# vectors - finding coordinates

• Feb 20th 2010, 07:50 AM
stealthmaths
vectors - finding coordinates
Hi:
I have this problem:
The vector equation of the line Q is 2i+j+3k+λ(i+3j-5k) and the point P has coordinates (2, 3, -1). Find the coordinates of the point R, where R is on the line Q and the vector PR is perpendicular to Q.

So:
Q=2i+j+3k+λ(i+3j-5k)
P=(2, 3, -1)
R=( )

If R is on the line Q, and PR is perpendicular to Q, doesn't this mean that R is at the point of intersection between Q and PR?
• Feb 20th 2010, 10:36 AM
Hello stealthmaths
Quote:

Originally Posted by stealthmaths
Hi:
I have this problem:
The vector equation of the line Q is 2i+j+3k+λ(i+3j-5k) and the point P has coordinates (2, 3, -1). Find the coordinates of the point R, where R is on the line Q and the vector PR is perpendicular to Q.

So:
Q=2i+j+3k+λ(i+3j-5k)
P=(2, 3, -1)
R=( )

If R is on the line Q, and PR is perpendicular to Q, doesn't this mean that R is at the point of intersection between Q and PR?

Yes, but that doesn't help much. You need to say:
$\displaystyle \vec{PR} = \vec r - \vec p$
$\displaystyle = \begin{pmatrix}2+\lambda\\1+3\lambda\\3-5\lambda\end{pmatrix} - \begin{pmatrix}2\\3\\-1\end{pmatrix}$

$\displaystyle = \begin{pmatrix}\lambda\\-2+3\lambda\\4-5\lambda\end{pmatrix}$
The direction of the line Q is the vector $\displaystyle \begin{pmatrix}1\\3\\-5\end{pmatrix}$. Now use the fact that the scalar product of this vector with $\displaystyle \vec{PR}$ is zero if the two vectors are perpendicular, to find the value of $\displaystyle \lambda$ at R.

Can you complete it now?

• Feb 20th 2010, 11:40 AM
stealthmaths
So: multiplying the vector Q with coordinates of P = vector PR
which is perpendicular to QP

So, if I find the point of intersection of PR and Q - this is the coordinates of R

Hopefully that's on the button:

To find the points of intersection, do I need to solve the position vectors of these lines simultaneously?

not sure how to do that with vectors
• Feb 20th 2010, 12:57 PM
Hello stealthmaths
Quote:

Originally Posted by stealthmaths
So: multiplying the vector Q with coordinates of P = vector PR
which is perpendicular to QP

So, if I find the point of intersection of PR and Q - this is the coordinates of R

Hopefully that's on the button:

To find the points of intersection, do I need to solve the position vectors of these lines simultaneously?

not sure how to do that with vectors

It sounds as if you haven't covered the basic things you need to know to solve this problem. You'll find some information about scalar (dot) products just here.

• Feb 20th 2010, 03:18 PM
stealthmaths
I'm aware of and can calculate the dot product. And that u.v=0 when two lines are perpendicular.
But I think I'm having trouble visualizing this problem. It just seems a little abstract to me.
We also use a different notation from you. We don't use the coordinates in bracket/parenthesis. I think this may also be throwing me a bit.
But to my knowledge, if something multiplied by something is equal to zero, then one of them must be zero
• Feb 20th 2010, 04:51 PM
stealthmaths
I think I may have finally got your message:
Q.PR=(1+3-5)(1+3-5)=1+9+25
=35

|Q|=√1²+3²+5²=√35
|PR|=√1²+3²+5²=√35

cosθ=Q.PR/|Q||PR|=35/√35.√35
=1

θ=0

No I didn't. This just proves the lines are perpendicular.

Is the answer (2, 1, 3)?
• Feb 21st 2010, 12:56 AM
Hello stealthmaths

OK, here's the working using the other notation.

