Question

The reflection of the points (3, 1, 2) in the plane
x + 2y + z = 1 is:​

Answer 1

Answer:

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Step-by-step explanation:

please

Answer 2

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

Step-by-step explanation:

Given:

The equation of the plane

x + 2y + z = 1

And the point (3, 1, 2)

To find out:

The reflection of point (3, 1, 2) in the plane

Solution:

The equation of plane is : x + 2y + z = 1

The vector normal to the plane will be (1, 2, 1)

Now a  line perpendicular to the given plane and passing through the point (3, 1, 2) can be written as

$\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{1}$

Let $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{1}=\lambda$

Any point lying on the line will be $(\lambda+3, 2\lambda+1, \lambda+2)$

To find the point where this line meets the plane:

The point will satisfy the equation of the plane

Therefore,

$\lambda+3+2(2\lambda+1)+\lambda+2=1$

$6\lambda+7=1$

$\implies \lambda=-\frac{6}{6}=-1$

Therefore, the coordinate of the point where line meets the plane is

$(2, -1, 1)$

This point will be the mid point of the given point (3, 1, 2) and the reflection point

If the coordinate of reflection points are (x', y', z') then

$\frac{x'+3}{2}=2\implies x'=1$

$\frac{y'+1}{2}=-1\implies y'=-3$

$\frac{z'+2}{2}=1\implies z'=0$

Therefore, the reflection of the point (3, 1, 2) is (1, -3, 0)

Hope this answer is helpful.

Know More:

Q: Prove that the image of the point (3,-2,1) in the plane 3x−y+4z=2 lies on the plane x+y+z+4=0.

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