Question

Find the image of the point (-1,3,4) in the plane x - 2y = 0.​

Answer 1

Let the point (x, y, z) be the foot of the perpendicular drawn from the point (-1, 3, 4) to the plane $x-2y=0.$ So the vector $\left<x+1,\ y-3,\ z-4\right>$ is perpendicular to the plane.

From the equation of the plane, $x-2y=0,$ it is clear that the vector $\left<1,\ -2,\ 0\right>$ is perpendicular to the plane.

[Equation of a plane perpendicular to the vector $\left<a,\ b,\ c\right>$ is $ax+by+cz=d.$]

The vectors $\left<x+1,\ y-3,\ z-4\right>$ and $\left<1,\ -2,\ 0\right>$ each are perpendicular to the plane, that means they're parallel to each other, i.e.,

$\longrightarrow\left<x+1,\ y-3,\ z-4\right>=\lambda\left<1,\ -2,\ 0\right>$

for some non - zero real number $\lambda.$

Equating corresponding components we get,

• $x=\lambda-1$

• $y=3-2\lambda$

• $z=4$

Putting these values in the plane equation,

$\longrightarrow (\lambda-1)-2(3-2\lambda)=0$

$\longrightarrow5\lambda-7=0$

$\longrightarrow\lambda=\dfrac{7}{5}$

Then,

• $x=\dfrac{2}{5}$

• $y=\dfrac{1}{5}$

• $z=4$

So the foot is $\left(\dfrac{2}{5},\ \dfrac{1}{5},\ 4\right)$ and the perpendicular vector from the point (-1, 3, 4) to the plane is $\left<\dfrac{7}{5},\ -\dfrac{14}{5},\ 0\right>.$

Now the position vector of the image will be,

$\longrightarrow\vec{r}=\left<\dfrac{2}{5},\ \dfrac{1}{5},\ 4\right>+\left<\dfrac{7}{5},\ -\dfrac{14}{5},\ 0\right>$

$\longrightarrow\vec{r}=\left<\dfrac{9}{5},\ -\dfrac{13}{5},\ 4\right>$

Hence the image is $\bf{\left(\dfrac{9}{5},\ -\dfrac{13}{5},\ 4\right)}.$

Answer 2

Answer:

Find the image of the point (-1,3,4) in the plane x - 2y = 0.

THE ABOVE ANS IS CORRECT ⬆️⬆️

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