Wow, so it is late and I can't think! How do I put this into scalar form: r=(6,0,1) + t(3,1,2)? thanks!
3 answers
wait actually I don't think you can do that...but then I have a problem. I am supposed to find the distance of the point to the line or something like that of the point (1,-5,2) and the line r=(6,0,1) + t(3,1,2) but I thought you needed the scalar equation in order to find the distance because of the equation! Wow, I'm confused!
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.
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.
One way to do it would be to write an equation for the distance from the point(1,-5,2) to any point on the line in terms of t, and then find the minimum distance by setting the derivatve with repsect to t equal to zero.
0 answers