Stating if lines are perpendicular or parallel.

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burt

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I was asked to "state whether the lines are parallel or perpendicular and find the angle between the lines" for the following lines: 1584072460314.png

I'm not quite sure how to do that. I know that with vectors you can use the dot product and cross product to figure out if the lines are perpendicular or parallel. I've been working at this problem, but I'm completely stuck. I know it's better if I show my work so far, but so far I haven't done very well. I tried to just plug in zero and see what happens - but thtat didn't work. Any help?
Thank you!
 

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[MATH]\text{Two vectors $v_1, v_2$ are parallel if $|v_1 \cdot v_2| = |v_1||v_2|$}[/MATH]
[MATH]\text{Two vectors are perpendicular if $v_1 \cdot v_2 = 0$}[/MATH]
[MATH]\text{The angle between two vectors is given by $\theta = \cos^{-1}\left(\dfrac{v_1 \cdot v_2}{|v_1||v_2|}\right)$}[/MATH]
[MATH]\text{You should be able to see by inspection that the direction vectors for your two lines are}\\ v_1 = (-2,3,2),~v_2 = (1,-2,3)\\~\\ \text{I leave it to you to determine whether they are parallel, perpendicular, or what the angle between them is} [/MATH]
 
burt said:
I was asked to "state whether the lines are parallel or perpendicular and find the angle between the lines" for the following lines: View attachment 17154

I'm not quite sure how to do that. I know that with vectors you can use the dot product and cross product to figure out if the lines are perpendicular or parallel. I've been working at this problem, but I'm completely stuck. I know it's better if I show my work so far, but so far I haven't done very well. I tried to just plug in zero and see what happens - but thtat didn't work. Any help?
Thank you!
Click to expand...
Plugging in one parameter value gives you one point. To get vectors you need 2 points.
 
burt said:
I was asked to "state whether the lines are parallel or perpendicular and find the angle between the lines" for the following lines: View attachment 17154

I'm not quite sure how to do that. I know that with vectors you can use the dot product and cross product to figure out if the lines are perpendicular or parallel. I've been working at this problem, but I'm completely stuck. I know it's better if I show my work so far, but so far I haven't done very well.
Click to expand...
Name the lines \(\ell_1~\&~\ell_2\). The line \(\ell_1\) has direction vector \(\left<-2,3,2\right>\)
The line \(\ell_2\) has direction vector \(\left<1,-2,3\right>\).
The two vectors are not multiples of each other therefore their lines are not parallel.
If the dot product of those vectors is not zero then the lines cannot be perpendicular.
If the lines do not intersect there is no angle between them. Do the lines intersect?
 
burt said:
Is the direction vector simply the coefficients?
Click to expand...
Good lord! I thought you were studying vector geometry. Pardon me!
If you have to ask that question then I don't know how to help.
I hope someone else know how.
 
Please tell us what you have learned about equations of lines and vectors. Have you been taught how the direction vector is related to the equations of a line, in any form? We need to see what we can assume you know, in order to help you appropriately.

But, yes, in this situation the coefficients of the parameter are the components of the direction vector.
 
There is a formula to angle:
m = (y2-y1)(x2-x1)
m = tan(angle)
There are formula to put the m in get the angle.
But, If you need to slope can use:
m = (y2 - y1)(x2 - x1)
Now:
we can write it explicit equation:
y = mx + n
Now we find the patten in the equations (notice the equation are explicit!):
y1 = m1x2 + n1
y2 = m2x2 + n2
Now when the angle in 90 degree, we have:
-1 = m 1* m2

Suppose, we have:
y = 2x + 5
y2 = 6x + 4
m1 = 2
m2 = 6
m1 * m2 not equal to -1. -> m1 & m2 not prependicular.
Now try the equation you have!
 
P.S. Is there a way to solve it by my (way of thinking)?
Can it facilitate solving the question or make it difficult and harder to solve?
 
shahar said:
P.S. Is there a way to solve it by my (way of thinking)?
Can it facilitate solving the question or make it difficult and harder to solve?
Click to expand...
Why would you want to? You are trying to use 2D methods on a 3D problem. Sorry, but it isn't even worth trying to figure out what you are thinking.
 
Actually the equation Shahar has is wrong.
It should be m= tan(angle)= (y2-y1)/(x2-x1),
not "m= (y2-y1)/(x2-x1)"!
 
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