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Vector Distance QUestion

From a past paper. Below are the answers I have already found

Equation of place ABC
2x+2y+z=2

Vector normal to that (2,2,1)

The vector equation of perpendicular Q from point D to the plane ABC
r = (-2,0,0) + LAMBDA (2,2,1)

Where l meets the plane (-2/3, 4/3, 2/3) During this I found lambda = 2/3

I am stuck on the very last part of the question which is, find the distance D to the plane ABC

Thanks

Reply 1

ghostwalker

Original post by Axion

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I presume (-2,0,0) is D. You have the point where it meets the plane.

So what's the vector from D to the point where it meets the plane?

And it's length is?

Reply 2

Axion

Original post by ghostwalker

I presume (-2,0,0) is D. You have the point where it meets the plane.

So what's the vector from D to the point where it meets the plane?

And it's length is?

Legend. Thanks.

Square root of [(-2-x)sq + (-y)sq + (-z)sq]

where x, y and z are the respective coordinates where the line meets the plane

Comes up as 2 on the calc

Thanks!