Q. 265.0( 3 Votes )

# Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3, 2, 1) from the plane 2x - y + z + 1 = 0. Find also, the image of the point in the plane.

Answer :

**Given****,** Equation of the plane is given, 2x - y + z + 1 = 0

**To Find:** Find the image of the point in the plane

**Explanation:** We have the equation of the plane is given

2x - y + z + 1 = 0 - - - (i)

And, A coordinate is given on Point P (3, 2, 1)

Let us consider a point on the foot of the perpendicular, and the coordinates of this point are

Q(p, q, r)

So, The distance of the line PQ is (p - 3 , q - 2 , r - 1)

And, The distance of the line PQ which is perpendicular to the given plane will be

By solving this, we get

p = 2k + 3

q = - k + 2

r = k + 1

Now, Q(p, q, r) lies on the plane, then

2p - q + r + 1 = 0

Put the value of p, q and r we get

2(2k + 3) - ( - k + 2) + (k + 1) + 1 = 0

6k + 6 = 0

6k = - 6

K = - 1

Since, Foot of the perpendicular from P is given by Q(1, 3, 0)

Now, Let us consider R(a, b, c) be the image of point P

So, Q is the midpoint of PR

Using midpoint formula, we get

(a, b, c) = (2×1 - 3, 2×3 - 2, 2×0 - 1)

(a, b, c) = ( - 1, 4, - 1)

And, The perpendicular distance PQ is given by

PQ =

PQ =

PQ =

**Hence, The distance of the mirror image of a point is**

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