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FP3 Vectors question: where did I go wrong?

Untitled.png

I let C = x, y, x

then did

(5, 2, 6) dot [(2, -1, 3) x (x, y, z)] = 30

-5z + 15y - 6x - 4z + 12y + 6x = 30

27y - 9z = 30
Reply 1
Original post by creativebuzz
Untitled.png

I let C = x, y, x

then did

(5, 2, 6) dot [(2, -1, 3) x (x, y, z)] = 30

-5z + 15y - 6x - 4z + 12y + 6x = 30

27y - 9z = 30


1. (5, 2, 6)? It should be (5, 2, 0).

2. Your cross product is wrong as well. It should be i(z3y)j(2z3x)+k(2y+x)i(-z-3y) - j(2z-3x) + k(2y+x), the k component will disappear after dot product.
Original post by Zacken
1. (5, 2, 6)? It should be (5, 2, 0).

2. Your cross product is wrong as well. It should be i(z3y)j(2z3x)+k(2y+x)i(-z-3y) - j(2z-3x) + k(2y+x), the k component will disappear after dot product.


What a clumsy mistake *facepalm* Thank you Zacken! I redid the question with 0 instead of 6 and got the right answer :smile:
Reply 3
Original post by creativebuzz
What a clumsy mistake *facepalm* Thank you Zacken! I redid the question with 0 instead of 6 and got the right answer :smile:


Awesome - first class work.
Original post by Zacken
Awesome - first class work.


Thank you! (I would rep but SR won't let me) :P

Last question,

where did I go wrong in part d?

Untitled.png

As P and M are perpendicular to each other I made P into a plane that is parallel to M and then put it in the form r.n(unit vector) = d(unit vector)

for M: r . 1/root14(2, 1, 3) = 21/root14

for P: r . 1/root14(2, 1, 3) = (1, 2, 1) . (2, 1, 3)

= 7/root14

so it's 21-7/root14 = root14
Reply 5
Original post by creativebuzz
Thank you! (I would rep but SR won't let me) :P

Last question,

where did I go wrong in part d?

Untitled.png

As P and M are perpendicular to each other I made P into a plane that is parallel to M and then put it in the form r.n(unit vector) = d(unit vector)

for M: r . 1/root14(2, 1, 3) = 21/root14

for P: r . 1/root14(2, 1, 3) = (1, 2, 1) . (2, 1, 3)

= 7/root14

so it's 21-7/root14 = root14


I'm not sure why you think "P and M are perpendicular to one another" - it doesn't even really make sense to say that two points are perpendicular to one another nor to say that a plane is parallel to a point. I understand what you're saying but it's not the case that P and M are "perpendicular" to one another. You'll need to draw a triangle OMP and use the area and height (which will be the perpendicular distance) to find the perpendicular distance since you know the base and the area.
Original post by Zacken
I'm not sure why you think "P and M are perpendicular to one another" - it doesn't even really make sense to say that two points are perpendicular to one another nor to say that a plane is parallel to a point. I understand what you're saying but it's not the case that P and M are "perpendicular" to one another. You'll need to draw a triangle OMP and use the area and height (which will be the perpendicular distance) to find the perpendicular distance since you know the base and the area.


The reason why I thought P and M were perpendicular was because the line l (which contains P) is perpendicular to the plane (which contains M)
Reply 7
Original post by creativebuzz
The reason why I thought P and M were perpendicular was because the line l (which contains P) is perpendicular to the plane (which contains M)


Yes, but it doesn't make sense to talk about points being perpendicular. I can't give you a point on the y-axis and a point on the x-axis and say they are perpendicular, that doesn't make sense. I can say that the y-axis is perpendicular to the x-axis, though.
Original post by Zacken
Yes, but it doesn't make sense to talk about points being perpendicular. I can't give you a point on the y-axis and a point on the x-axis and say they are perpendicular, that doesn't make sense. I can say that the y-axis is perpendicular to the x-axis, though.


Yup, that's a very valid point! So how would you advise that I draw vector situations like this one? Because I would have use the whole area of a triangle if I had drawn it right in the first place to see the triangle clearly
Reply 9
Original post by creativebuzz
Yup, that's a very valid point! So how would you advise that I draw vector situations like this one? Because I would have use the whole area of a triangle if I had drawn it right in the first place to see the triangle clearly


The solution bank (for this question) has a nice drawing that showcases how to do this - I will admit that this part of the question was very difficult and required some insight.
Original post by Zacken
The solution bank (for this question) has a nice drawing that showcases how to do this - I will admit that this part of the question was very difficult and required some insight.


Yeah that's what I was checking for my answer! I tried drawing it again but it still doesn't work out as nicely as the one in the solution bank.

WIN_20160326_155704.JPG


But I don't see how my drawing is technically wrong? (I know I could just copy the one in the solution bank but I'm more concerned about whether my method for drawing vectors is wrong because I can't seem to get what they go on my own)

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