Answer :

Given :

Equation of plane : 3x + 2y + 2z + 5 = 0

Equation of line :

Point : P = (2, 3, 4)

To Find : Distance of point P from the given plane parallel to the given line.

Formula :

1) Equation of line :

Equation of line passing through A = (x_{1}, y_{1}, z_{1}) & having direction ratios (a, b, c) is

2) Distance formula :

The distance between two points A = (a_{1}, a_{2}, a_{3}) & B = (b_{1}, b_{2}, b_{3}) is

For the given line,

Direction ratios are (a, b, c) = (3, 6, 2)

Let Q be the point on the plane such that is parallel to the given line.

Therefore direction ratios of given line and line PQ will be same.

Therefore equation of line PQ with point P = (2, 3, 4) and with direction ratios (3, 6, 2) is

Let co-ordinates of Q be (u, v, w)

As point Q lies on the line PQ, we can write,

⇒u = 3k+2, v = 6k+3, w = 2k+4 ………(1)

Also point Q lies on the plane

3u + 2v + 2w = -5

⇒3(3k+2) + 2(6k+3) + 2(2k+4) = -5 …..from (1)

⇒9k + 6 + 12k + 6 + 4k + 8 = -5

⇒25k = -25

⇒k = -1

,

Therefore, co-ordinates of point Q are

Q =

Now distance between points P and Q by distance formula is

= 7

Therefore distance of point P from the given plane measured parallel to the given line is

d = 7 units

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