# Thread: Find the angle between two straight lines

1. ## Find the angle between two straight lines

Hi

can someone tell me if they find anything wrong with my solution for the following question:

1) Find the angle between the straight lines L and M, where L has symmetric equations

$\frac{x-1}{4}=\frac{y+3}{4}=\frac{z-2}{3}$

And M parametric equations

$x=3+2t$
$y=5-4t$
$z=-7+2t$

$cos\theta = \frac{8-16+6}{\sqrt{41}\sqrt{24}} = \frac{-2}{\sqrt{41}\sqrt{24}} = 93.66$

P.S

2. Originally Posted by Paymemoney
Hi

can someone tell me if they find anything wrong with my solution for the following question: Mr F says: Yes. It's sloppy. See below.

1) Find the angle between the straight lines L and M, where L has symmetric equations

$\frac{x-1}{4}=\frac{y+3}{4}=\frac{z-2}{3}$

And M parametric equations

$x=3+2t$
$y=5-4t$
$z=-7+2t$

$cos\theta = \frac{8-16+6}{\sqrt{41}\sqrt{24}} = \frac{-2}{\sqrt{41}\sqrt{24}} = 93.66$

P.S
1. There is not enough working for the assessor to give you full marks. You should show all working so that it's clear where you answer comes from.

2. How can the cosine of an angle be equal to 93.66, which is what you have written. Sloppy.

3. My full workings

$L = i-3j+2k+t(4i+4j+3k)$

$M= 3i+5j-7k+t(2i-4j+2k)$

$cos\theta = \frac{8-16+6}{\sqrt{41}\sqrt{24}}$

$cos\theta = \frac{-2}{\sqrt{41}\sqrt{24}}$

$arccos\theta = 93.66$

4. Almost full marks! If I were your teacher I would insist that you show where you got that $\frac{8- 16+ 6}{\sqrt{41}\sqrt{24}}$ from.

More importantly, your final answer is $\theta$, not " $arccos(\theta)$" and you do not have units on the angle. Is that degrees, radians, cubic liters?

5. however is the answer correct?

93.66 degrees

book says its 99.974 degrees

6. Originally Posted by Paymemoney