# Is this all correct? (more bl dy vectors)

• Dr Zoidburg
I can't imagine why I wrote "4/3". I guess I just saw "1/3" and "2/3" and thought "3/3"! Thank you.
Dr Zoidburg
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)?

## 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.

Dr Zoidburg said:
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.
Yes, it really is that simple!

(b) parametric equations:
x=2t-3
y=t
z=t-1
I presume you just set (x+3)/2 = y = z-1= t. Yes, that's correct.

(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!
u= (2 1 2) is the "position vector" of Q but where did you get (-3 0 1)? It is NOT the direction vector of P, which is (2, 1, 1). It is the position vector of a point on the line (when t= 0) but that is pretty much an arbitrary vector. In fact, it is easy to check that you "R" does not even satisfy the equations to lie on P!

I would have done this differently. Let R be the point (x, y, z)= (2t- 3, t, t-1), on P. Then the vector from Q to R is given by (2t-3- 2, t- 1, t-1- 2)= (2t- 5, t-1, t- 3). The line from R to Q will be perpendicular to P as long as that vector is perpendicular to the direction vector of P: we must have (2t-5, t-1, t-3)$\cdot$(2, 1, 1)= 0. That gives 4t- 10+ t- 1+ t- 3= 0 or 6t= 14. t= 14/6= 7/3. Putting that back into the equations for R, R= (5/3, 7/3, 4/3).

As a check on the arithmetic, the vector from Q to R is (4/3- 2, 7/3- 1, 4/3- 2)= (-1/3, 4/3, -2/3). It's dot product with (2, 1, 1) is -2/3+ 4/3- 2/3= 0.

many thanks in advance! I'll be checking again in the am.

thanks a lot for that.
I have this problem of always thinking the solution has to be more complex than it appears to be!
I really must go out and buy another Occam's razor. Mine's obviously too blunt and ain't working as well as it should.

HallsofIvy said:
Putting that back into the equations for R, R= (5/3, 7/3, 4/3).

As a check on the arithmetic, the vector from Q to R is (4/3- 2, 7/3- 1, 4/3- 2)= (-1/3, 4/3, -2/3). It's dot product with (2, 1, 1) is -2/3+ 4/3- 2/3= 0.
very minor quibble here: shouldn't that 4/3 be 5/3 ?

Yes, of course it should.

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