Question:
A perpendicular is drawn from the point \( P(2,4,-1) \) to the line \( \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} \). The equation of the perpendicular from \( P \) to the given line is
A perpendicular is drawn from the point \( P(2,4,-1) \) to the line \( \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} \). The equation of the perpendicular from \( P \) to the given line is
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For perpendicular lines in 3D, use dot product = 0 condition.
Updated On: Apr 23, 2026
- $\frac{x-2}{6}=\frac{y-4}{3}=\frac{z+1}{2}$
- $\frac{x+2}{6}=\frac{y-4}{3}=\frac{z+1}{2}$
- $\frac{x+2}{-6}=\frac{y-4}{3}=\frac{z+1}{2}$
- $\frac{x+2}{6}=\frac{y+4}{3}=\frac{z+1}{2}$
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The Correct Option is A
Solution and Explanation
Concept:
Direction vector of perpendicular is perpendicular to direction vector of given line.
Step 1: Extract direction vector of given line.
\[ \vec{d} = \langle 1,\,4,\,-9 \rangle \]
Step 2: Assume required line passes through $P(2,4,-1)$.
Step 3: Find perpendicular direction vector.
Required direction $\vec{D}$ must satisfy: \[ \vec{D} \cdot \vec{d} = 0 \] Check option (A): \[ \vec{D} = \langle 6,3,2 \rangle \]
Step 4: Verify perpendicularity.
\[ 6(1) + 3(4) + 2(-9) = 6 + 12 - 18 = 0 \]
Step 5: Confirm passing point.
Line passes through $(2,4,-1)$ \checkmark Conclusion:
Correct equation = Option (A)
Step 1: Extract direction vector of given line.
\[ \vec{d} = \langle 1,\,4,\,-9 \rangle \]
Step 2: Assume required line passes through $P(2,4,-1)$.
Step 3: Find perpendicular direction vector.
Required direction $\vec{D}$ must satisfy: \[ \vec{D} \cdot \vec{d} = 0 \] Check option (A): \[ \vec{D} = \langle 6,3,2 \rangle \]
Step 4: Verify perpendicularity.
\[ 6(1) + 3(4) + 2(-9) = 6 + 12 - 18 = 0 \]
Step 5: Confirm passing point.
Line passes through $(2,4,-1)$ \checkmark Conclusion:
Correct equation = Option (A)
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