The line $\displaystyle Q$ has equation:
$\displaystyle \textbf{r} = 2\textbf i + \textbf j+3\textbf k+\lambda(\textbf i + 3\textbf j-5\textbf k)$
This means that, for any value of $\displaystyle \lambda$, the point $\displaystyle R$ with position vector $\displaystyle \textbf r$ is any general point on the line $\displaystyle Q$.

The line joining the point $\displaystyle P$ to the point $\displaystyle R$ has vector given by:
$\displaystyle \textbf{PR} = \lambda\textbf i + (-2+3\lambda)\textbf j+(4-5\lambda)\textbf k$
The vector that determines the direction of $\displaystyle Q$, $\displaystyle \textbf q$, say, is:
$\displaystyle \textbf q =\textbf i + 3\textbf j-5\textbf k$
So PR is perpendicular to $\displaystyle Q$ if:
$\displaystyle \textbf q.\textbf{PR} = 0$

$\displaystyle \Rightarrow (\textbf i + 3\textbf j-5\textbf k).\Big(\lambda\textbf i + (-2+3\lambda)\textbf j+(4-5\lambda)\textbf k\Big) = 0$

$\displaystyle \Rightarrow \lambda+3(-2+3\lambda)-5(4-5\lambda)=0$
Solve for $\displaystyle \lambda$ and substitute back into the equation for $\displaystyle \textbf r$ to find the coordinates of $\displaystyle R$.

Can you complete it now?

• Feb 21st 2010, 04:52 AM
stealthmaths
Thank you so much for putting that into the form we are using at my school. (Bow). Although, I have to say that I prefer your method.

λ+3(-2+3λ)-5(4-5λ)=0
λ-6+9λ-20+20λ=0
λ=-26+29λ=0
-26+30λ=0
30λ=26
λ=13/15

r=2i+j+3k+13/15(i+3j-5k)
r=2i+j+3k+13/15i+2.6j-13/3k

r=(43/15, 3.6, -4/3)

Did I succeed in following correctly?
• Feb 21st 2010, 08:26 AM
Hello stealthmaths

Your method is now OK, but the arithmetic isn't quite correct. See below.

Quote:

Originally Posted by stealthmaths
Thank you so much for putting that into the form we are using at my school. (Bow). Although, I have to say that I prefer your method.

λ+3(-2+3λ)-5(4-5λ)=0
λ=-26+29λ=0
-26+30λ=0
30λ=26

r=2i+j+3k+13/15(i+3j-5k)
r=2i+j+3k+13/15i+2.6j-13/3k

r=(43/15, 3.6, -4/3) From Grandad....r = (96/35, 113/35, -5/7) Not very nice numbers, are they?

Did I succeed in following correctly?

• Feb 21st 2010, 08:46 AM
stealthmaths
yes. It seems that inputting values is not one of my skills. I will make more effort to check my work. It's just a little crazy with projects this week. Sorry(Bow).

λ+3(-2+3λ)-5(4-5λ)=0
λ-6+9λ-20+25λ=0
-26+35λ=0
35λ=26

r=2i+j+3k+26/35(i+3j-5k)
r=2i+j+3k+26/35i+78/35j-26/7k

r = (96/35, 113/35, -5/7) No. they are not very nice numbers. I think our tutor generates random numbers. She hands us assignments with the same questions for each student, but with different numbers. It does make things hard going and easy to mistake, actually. I hope she has more consideration when she sets the exam.

Thank you for your guidance on this question Grandad. I will write up my assignment now. Though, I don't think I am comfortable with the process to come to this answer here. So, I would like to analyze it a little later and return to this post with a few question for you, no doubt. Hope that is ok?
thanks again(Happy)

I have put my last post needed today in the geometry.
"2(cd) intersecting lines"
It is also a vector question. I recently received a warning from a moderator about putting vector questions in trig (i think trig). So, is geometry a better place for them on this site Grandad?
• Feb 21st 2010, 11:56 AM