Question:
If the shortest distance between the lines $\frac{x - k}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{\sqrt{29}}$, then $k =$}
If the shortest distance between the lines $\frac{x - k}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{\sqrt{29}}$, then $k =$}
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The two lines are parallel, so use the distance between parallel lines with the cross product of the connecting vector and the direction.
Updated On: Nov 13, 2026
- $1$
- $-1$
- $2$
- $-2$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
Direction \(\vec d=(2,3,4)\) for both lines, since \((4,6,8)=2(2,3,4)\).
Points: \(A(k,4,3)\) and \(B(2,4,7)\).
Step 2: Distance formula
\(\text{distance}=\dfrac{|\overrightarrow{AB}\times\vec d|}{|\vec d|}\), \(\overrightarrow{AB}=(2-k,0,4)\).
Step 3: Calculation
\(\overrightarrow{AB}\times\vec d=(-12,\,4k,\,6-3k)\)
\(|\cdot|^2=144+16k^2+(6-3k)^2=25k^2-36k+180\), and \(|\vec d|^2=29\).
Set \(\dfrac{25k^2-36k+180}{29}=\dfrac{169}{29}\), so \(25k^2-36k+11=0\).
Step 4: Compare the options
\((k-1)(25k-11)=0\), so \(k=1\) (the other root \(\tfrac{11}{25}\) is not offered). The first option is \(1\).
Final Answer:
The value of k is 1. \[ \boxed{\text{(A) }1} \]
Direction \(\vec d=(2,3,4)\) for both lines, since \((4,6,8)=2(2,3,4)\).
Points: \(A(k,4,3)\) and \(B(2,4,7)\).
Step 2: Distance formula
\(\text{distance}=\dfrac{|\overrightarrow{AB}\times\vec d|}{|\vec d|}\), \(\overrightarrow{AB}=(2-k,0,4)\).
Step 3: Calculation
\(\overrightarrow{AB}\times\vec d=(-12,\,4k,\,6-3k)\)
\(|\cdot|^2=144+16k^2+(6-3k)^2=25k^2-36k+180\), and \(|\vec d|^2=29\).
Set \(\dfrac{25k^2-36k+180}{29}=\dfrac{169}{29}\), so \(25k^2-36k+11=0\).
Step 4: Compare the options
\((k-1)(25k-11)=0\), so \(k=1\) (the other root \(\tfrac{11}{25}\) is not offered). The first option is \(1\).
Final Answer:
The value of k is 1. \[ \boxed{\text{(A) }1} \]
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