# Vector - straight lines

• Nov 12th 2008, 07:56 AM
geton
Vector - straight lines
The straight lines $\displaystyle l_1$ and $\displaystyle l_2$ have vector equations r = (i + 4j + 2k) + t(8i + 5j + k) and r = (i + 4j + 2k) + s(3i + j) respectively. P is the point with coordinate (1, 4, 2) & Q is the point with coordinate (9, 9, 3).

Find the possible coordinates of the point R, such that R lies on $\displaystyle l_2$ and PQ = PR.

How could I do this?
• Nov 12th 2008, 09:35 AM
Soroban
Hello, geton!

Are there more parts to this problem?
There is too much information . . .

Quote:

The line $\displaystyle L$ has vector equation: .$\displaystyle r \;=\;(i + 4j + 2k) + s(3i + j)$

$\displaystyle P$ is the point $\displaystyle (1, 4, 2)$ and $\displaystyle Q$ is the point $\displaystyle (9, 9, 3).$

Find the possible coordinates of the point $\displaystyle R$ such that $\displaystyle R$ lies on $\displaystyle L$ and $\displaystyle PQ = PR.$

Distance $\displaystyle PQ$ is: .$\displaystyle \sqrt{(9-1)^2 + (9-4)^2 + (3-2)^2} \;=\; \sqrt{8^2+ 5^2 + 1^2} \:=\:\sqrt{90}$ .[1]

Line $\displaystyle L$ has parametric equations: .$\displaystyle \begin{array}{c}x \:=\:1 + 3s \\ y \:=\:4 + s \\ z \:=\:2 \end{array}$ .(a)

Point $\displaystyle R$ has coordinates: .$\displaystyle (1+3s,\:4+s,\:2)$

Distance $\displaystyle QR$ is: .$\displaystyle \sqrt{(1+3s-4)^2 + (4+s-9)^2 + (2-3)^2} \;=\;\sqrt{10s^2 - 58s + 90}$ .[2]

Equate [2] and [1]: .$\displaystyle \sqrt{10s^2 - 58s + 90} \;=\;\sqrt{90} \quad\Rightarrow\quad 10s^2 - 58s + 90 \;=\;90$

. . $\displaystyle 10s^2 - 58s \:=\:0 \quad\Rightarrow\quad 2s(5s-29) \:=\:0 \quad\Rightarrow\quad s \;=\;0,\:\tfrac{29}{5}$

Substitute into (a): . $\displaystyle \begin{array}{c|c} s=0 & s = \frac{29}{5} \\ \\[-4mm] \hline \\[-3mm] x = 1 & x = \frac{92}{5} \\ \\[-3mm] y = 4 & y = \frac{49}{5} \\ \\[-4mm] z = 2 & z = 2 \end{array}$

Therefore: . $\displaystyle R \;=\;(1,4,2)\:\text{ or }\:\left(\tfrac{92}{5},\:\tfrac{49}{5},\:2\right)$

• Nov 12th 2008, 10:44 AM
geton
Quote:

Originally Posted by Soroban
Hello, geton!

Are there more parts to this problem?
There is too much information . . .

Yes, I tried to short the question. But ...

Quote:

Distance $\displaystyle PQ$ is: .$\displaystyle \sqrt{(9-1)^2 + (9-4)^2 + (3-2)^2} \;=\; \sqrt{8^2+ 5^2 + 1^2} \:=\:\sqrt{90}$ .[1]

Line $\displaystyle L$ has parametric equations: .$\displaystyle \begin{array}{c}x \:=\:1 + 3s \\ y \:=\:4 + s \\ z \:=\:2 \end{array}$ .(a)

Point $\displaystyle R$ has coordinates: .$\displaystyle (1+3s,\:4+s,\:2)$

Distance $\displaystyle QR$ is: .$\displaystyle \sqrt{(1+3s-4)^2 + (4+s-9)^2 + (2-3)^2} \;=\;\sqrt{10s^2 - 58s + 90}$ .[2]

Equate [2] and [1]: .$\displaystyle \sqrt{10s^2 - 58s + 90} \;=\;\sqrt{90} \quad\Rightarrow\quad 10s^2 - 58s + 90 \;=\;90$

. . $\displaystyle 10s^2 - 58s \:=\:0 \quad\Rightarrow\quad 2s(5s-29) \:=\:0 \quad\Rightarrow\quad s \;=\;0,\:\tfrac{29}{5}$

Substitute into (a): . $\displaystyle \begin{array}{c|c} s=0 & s = \frac{29}{5} \\ \\[-4mm] \hline \\[-3mm] x = 1 & x = \frac{92}{5} \\ \\[-3mm] y = 4 & y = \frac{49}{5} \\ \\[-4mm] z = 2 & z = 2 \end{array}$

Therefore: . $\displaystyle R \;=\;(1,4,2)\:\text{ or }\:\left(\tfrac{92}{5},\:\tfrac{49}{5},\:2\right)$

Thank you so much for giving the method to do. But the distance QR will be PR.

Thanks a lot.