Question

Reflection of the line $\frac{x-1}{-1}=\frac{y-2}{3}=\frac{z-4}{1}$ in the plane x +y +z =7 is :
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• A. $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-4}{1}$
• B. $\frac{x-1}{-3}=\frac{y-2}{-1}=\frac{z-4}{1}$
• C. $\frac{x-1}{-3}=\frac{y-2}{1}=\frac{z-4}{-1}$
• D. $\frac{x+1}{3}=\frac{y-2}{1}=\frac{z+4}{1}$

Answer 1

The correct option is C $\frac{x-1}{-3}=\frac{y-2}{1}=\frac{z-4}{-1}$

Given line passes through the point A(1, 2, 4) and this point also lies in the plane. To find the reflection of the line, we need one more point of the line. Clearly P(0, 5, 5) also lies in the line.
Let $Q(\alpha ,\beta ,\gamma )$be the reflection of P in the plane x + y + z = 7
Then $\frac{\alpha }{2}=\frac{\beta +5}{2}=\frac{\gamma +5}{2}=7\Rightarrow \alpha +\beta +\gamma =4$
Also $PQ\perp$ to the plane, i.e., parallel to the normal of the plane.
$\therefore \frac{\alpha -0}{1}=\frac{\beta -5}{1}=\frac{\gamma -5}{1}=\lambda \Rightarrow \alpha =\lambda ,\beta =\lambda +5,\gamma =\lambda +5\therefore \lambda +\lambda +5\lambda +5=4\Rightarrow \lambda =-2$
$\therefore$ Q is (-2,3,3)
Now AQ will be the reflection of the given line, equation of AQ is
$\frac{x-1}{-3}=\frac{y-2}{1}=\frac{z-4}{-1}$