- #1

mathlabrat

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## Homework Statement

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Find points P,Q which are closest possible with P lying on the line

x=8+1t

y=8+1t

z=7−3t

and Q lying on the line

x=231−6t

y=−10−17t

z=71−13t

**3. Attempt at solution**

Hi,

I am at loss as to how to do this.

I know that from the equations I can get point (8,8,7) and direction vector (1,1,-3) for the first line and similar for the second line. Then I found the normal to both lines as the shortest distance points should form a perpendicular.

I don't know how to proceed from there.

Any help will be appreciated

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