Question

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.

Answer 1

It is proved that the image of the point (3,−2, 1) in the plane 3x−y+4z=2 lies on the plane.

Step-by-step explanation:

We are given the equation of plane as:

• 3x-y+4z=2           ---(1)

• Let image of point P(3,−2,1) is P′. We assume the coordinate of P' as $(x_1, y_1, z_1)$

• Now, we can understand that PP' is perpendicular to plane $3x-y+4z=2.$

Now, we assume M is the mid point of line PP'. it would lie on the plane.

Therefore, we can write the equation of line PP' as:

• $\dfrac{x-3}{3} = \dfrac{y+2}{-1} = \dfrac{z-1}{4} = k$

• Any point that lies on this line is:

                         (3k+3, -k-2, 4k+1)

We can understand that this point also lies on the plane.

•              $3(3k+3)-(-k-2)+4(4k+1)=2$

    So, we get k = 2

Now, we put the value of k to find out the coordinates of M.

• $(3(\dfrac{-1}{2})+3, \dfrac{1}{2}-2, \ 4 \ .\dfrac{-1}{2} +1)$

        = $(\dfrac{3}{2}, \dfrac{-3}{2} ,-1)$

We know that M is the mid point of PP'. So, we can write as;

• $\dfrac{3+x_1}{2} = \dfrac{3}{2}$

$x_1 = 0$

• $\dfrac{-2+y_1}{2} = \dfrac{-3}{2}$

$y_1 = -1$

• $\dfrac{1+z_1}{2} = -1$

 $z_1 = -3$

Therefore, we have got the coordinates of M as (0,-1,-3). These are the values of x, y and z. Put these values of x, y, z in the equation of plane x+y+z+4=0.

We get,

       0-1-3+4 = 0

Hence, it is proved that the image of the point (3,−2, 1) in the plane 3x-y+4z=2 lies on the plane.

Answer 2

Answer:

The answer to the question happens to be the solution