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 \(7d|l+m+n|=\)

• A. 3
• B. 2
• C. 5
• D. 4

Hint:

For plane \(ax+by+cz+d=0\), the direction cosines of the normal are proportional to \((a,b,c)\).

Answer 1

The Correct Option is C

Step 1: Identify the normal.
For the plane $2x-3y+6z-7=0$, the normal vector is $\vec n=(2,-3,6)$.
Step 2: Find its length.
\[ |\vec n|=\sqrt{2^2+(-3)^2+6^2}=\sqrt{49}=7. \]
Step 3: Direction cosines.
Divide each component by $7$: $l=\dfrac27,\;m=-\dfrac37,\;n=\dfrac67$.
Step 4: Combine the direction cosines.
\[ l+m+n=\frac{2-3+6}{7}=\frac57,\qquad |l+m+n|=\frac57. \]
Step 5: Distance from the origin.
\[ d=\frac{|-7|}{\sqrt{49}}=\frac{7}{7}=1. \]
Step 6: Put it together.
\[ 7d\,|l+m+n|=7(1)\left(\frac57\right)=5. \]
\[ \boxed{5} \]