Question:
The perpendicular distance from the origin to the plane containing the two lines, \(\frac{x+2}{3} = \frac{y-2}{5} = \frac{z+5}{7}\) and \(\frac{x-1}{1} = \frac{y-4}{4} = \frac{z+4}{7}\), is :
The perpendicular distance from the origin to the plane containing the two lines, \(\frac{x+2}{3} = \frac{y-2}{5} = \frac{z+5}{7}\) and \(\frac{x-1}{1} = \frac{y-4}{4} = \frac{z+4}{7}\), is :
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The plane's normal is the cross product of the two direction vectors; use one point on a line to get the plane equation.
Updated On: Nov 5, 2026
- \(6\sqrt{11}\)
- \(\frac{11}{\sqrt{6}}\)
- \(11\)
- \(11\sqrt{6}\)
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The Correct Option is B
Solution and Explanation
Direction vectors: \(\vec{d_1} = (3,5,7)\), \(\vec{d_2} = (1,4,7)\).
Normal \(\vec{n} = \vec{d_1}\times\vec{d_2} = (35-28,\ 7-21,\ 12-5) = (7,-14,7) \parallel (1,-2,1)\).
Plane through \((-2,2,-5)\): \(1(x+2) - 2(y-2) + 1(z+5) = 0 \Rightarrow x - 2y + z + 11 = 0\).
Check with the point \((1,4,-4)\) of the second line: \(1 - 8 - 4 + 11 = 0\). It lies on the plane, so the lines are coplanar.
Distance from origin \(= \frac{|11|}{\sqrt{1+4+1}} = \frac{11}{\sqrt{6}}\).
Answer: (2) \(\frac{11}{\sqrt{6}}\)
Normal \(\vec{n} = \vec{d_1}\times\vec{d_2} = (35-28,\ 7-21,\ 12-5) = (7,-14,7) \parallel (1,-2,1)\).
Plane through \((-2,2,-5)\): \(1(x+2) - 2(y-2) + 1(z+5) = 0 \Rightarrow x - 2y + z + 11 = 0\).
Check with the point \((1,4,-4)\) of the second line: \(1 - 8 - 4 + 11 = 0\). It lies on the plane, so the lines are coplanar.
Distance from origin \(= \frac{|11|}{\sqrt{1+4+1}} = \frac{11}{\sqrt{6}}\).
Answer: (2) \(\frac{11}{\sqrt{6}}\)
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