Question

A line passing through \(A(1,3,2)\) and parallel to the line \(\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}\) intersects the plane \(x-y+3z-6=0\) at point B. Another line through B and perpendicular to the plane \(x-y+3z-6=0\) intersects the plane \(2x-y+z+4=0\) at the point C. Then the equation of median of \(\Delta ABC\) through the vertex A is ...

• A. \(\frac{x-1}{5}=\frac{y-3}{1}=\frac{z-2}{-1}\)
• B. \(\frac{x-1}{1}=\frac{y-3}{5}=\frac{z-2}{-1}\)
• C. \(\frac{x-1}{2}=\frac{y-3}{3}=\frac{z-2}{5}\)
• D. \(\frac{x-1}{3}=\frac{y-3}{-5}=\frac{z-2}{2}\)

Hint:

Find B and C by intersecting the lines with the planes, then take the midpoint of BC.

Answer 1

The Correct Option is B

Step 1: Find B
The line through \(A(1,3,2)\) with direction \((1,2,1)\) is \((1+t,\ 3+2t,\ 2+t)\).
In the plane \(x-y+3z=6\): \(2t-2=0\), so \(t=1\) and \(B=(2,5,3)\).

Step 2: Find C
The normal line through \(B\) is \((2+s,\ 5-s,\ 3+3s)\).
In the plane \(2x-y+z+4=0\): \(6s+6=0\), so \(s=-1\) and \(C=(1,6,0)\).

Step 3: Midpoint of BC
\(M=\left(\dfrac{3}{2},\dfrac{11}{2},\dfrac{3}{2}\right)\).

Step 4: Median from A
\(\overrightarrow{AM}=\left(\dfrac{1}{2},\dfrac{5}{2},-\dfrac{1}{2}\right)\propto (1,5,-1)\).
The line is \(\dfrac{x-1}{1}=\dfrac{y-3}{5}=\dfrac{z-2}{-1}\).

Final Answer:
The median is \(\dfrac{x-1}{1}=\dfrac{y-3}{5}=\dfrac{z-2}{-1}\). \[ \boxed{\text{(B) }\dfrac{x-1}{1}=\dfrac{y-3}{5}=\dfrac{z-2}{-1}} \]