Question

If the direction cosines of the line common to the planes \[ x+2y-z-1=0 \] and \[ 3x-4y+z-5=0 \] are \((l,m,n)\), then \(|l+m-n|=\)

• A. \(\frac6{\sqrt{30}}\)
• B. \(\frac4{\sqrt{30}}\)
• C. \(\frac2{\sqrt{30}}\)
• D. \(\frac8{\sqrt{30}}\)

Hint:

The line of intersection of two planes is perpendicular to both normals, so its direction vector is their cross product.

Answer 1

The Correct Option is C

Concept: Direction ratios of the line of intersection of two planes are obtained from the cross product of their normal vectors.

Step 1: Find normals.
\[ \vec n_1=(1,2,-1), \] \[ \vec n_2=(3,-4,1). \]

Step 2: Find cross product.
\[ \vec n_1\times \vec n_2 = (-2,-4,-10). \] Direction ratios \[ (1,2,5). \]

Step 3: Find direction cosines.
\[ \sqrt{1^2+2^2+5^2} = \sqrt{30}. \] Hence \[ l=\frac1{\sqrt{30}}, \quad m=\frac2{\sqrt{30}}, \quad n=\frac5{\sqrt{30}}. \]

Step 4: Compute required quantity.
\[ |l+m-n| = \left| \frac{1+2-5}{\sqrt{30}} \right|. \] \[ = \frac2{\sqrt{30}}. \] Thus \[ \boxed{\frac2{\sqrt{30}}}. \]