Question:
Let P be a plane \(lx + my + nz = 0\) containing the line, \(\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}\). If plane P divides the line segment AB joining points A(\(-3, -6, 1\)) and B(\(2, 4, -3\)) in ratio k : 1 then the value of k is equal to :
Let P be a plane \(lx + my + nz = 0\) containing the line, \(\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}\). If plane P divides the line segment AB joining points A(\(-3, -6, 1\)) and B(\(2, 4, -3\)) in ratio k : 1 then the value of k is equal to :
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The plane passes through the origin and a point of the line, and its normal is perpendicular to the line's direction. Then use the section formula so that the dividing point satisfies the plane equation.
Updated On: Nov 4, 2026
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The Correct Option is A
Solution and Explanation
The line is \(\frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3}\): it passes through \((1,-4,-2)\) with direction ratios \((-1,2,3)\).
The plane contains the line, so \(l(1)+m(-4)+n(-2)=0\) and \(-l+2m+3n=0\).
Adding: \(-2m+n=0 \Rightarrow n=2m\). Then \(l = 4m+2n = 8m\). So \((l,m,n) \propto (8,1,2)\) and the plane is \(8x+y+2z=0\).
Value of the plane expression at A: \(8(-3)+(-6)+2(1) = -28\); at B: \(8(2)+4+2(-3) = 14\).
The point dividing AB in ratio k : 1 is \(\frac{kB+A}{k+1}\); it lies on the plane when \(k(14) + (-28) = 0\), so \(k = 2\).
Answer: (1) 2
The plane contains the line, so \(l(1)+m(-4)+n(-2)=0\) and \(-l+2m+3n=0\).
Adding: \(-2m+n=0 \Rightarrow n=2m\). Then \(l = 4m+2n = 8m\). So \((l,m,n) \propto (8,1,2)\) and the plane is \(8x+y+2z=0\).
Value of the plane expression at A: \(8(-3)+(-6)+2(1) = -28\); at B: \(8(2)+4+2(-3) = 14\).
The point dividing AB in ratio k : 1 is \(\frac{kB+A}{k+1}\); it lies on the plane when \(k(14) + (-28) = 0\), so \(k = 2\).
Answer: (1) 2
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