if you have two equations that do not cancel out a number, can you multiply one equation by a constant (on both sides) to cancel out a variable? Ex: eq1 --> -x-3y-4z = 2 eq2 --> 2x + 2y +2z = 5
i used a random y value to get point B and when i subtracted B-A to get the vector, it did not match to method 2. I used fractions and i am certain i havent miscalculated. What may be the issue then? great video by the way
hello, thank you for your video. I had a question regarding your first approach. I already used your second approach and then attempted to check my answer for V by using the first approach. However, the Z did not cancel out so I had to multiply the equation of the second plane by -2 to get the Z to cancel. Next, I proceeded with your approach and got 2 points and subtracted them from each other. However, the answer did not end up the same as the answer of V with the first approach. Do you have any idea why this is? I could send you a picture of my work so you could check it out, if you would like. Thanks for your time.
The direction vector we get for the line is not unique. The one that you got should be parallel to the one in the other approach. That is, for example, if the direction vector is , then another possible direction vector can be .
If they aren't parallel, they can go in different directions and not touch each other at all. E.g. imagine you had two pens, both infinitely long, one in each hand. You could orient them in a way in which they wouldn't intersect. This wouldn't work for planes.@@marthamaywhovier5962
hey I have a question so I understand approach two how when you cross them you get a vector perpendicular to the two norm vectors of my plane which result in the direction vector of the line of intersection, however where do I get my point from? since you need a direction vector and a point on the line to create the vector equation. do I just use the norm vector as one of my points so norm vector 1 was so my equation would be + t then obviously multiply it out if you want the parametric form
@@GlassofNumbers so is using the norm vector 1 as position vector correct? or do we do as you said it, substitute 2 variable and solve for third and get the point?
Didn't much like the video, not enough explanation. Needed approach 2, but everything was super fast forwards, expecially with the n1 (multiplication sign) n2. I mean I understood it because I knew how to do it, but for the people that don't?
Thank you for the feedback! This video assumes knowledge of computing the cross product of two vectors. If one needs to review how to do a cross product, I am happy to make another video just for that. This way, audience who already knows how to do cross product won't be bogged down by calculation.
@GlassofNumbers you're okay, I was just in a bad mood for not understanding this even though I've been trying for hours. Disregard that comment. You're doing great.
