Find perpendicular vectors and angles

1. Nov 28, 2008

thomas49th

1. The problem statement, all variables and given/known data
I'm awful with vectors. I know the dot product rule and the uses to find perpendicular vectors and angles. I know about multiplying vectors (a1b1 + a2b2 + a3b3). I know vector arithmetic.

But i cant do:

r = 8i + 12j + 14k + t(i + j - k)

where t is a parameter

The point A has co-ordinates (4,8,a), where a is a constant. The point B has co-ordinates (b,13,13), where b is a constant. Points A and B lie on the line l1.

a) Find the values a and b

2. Relevant equations

3. The attempt at a solution

Well i presume i need to find the parameter t first... how? I dont have enough information do i?

thanks :)

2. Nov 28, 2008

Staff: Mentor

Can't do what?

What line? And what does this have to do with r above?

3. Nov 28, 2008

thomas49th

sorry the question is

The line l1 has vector equation:

r = 8i + 12j + 14k + t(i + j - k)

where t is a parameter

The point A has co-ordinates (4,8,a), where a is a constant. The point B has co-ordinates (b,13,13), where b is a constant. Points A and B lie on the line l1.

a) Find the values a and b

Thanks :)

4. Nov 28, 2008

Staff: Mentor

My bad. (I realized what you meant shortly after I posted.)

Think of it this way: All points on that line must satisfy the parametric equations:
x = 8 + t
y = 12 + t
z = 14 - t
(where t is just a parameter that takes on all real values.)

Since points A and B are on that line, they must satisfy those three equations. See what you can deduce from that. Hint: Start by finding the value of the parameter t for each point. (You have all the information needed.)

5. Nov 29, 2008

thomas49th

4 = 8 + t
8 = 12 + t
a = 14 + t

b = 8 + t
13 = 12 + t
13 = 14 + t

point a t = -4
point b t = 9

a = 18, b = 9

Okay cheeers :)

Next question:

Given that point O is the origin, and that the point P lies on l1 such that OP is perpendicular to l1

b) Find the coordinates of P

i've sketched a graph of it. using the dot product rule, a.b =0 (0 as it's cos(90))

so does that mean vector a x vector b = 0?

Where exactly now?

Thanks :)

6. Nov 30, 2008

Staff: Mentor

Sorry for the delay--I forgot that you had a second question.

No. The cross product will not be zero, only the dot product.

Call the point P (x, y, z). Thus the vector OP is (x, y, z). That point must satisfy two conditions:
(1) The vector OP must be perpendicular to the line, thus its dot product with a vector parallel to the line must be zero. (Find a vector parallel to the line--any one will do.)
(2) The point P must be on line, thus satisfy the same parametric equations from before. (Solve for t.)