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

Concept: Reflection of point about plane \[ ax+by+cz+d=0 \] is given by \[ P'=P-\frac{2(ax_0+by_0+cz_0+d)}{a^2+b^2+c^2}(a,b,c) \]

Step 1: Write plane in standard form.
\[ x-2y+3z-4=0 \] Thus \[ a=1,\qquad b=-2,\qquad c=3,\qquad d=-4 \] Point \[ P=(1,-1,1) \]

Step 2: Substitute into numerator.
\[ ax_0+by_0+cz_0+d = 1+2+3-4 \] \[ =2 \] Denominator \[ a^2+b^2+c^2 = 1+4+9 = 14 \]

Step 3: Find reflected point.
\[ P' = (1,-1,1)-\frac{4}{14}(1,-2,3) \] \[ = (1,-1,1)-\frac27(1,-2,3) \] \[ = \left( 1-\frac27, -1+\frac47, 1-\frac67 \right) \] \[ = \left( \frac57,-\frac37,\frac17 \right) \]

Step 4: Compute required expression.
\[ x_1-y_1-z_1 = \frac57-\left(-\frac37\right)-\frac17 \] \[ = \frac57+\frac37-\frac17 = \frac77 = 1 \] Hence \[ \boxed{1} \]