The image point of (1, 3, 4) in the plane 2x | vedclass

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(A) Let $Q$ be the image of the point $P(1, 3, 4)$ in the given plane $2x - y + z + 3 = 0$. The line $PQ$ is normal to the plane.The direction ratios

Vedclass · Accepted Answer

$(-3, 5, 2)$

Vedclass · Answer

(A) Let $Q$ be the image of the point $P(1, 3, 4)$ in the given plane $2x - y + z + 3 = 0$. The line $PQ$ is normal to the plane.The direction ratios of the normal to the plane are $2, -1, 1$.Since $PQ$ passes through $P(1, 3, 4)$ and is parallel to the normal vector,the equation of the line $PQ$ is $\frac{x - 1}{2} = \frac{y - 3}{-1} = \frac{z - 4}{1} = r$.Any point on this line is given by $(2r + 1, -r + 3, r + 4)$. Let this be $Q$.The midpoint $R$ of $PQ$ is $\left( \frac{2r + 1 + 1}{2}, \frac{-r + 3 + 3}{2}, \frac{r + 4 + 4}{2} \right) = \left( r + 1, \frac{-r + 6}{2}, \frac{r + 8}{2} \right)$.Since $R$ lies on the plane $2x - y + z + 3 = 0$,we substitute the coordinates of $R$ into the plane equation:$2(r + 1) - \left( \frac{-r + 6}{2} \right) + \left( \frac{r + 8}{2} \right) + 3 = 0$Multiplying by $2$:$4(r + 1) - (-r + 6) + (r + 8) + 6 = 0$$4r + 4 + r - 6 + r + 8 + 6 = 0$$6r + 12 = 0 \implies r = -2$.Substituting $r = -2$ into the coordinates of $Q$:$x = 2(-2) + 1 = -3$$y = -(-2) + 3 = 5$$z = -2 + 4 = 2$Thus,the image point is $(-3, 5, 2)$.

The image point of $(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$ is

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