Question

If the image of point \((1,-1,1)\) in the plane \[ x-2y+3z=4 \] is \((x_1,y_1,z_1)\), then \(x_1-y_1-z_1=\)

• A. \(0\)
• B. \(1\)
• C. \(4\)
• D. \(3\)

Hint:

Reflection about a plane can be solved directly by vector formula instead of finding foot of perpendicular separately.

Answer 1

The Correct Option is B

Step 1: Reflection formula.
The image of $P(x_0,y_0,z_0)$ in the plane $ax+by+cz+d=0$ is \[ P'=P-\frac{2(ax_0+by_0+cz_0+d)}{a^2+b^2+c^2}(a,b,c). \]
Step 2: Standard form.
Write $x-2y+3z=4$ as $x-2y+3z-4=0$, so $a=1$, $b=-2$, $c=3$, $d=-4$, and $P=(1,-1,1)$.
Step 3: Numerator.
$ax_0+by_0+cz_0+d=1(1)-2(-1)+3(1)-4=1+2+3-4=2$.
Step 4: Denominator.
$a^2+b^2+c^2=1+4+9=14$, so the scalar factor is $\dfrac{2(2)}{14}=\dfrac{4}{14}=\dfrac27$.
Step 5: Compute the image.
\[ P'=(1,-1,1)-\tfrac27(1,-2,3)=\left(1-\tfrac27,\ -1+\tfrac47,\ 1-\tfrac67\right)=\left(\tfrac57,-\tfrac37,\tfrac17\right). \]
Step 6: Required expression.
$x_1-y_1-z_1=\tfrac57-\left(-\tfrac37\right)-\tfrac17=\tfrac{5+3-1}{7}=\tfrac77=1$, option (2).
\[ \boxed{x_1-y_1-z_1=1\ \text{(option 2)}} \]