Question:
If the lines \(L_1:\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2},L_2:\frac{x-1}{3k} = \frac{y-5}{1} = \frac{z-6}{-5}\) are perpendicular to each other, then the equation of a plane containing the line \(L_1\) and parallel to the line \(L_2\) for the value of \(k\) that satisfies this condition is ...
If the lines \(L_1:\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2},L_2:\frac{x-1}{3k} = \frac{y-5}{1} = \frac{z-6}{-5}\) are perpendicular to each other, then the equation of a plane containing the line \(L_1\) and parallel to the line \(L_2\) for the value of \(k\) that satisfies this condition is ...
Show Hint
Find $k$ from the perpendicular condition, then take the cross product of the directions as the normal.
Updated On: Oct 1, 2026
- \(31x+21y+46z-211 = 0\)
- \(602x-1155y-747z+3949 = 0\)
- \(43x+105y-154z+209 = 0\)
- \(2x-3y+5z-11 = 0\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Find \(k\)
Direction ratios: \(L_1=(-3,2k,2)\), \(L_2=(3k,1,-5)\). Perpendicular: \(-9k+2k-10=0\), so \(k=-\frac{10}{7}\).
Step 2: Directions
\(L_1:(-3,-\frac{20}{7},2)\propto(-21,-20,14)\). \(L_2:(-\frac{30}{7},1,-5)\propto(-30,7,-35)\).
Step 3: Normal
\[ \vec n=(-21,-20,14)\times(-30,7,-35)=(602,-1155,-747) \]
Step 4: Plane
It passes through \((1,2,3)\), which lies on \(L_1\). Constant: \(602-2310-2241=-3949\).
So \(602x-1155y-747z+3949=0\), option (B).
Final Answer:
The plane is \(602x-1155y-747z+3949=0\), option (B). \[ \boxed{\text{(B)}} \]
Direction ratios: \(L_1=(-3,2k,2)\), \(L_2=(3k,1,-5)\). Perpendicular: \(-9k+2k-10=0\), so \(k=-\frac{10}{7}\).
Step 2: Directions
\(L_1:(-3,-\frac{20}{7},2)\propto(-21,-20,14)\). \(L_2:(-\frac{30}{7},1,-5)\propto(-30,7,-35)\).
Step 3: Normal
\[ \vec n=(-21,-20,14)\times(-30,7,-35)=(602,-1155,-747) \]
Step 4: Plane
It passes through \((1,2,3)\), which lies on \(L_1\). Constant: \(602-2310-2241=-3949\).
So \(602x-1155y-747z+3949=0\), option (B).
Final Answer:
The plane is \(602x-1155y-747z+3949=0\), option (B). \[ \boxed{\text{(B)}} \]
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