Question

The coordinates of the point where the line $\dfrac{x-1}{2}=\dfrac{y-2}{-3}=\dfrac{z+3}{4}$ meets the plane $2x+4y-z=1$ are

• A. $(3,-1,-1)$
• B. $(3,-1,1)$
• C. $(3,1,-1)$
• D. $(-2,1,-1)$

Hint:

Always convert symmetric equations of a line into parametric form before substitution.

Answer 1

The Correct Option is B

Step 1: Writing the parametric form of the line.
Let \[ \frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}=t \] Then \[ x=1+2t,\quad y=2-3t,\quad z=-3+4t \]
Step 2: Substituting in the plane equation.
\[ 2(1+2t)+4(2-3t)-(-3+4t)=1 \] \[ 2+4t+8-12t+3-4t=1 \] \[ 13-12t=1 \Rightarrow t=1 \]
Step 3: Finding the point of intersection.
\[ x=3,\quad y=-1,\quad z=1 \]
Step 4: Conclusion.
The point of intersection is $(3,-1,1)$.