

\[ 2x - 3y - z = 1 \]

The normal vector of the plane is:

\[ \overrightarrow{N} = 2\hat{i} - 3\hat{j} - \hat{k} \]

The angle \( \theta \) between the line and the plane is given by:

\[ \sin\theta = \frac{|\overrightarrow{d} \cdot \overrightarrow{N}|}{|\overrightarrow{d}| |\overrightarrow{N}|} \]

Computing the dot product:

\[ (3\hat{i} + 2\hat{j} - \hat{k}) \cdot (2\hat{i} - 3\hat{j} - \hat{k}) \] \[ = (3 \times 2) + (2 \times -3) + (-1 \times -1) \] \[ = 6 - 6 + 1 = 1 \]

Computing the magnitudes:

\[ |\overrightarrow{d}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \] \[ |\overrightarrow{N}| = \sqrt{2^2 + (-3)^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \]

Computing \( \sin\theta \):

\[ \sin\theta = \frac{|1|}{\sqrt{14} \times \sqrt{14}} = \frac{1}{14} \]

Final Answer:

\[ \theta = \sin^{-1} \left( \frac{1}{14} \right) \]