Posted by **Derek** on Thursday, July 10, 2008 at 10:33am.

this can be done by projections of two vector.

recall that the scalar projection of vector b on vector a is a∙b/│a│

so let's find a point on the give line,

e.g. the point B(6,0,1) (I let t=0)

draw a perpendicular from your given point P(1,-5,2) to meet the line at Q

So PQB forms a righ-angled triangle with │vector BP│ the hypotenuse.

vector BP, lets call it vector b, is (5,5,-1) and vector a is (3,1,2), (from your given line)

so the projection of vector b on vector a

= (5,5,-1)∙(3,1,2)/│(3,1,2)│

= (15+5-2)/√(9+1+4)

= 18/√14

also │b│ = √(25+25+1) = √51

You now have 2 sides of a right-angled triangle, with your required distance as the third side.

I will let you finish the arithmetic.

What do you with the triangle? How do you know which side is which, are you finding the hypotenuse, or just a side? This changes the math done. Like a^2+b^2=c^2, but if you are finding a side then it would be c^2-a^2=b^2...so can you please explain, thanks.

- Calc. for Reiny -
**Reiny**, Thursday, July 10, 2008 at 9:22pm
Since you posted my reply, I will use my letters

I found │vector BP│ to be √51 , the hypotenuse

I found │vector bQ│ to be 18/√14

You want │vector PQ│

so (PQ)^2 + (BQ)^2 = (BP)^2

(PQ)^2 = (BP)^2 - (BQ)^2

= 51 - 324/14

= 390/14

BP = √(390/14)

= 5.278

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