Question:
The length of the perpendicular from the point \((3,2,1)\) to the line \(\frac{x-7}{-2}=\frac{y-7}{2}=\frac{z-6}{3}\) is \(k\) units. then the value of \(k\)....
The length of the perpendicular from the point \((3,2,1)\) to the line \(\frac{x-7}{-2}=\frac{y-7}{2}=\frac{z-6}{3}\) is \(k\) units. then the value of \(k\)....
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Take a general point on the line, make the joining vector perpendicular to the direction ratios, and find the distance.
Updated On: Nov 12, 2026
- \(3\)
- \(5\)
- \(7\)
- \(9\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
A general point on the line is \(P(7-2t,\,7+2t,\,6+3t)\). Let \(A=(3,2,1)\).
Step 2: Join A and P
\(\overrightarrow{AP}=(4-2t,\;5+2t,\;5+3t)\).
Step 3: Perpendicular condition
The direction ratios of the line are \((-2,2,3)\).
\(-2(4-2t)+2(5+2t)+3(5+3t)=17+17t=0\Rightarrow t=-1\).
Step 4: Distance
For \(t=-1\): \(\overrightarrow{AP}=(6,3,2)\), so \(k=\sqrt{36+9+4}=\sqrt{49}=7\).
Final Answer:
The length of the perpendicular is 7 units. \[ \boxed{\text{(C) }7} \]
A general point on the line is \(P(7-2t,\,7+2t,\,6+3t)\). Let \(A=(3,2,1)\).
Step 2: Join A and P
\(\overrightarrow{AP}=(4-2t,\;5+2t,\;5+3t)\).
Step 3: Perpendicular condition
The direction ratios of the line are \((-2,2,3)\).
\(-2(4-2t)+2(5+2t)+3(5+3t)=17+17t=0\Rightarrow t=-1\).
Step 4: Distance
For \(t=-1\): \(\overrightarrow{AP}=(6,3,2)\), so \(k=\sqrt{36+9+4}=\sqrt{49}=7\).
Final Answer:
The length of the perpendicular is 7 units. \[ \boxed{\text{(C) }7} \]
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