ttpp1124
- 110
- 4
- Homework Statement
- Calculus and Vectors - Vector and Parametric Equations
I solved it, can someone concur?
- Relevant Equations
- Not Available
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ttpp1124 said:When I sub in 0 for t, I get the values corresponding to the direction vector. That's how I know it's correct
how is it wrong?PeroK said:Looks wrong to me!
Is that ##x = -2t, \ y = -1 + 3t, \ z = 6 + 5t##?ttpp1124 said:how is it wrong?
YesPeroK said:Is that ##x = -2t, \ y = -1 + 3t, \ z = 6 + 5t##?
That's not right. What values of ##t## give you ##P## and ##Q##?ttpp1124 said:Yes
Is it the direction vector, 0,-1,6?PeroK said:That's not right. What values of ##t## give you ##P## and ##Q##?
The vector equation is correct. That's the direction ##\vec{QP}##.ttpp1124 said:Is it the direction vector, 0,-1,6?
Right, so I think I didn't multiply the t value and do vector addition. When I do multiply, I get r = (-2,3,5)+(0t,-t,6t).PeroK said:The vector equation is correct. That's the direction ##\vec{QP}##.
So, what value of ##t## gives you ##Q##?ttpp1124 said:Right, so I think I didn't multiply the t value and do vector addition. When I do multiply, I get r = (-2,3,5)+(0t,-t,6t).
Then I do vector addition by components, so then my parametric equations should be -2+0t, 3-t, and 5+6t
-1PeroK said:So, what value of ##t## gives you ##Q##?
My question is, how does the value of t have anything to do with the actual question? They just wanted to know the vector and parametric equations.ttpp1124 said:-1
ttpp1124 said:My question is, how does the value of t have anything to do with the actual question? They just wanted to know the vector and parametric equations.
Also, are my new parametric equations correct?