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Question

Find the distance of the point (-2,3,-4) from the line (x+2)3=(2y+3)4=(3z+4)5 measured parallel to the plane 4x+12y-3z+1=0


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Solution

Step 1:Find the general co-ordinates of a point on the given line

Let (x+2)3=(2y+3)4=(3z+4)5=k (say)

x=3k-2,y=4k-32,z=5k-43………………………(1)

Let us considerP=(-2,3,-4).

Let πbe the plane parallel to the given plane throughP

let Q be the point where the given line intersectsπ.

Step 2:Solve for the equation of π

We know that πis parallel to the given plane so the equation of π which differs only in the constant term.

4x+12y3z+c=0

Where c is a constant

Since point P lies on the plane π, its co-ordinates satisfy the equation of the plane

4(-2)+12(3)3(-4)+c=0

-8+36+12+c=0

c=-40

So the equation forπ is

4x+12y3z40=0……………………..(2)

Step 3: Solve for point Q which lies on the plane and the given line

General co-ordinates of a point on the given line are

x=3k-2,y=4k-32,z=5k-43 …[From(1)]

These co-ordinates satisfy the equation of plane π

Hence, substituting the values of x,y,z from (1) in (2) we get

43k-2+124k-32-35k-43-40=0

12k-8+24k-18-5k+4-40=0

k=2

Resubstituting the value of k we get

Q=32-2,42-32,52-43

Q=4,52,2

Step 4: Find the length ofPQ:

We know P=(-2,3,-4) and Q=4,52,2

Length of PQ=-2-42+3-522+-4-22

=-62+122+-62

=36+14+36

=2894

Length of PQ=172

The distance between coordinates(4,52,2) and (2,3,4) is 172 unit.

Hence, the distance between the point (2,3,4) and the given line passing parallel to the given plane is 172 unit


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