The image of the point (-1, 3, 4) in the plan | vedclass

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(D) Let the image of the point $P(-1, 3, 4)$ in the plane $x - 2y = 0$ be $P'(\alpha, \beta, \gamma)$.The line passing through $P$ and $P'$ is perpend

Vedclass · Accepted Answer

None of these

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(D) Let the image of the point $P(-1, 3, 4)$ in the plane $x - 2y = 0$ be $P'(\alpha, \beta, \gamma)$.The line passing through $P$ and $P'$ is perpendicular to the plane $x - 2y = 0$. The normal vector to the plane is $\vec{n} = (1, -2, 0)$.The equation of the line passing through $P(-1, 3, 4)$ with direction ratios $(1, -2, 0)$ is $\frac{x + 1}{1} = \frac{y - 3}{-2} = \frac{z - 4}{0} = k$.Thus,any point on this line is $(k - 1, -2k + 3, 4)$.The midpoint of $PP'$ is $M = (\frac{\alpha - 1}{2}, \frac{\beta + 3}{2}, \frac{\gamma + 4}{2})$.Since $M$ lies on the plane $x - 2y = 0$,we have $\frac{\alpha - 1}{2} - 2(\frac{\beta + 3}{2}) = 0$,which simplifies to $\alpha - 1 - 2\beta - 6 = 0$,or $\alpha - 2\beta = 7$.Also,the line $PP'$ is perpendicular to the plane,so the direction ratios of $PP'$ are proportional to the normal of the plane: $\frac{\alpha - (-1)}{1} = \frac{\beta - 3}{-2} = \frac{\gamma - 4}{0} = \lambda$.From $\frac{\gamma - 4}{0} = \lambda$,we get $\gamma = 4$.From $\alpha + 1 = \lambda$ and $\beta - 3 = -2\lambda$,we have $\alpha = \lambda - 1$ and $\beta = -2\lambda + 3$.Substituting into $\alpha - 2\beta = 7$: $(\lambda - 1) - 2(-2\lambda + 3) = 7 \Rightarrow \lambda - 1 + 4\lambda - 6 = 7 \Rightarrow 5\lambda = 14 \Rightarrow \lambda = \frac{14}{5}$.Then $\alpha = \frac{14}{5} - 1 = \frac{9}{5}$ and $\beta = -2(\frac{14}{5}) + 3 = -\frac{28}{5} + \frac{15}{5} = -\frac{13}{5}$.The image is $(\frac{9}{5}, -\frac{13}{5}, 4)$. Since this is not among the options,the correct choice is $D$.

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

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