Question:
If $\theta$ is the acute angle between a line $\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-1}{1}$ and a normal to the plane $2x+3y+4z=0$ then $\tan^{2}\theta+sec^{2}\theta=$
If $\theta$ is the acute angle between a line $\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z-1}{1}$ and a normal to the plane $2x+3y+4z=0$ then $\tan^{2}\theta+sec^{2}\theta=$
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Be careful with the phrasing: the angle between a line and a *normal* to a plane uses the direct cosine dot product formula, not the sine formula used for the angle between a line and the plane surface.
Updated On: Jun 3, 2026
- 1
- $\frac{16}{13}$
- $\frac{55}{3}$
- $\frac{22}{7}$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
The angle $\theta$ between a line with direction vector $\vec{d}$ and the normal vector $\vec{n}$ of a plane is given by $\cos\theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}||\vec{n}|}$.
Step 2: Meaning
From the given equations, the direction vector of the line is $\vec{d} = \hat{i} - \hat{j} + \hat{k}$, and the normal vector to the plane is $\vec{n} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Step 3: Analysis
Compute the dot product and magnitudes: $\vec{d} \cdot \vec{n} = 1(2) + (-1)(3) + 1(4) = 2 - 3 + 4 = 3$. $|\vec{d}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$. |n| = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29$. Thus, $ = 3329 = 329 ^2 = 329$. Then $^2 = 293$ and $^2 = ^2 - 1 = 293 - 1 = 263$. \textcolor{red}{\textbf{Step 4: Conclusion}} \\Substitute into the required expression: $^2 + ^2 = 263 + 293 = 553$. This matches option (C).
Final Answer: (C)
The angle $\theta$ between a line with direction vector $\vec{d}$ and the normal vector $\vec{n}$ of a plane is given by $\cos\theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}||\vec{n}|}$.
Step 2: Meaning
From the given equations, the direction vector of the line is $\vec{d} = \hat{i} - \hat{j} + \hat{k}$, and the normal vector to the plane is $\vec{n} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Step 3: Analysis
Compute the dot product and magnitudes: $\vec{d} \cdot \vec{n} = 1(2) + (-1)(3) + 1(4) = 2 - 3 + 4 = 3$. $|\vec{d}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$. |n| = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29$. Thus, $ = 3329 = 329 ^2 = 329$. Then $^2 = 293$ and $^2 = ^2 - 1 = 293 - 1 = 263$. \textcolor{red}{\textbf{Step 4: Conclusion}} \\Substitute into the required expression: $^2 + ^2 = 263 + 293 = 553$. This matches option (C).
Final Answer: (C)
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