Topic: Engineering Mathematics
The point where the plane $x+3y+2z-6=0$ meets the line joining P (2, 1, 1) and Q (1, 3, -2) is,
- A. (2, 1, 1)
- B. (1, 1, 1)
- C. (1, 3, -2)
- D. (-1, 3, 2)
Answer: C. (1, 3, -2)
Explanation: Equation of the line passing through P and Q is $\frac{x-2}{1-2}=\frac{y-1}{3-1}=\frac{z-1}{-2-1} \Rightarrow \frac{x-2}{-1}=\frac{y-1}{2}=\frac{z-1}{-3}=k$. (This can be rewritten as $\frac{x-2}{1}=\frac{y-1}{-2}=\frac{z-1}{3}=k$). Any point on this line is $x=2+k$, $y=1-2k$, $z=1+3k$. This point must lie on the plane $x+3y+2z-6=0$. Substituting the coordinates into the plane equation: $(2+k)+3(1-2k)+2(1+3k)-6=0 \Rightarrow 2+k+3-6k+2+6k-6=0 \Rightarrow k+1=0 \Rightarrow k=-1$. Substituting $k=-1$ into the point coordinates yields: $(2-1, 1-2(-1), 1+3(-1)) = (1, 3, -2)$.