Question:
The shortest distance between the two lines
\[
\vec r=(\hat i-\hat j)+s(\hat j+2\hat k)
\]
and
\[
\vec r=(2\hat i+\hat k)+t(\hat i-\hat j+\hat k)
\]
is
The shortest distance between the two lines
\[
\vec r=(\hat i-\hat j)+s(\hat j+2\hat k)
\]
and
\[
\vec r=(2\hat i+\hat k)+t(\hat i-\hat j+\hat k)
\]
is
Show Hint
For skew lines, use
$\displaystyle D=\frac{|(\vec b-\vec a)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}$.
Updated On: Jun 3, 2026
- $4$
- $5$
- \[ \sqrt{\frac{6}{5}} \]
- \[ \sqrt{\frac{8}{7}} \]
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Use the formula for the shortest distance between two skew lines.
Step 2: Meaning
Let \[ \vec a=(1,-1,0), \quad \vec b=(2,0,1), \] and direction vectors \[ \vec d_1=(0,1,2), \quad \vec d_2=(1,-1,1). \]
Step 3: Analysis
Compute \[ \vec d_1\times\vec d_2 = \begin{vmatrix} \hat i & \hat j & \hat k\\ 0 & 1 & 2\\ 1 & -1 & 1 \end{vmatrix} = (3,2,-1). \] Its magnitude is \[ |\vec d_1\times\vec d_2| = \sqrt{14}. \] Also, \[ \vec b-\vec a=(1,1,1). \] Hence \[ |(\vec b-\vec a)\cdot(\vec d_1\times\vec d_2)| = |3+2-1| = 4. \] Therefore, \[ D = \frac{4}{\sqrt{14}} = \sqrt{\frac{8}{7}}. \]
Step 4: Conclusion
Thus the shortest distance is \[ \sqrt{\frac{8}{7}}. \]
Final Answer: (D)
Use the formula for the shortest distance between two skew lines.
Step 2: Meaning
Let \[ \vec a=(1,-1,0), \quad \vec b=(2,0,1), \] and direction vectors \[ \vec d_1=(0,1,2), \quad \vec d_2=(1,-1,1). \]
Step 3: Analysis
Compute \[ \vec d_1\times\vec d_2 = \begin{vmatrix} \hat i & \hat j & \hat k\\ 0 & 1 & 2\\ 1 & -1 & 1 \end{vmatrix} = (3,2,-1). \] Its magnitude is \[ |\vec d_1\times\vec d_2| = \sqrt{14}. \] Also, \[ \vec b-\vec a=(1,1,1). \] Hence \[ |(\vec b-\vec a)\cdot(\vec d_1\times\vec d_2)| = |3+2-1| = 4. \] Therefore, \[ D = \frac{4}{\sqrt{14}} = \sqrt{\frac{8}{7}}. \]
Step 4: Conclusion
Thus the shortest distance is \[ \sqrt{\frac{8}{7}}. \]
Final Answer: (D)
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