Question:
If l, m, n are the direction cosines of a normal drawn to the plane \(2x-3y+6z-7=0\) and d is the length of the perpendicular drawn from origin to this plane then \(7d|l+m+n|=\)
If l, m, n are the direction cosines of a normal drawn to the plane \(2x-3y+6z-7=0\) and d is the length of the perpendicular drawn from origin to this plane then \(7d|l+m+n|=\)
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For plane \(ax+by+cz+d=0\), the direction cosines of the normal are proportional to \((a,b,c)\).
Updated On: Oct 17, 2026
- 3
- 2
- 5
- 4
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The Correct Option is C
Solution and Explanation
Step 1: Find the normal vector.
For the plane
\[
2x-3y+6z-7=0
\]
the normal vector is
\[
\vec n=(2,-3,6)
\]
Its magnitude is
\[
|\vec n|
=\sqrt{2^2+(-3)^2+6^2}
\]
\[
=\sqrt{4+9+36}
=\sqrt{49}
=7
\]
Step 2: Find the direction cosines. \[ l=\frac{2}{7}, \qquad m=-\frac{3}{7}, \qquad n=\frac{6}{7} \] Therefore, \[ l+m+n = \frac{2-3+6}{7} = \frac{5}{7} \] Hence, \[ |l+m+n| = \frac{5}{7} \]
Step 3: Find the perpendicular distance from origin. Distance from origin to the plane \[ 2x-3y+6z-7=0 \] is \[ d= \frac{|{-7}|}{\sqrt{2^2+(-3)^2+6^2}} \] \[ =\frac{7}{7} =1 \]
Step 4: Evaluate the expression. \[ 7d|l+m+n| = 7(1)\left(\frac{5}{7}\right) \] \[ =5 \] \[ \boxed{5} \]
Step 2: Find the direction cosines. \[ l=\frac{2}{7}, \qquad m=-\frac{3}{7}, \qquad n=\frac{6}{7} \] Therefore, \[ l+m+n = \frac{2-3+6}{7} = \frac{5}{7} \] Hence, \[ |l+m+n| = \frac{5}{7} \]
Step 3: Find the perpendicular distance from origin. Distance from origin to the plane \[ 2x-3y+6z-7=0 \] is \[ d= \frac{|{-7}|}{\sqrt{2^2+(-3)^2+6^2}} \] \[ =\frac{7}{7} =1 \]
Step 4: Evaluate the expression. \[ 7d|l+m+n| = 7(1)\left(\frac{5}{7}\right) \] \[ =5 \] \[ \boxed{5} \]
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