- #1
Dr Zoidburg
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Is this all correct? (more bl**dy vectors!)
I hope these are right, but I have the nagging suspicion that they're not? Where have I gone wrong (if I have indeed gone wrong)?
line P has equation (x+3)/2 = y = z-1
(a) Show that point Q (2,1,2) does not lie on line P.
(b) Write down the parametric equations for the line.
(c) Use dot product properties to find the co-ords of point R which is on P, such that QR is perpendicular to P.
(a) sub (2,1,2) into the equation:
(2+3)/2 = 1 = 2-1
5/2 = 1 = 1
they don't equal, therefore point Q does not lie on line P.
Is it that simple, or do I need to use [v(P) = (-3,0,1) + t(2,1,1)] and the parametric equations such that x=2t-3=2, y=t=1 & z=t+1=2 and attempt a solution. Here it is inconsistent.
(b) parametric equations:
x=2t-3
y=t
z=t-1
(c) dot product:
let u = (2,1,2) and v=(-3,0,1)
u.v = (2 1 2).(-3 0 1) |i 2 -3 |
= det |j 1 0 | = i - 8j + 3k
|k 2 1 |
point R = (1, -8, 3).
again, I feel I've skipped a step here, but it's 11.30pm and I'm too tired to think any further!
many thanks in advance! I'll be checking again in the am.
I hope these are right, but I have the nagging suspicion that they're not? Where have I gone wrong (if I have indeed gone wrong)?
Homework Statement
line P has equation (x+3)/2 = y = z-1
(a) Show that point Q (2,1,2) does not lie on line P.
(b) Write down the parametric equations for the line.
(c) Use dot product properties to find the co-ords of point R which is on P, such that QR is perpendicular to P.
The Attempt at a Solution
(a) sub (2,1,2) into the equation:
(2+3)/2 = 1 = 2-1
5/2 = 1 = 1
they don't equal, therefore point Q does not lie on line P.
Is it that simple, or do I need to use [v(P) = (-3,0,1) + t(2,1,1)] and the parametric equations such that x=2t-3=2, y=t=1 & z=t+1=2 and attempt a solution. Here it is inconsistent.
(b) parametric equations:
x=2t-3
y=t
z=t-1
(c) dot product:
let u = (2,1,2) and v=(-3,0,1)
u.v = (2 1 2).(-3 0 1) |i 2 -3 |
= det |j 1 0 | = i - 8j + 3k
|k 2 1 |
point R = (1, -8, 3).
again, I feel I've skipped a step here, but it's 11.30pm and I'm too tired to think any further!
many thanks in advance! I'll be checking again in the am.