Let S be the mirror image of the point Q(1,3, | vedclass

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(A) Since $R(3,5,\gamma)$ lies on the plane $2x-y+z+3=0$,we have:$2(3) - 5 + \gamma + 3 = 0$$6 - 5 + \gamma + 3 = 0$$4 + \gamma = 0 \Rightarrow \gamma

Vedclass · Accepted Answer

$72$

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(A) Since $R(3,5,\gamma)$ lies on the plane $2x-y+z+3=0$,we have:$2(3) - 5 + \gamma + 3 = 0$$6 - 5 + \gamma + 3 = 0$$4 + \gamma = 0 \Rightarrow \gamma = -4$.Thus,$R$ is $(3,5,-4)$.The normal vector to the plane is $\vec{n} = (2, -1, 1)$. The line $QS$ passes through $Q(1,3,4)$ and is parallel to $\vec{n}$.The equation of line $QS$ is $\frac{x-1}{2} = \frac{y-3}{-1} = \frac{z-4}{1} = \lambda$.Any point on this line is $F(2\lambda+1, -\lambda+3, \lambda+4)$.Since $F$ is the foot of the perpendicular from $Q$ to the plane,it lies on the plane:$2(2\lambda+1) - (-\lambda+3) + (\lambda+4) + 3 = 0$$4\lambda + 2 + \lambda - 3 + \lambda + 4 + 3 = 0$$6\lambda + 6 = 0 \Rightarrow \lambda = -1$.Substituting $\lambda = -1$ into $F$,we get $F(-1, 4, 3)$.Since $F$ is the midpoint of $QS$,let $S = (x_s, y_s, z_s)$:$\frac{x_s+1}{2} = -1 \Rightarrow x_s = -3$$\frac{y_s+3}{2} = 4 \Rightarrow y_s = 5$$\frac{z_s+4}{2} = 3 \Rightarrow z_s = 2$.So,$S = (-3, 5, 2)$.The square of the length $SR$ is:$SR^2 = (3 - (-3))^2 + (5 - 5)^2 + (-4 - 2)^2$$SR^2 = (6)^2 + (0)^2 + (-6)^2 = 36 + 0 + 36 = 72$.

Let $S$ be the mirror image of the point $Q(1,3,4)$ with respect to the plane $2x-y+z+3=0$ and let $R(3,5,\gamma)$ be a point on this plane. Then the square of the length of the line segment $SR$ is ..... .

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