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Vectors

Screen Shot 2015-05-10 at 8.16.48 pm.png

I don't have a clue about how to even start this question, would anyone be able to help? Thanks :smile:
Original post by r3l3ntl3ss
Screen Shot 2015-05-10 at 8.16.48 pm.png

I don't have a clue about how to even start this question, would anyone be able to help? Thanks :smile:


If a point lies on the intersection of both planes then it must satisfy both equations. Sub in (x,y,z) and solve.
Reply 2
Original post by ghostwalker
If a point lies on the intersection of both planes then it must satisfy both equations. Sub in (x,y,z) and solve.


I did the cross product of both of the direction vectors to get the direction vector of the line: (sin(2θ)00)\begin{pmatrix} sin(2\theta) \\ 0 \\ 0 \end{pmatrix}, and after subbing in I got to 2ysin(θ)=02y\sin(\theta) = 0.

Am I correct in saying the answer is λ(sin(2θ)00)\lambda\begin{pmatrix} sin(2\theta) \\ 0 \\ 0 \end{pmatrix}? It just seemed a bit odd.
Original post by r3l3ntl3ss
I did the cross product of both of the direction vectors to get the direction vector of the line: (sin(2θ)00)\begin{pmatrix} sin(2\theta) \\ 0 \\ 0 \end{pmatrix},


OK, that's useful - forgot about that


and after subbing in I got to 2ysin(θ)=02y\sin(\theta) = 0.


Not sure how you got that


Am I correct in saying the answer is λ(sin(2θ)00)\lambda\begin{pmatrix} sin(2\theta) \\ 0 \\ 0 \end{pmatrix}? It just seemed a bit odd.


That seems odd to me too, as that is a line through the origin, but the planes don't even necessarily go through the origin.
Reply 4
Original post by ghostwalker
OK, that's useful - forgot about that



Not sure how you got that



That seems odd to me too, as that is a line through the origin, but the planes don't even necessarily go through the origin.


Since dd was the same in both planes, I just equated one to the other and it all cancelled out leaving 2ysin(θ)=02y\sin(\theta) = 0; i.e. ysin(θ)+zcos(θ)=ysin(θ)+zcos(θ)y\sin(\theta) + z\cos(\theta) = -y\sin(\theta) + z\cos(\theta). I just thought it may be more difficult than that since it was part of a question which was allocated 45% of the marks.
Original post by r3l3ntl3ss
Since dd was the same in both planes, I just equated one to the other and it all cancelled out leaving 2ysin(θ)=02y\sin(\theta) = 0; i.e. ysin(θ)+zcos(θ)=ysin(θ)+zcos(θ)y\sin(\theta) + z\cos(\theta) = -y\sin(\theta) + z\cos(\theta). I just thought it may be more difficult than that since it was part of a question which was allocated 45% of the marks.


Yep, sorry, my brain is starting to go.

OK, ysinθ=0y\sin\theta = 0, so for sinθ0\sin\theta\not=0, we have y =0; but what about z? That's not zero.
Reply 6
Original post by ghostwalker
Yep, sorry, my brain is starting to go.

OK, ysinθ=0y\sin\theta = 0, so for sinθ0\sin\theta\not=0, we have y =0; but what about z? That's not zero.


zz can be any value can't it, since LHS cancels out RHS?
Original post by r3l3ntl3ss
zz can be any value can't it, since LHS cancels out RHS?


It may do in that equation, but not in the original equations of the planes.
Original post by r3l3ntl3ss

Am I correct in saying the answer is λ(sin(2θ)00)\lambda\begin{pmatrix} sin(2\theta) \\ 0 \\ 0 \end{pmatrix}? It just seemed a bit odd.


Since theta is fixed, that can be reduced to

λ(100)\lambda\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}

for a different lambda of couse, and sin(2θ)0\sin(2\theta)\not=0

Note - this is the direction vector
(edited 8 years ago)
Reply 9
Original post by ghostwalker
Since theta is fixed, that can be reduced to

λ(100)\lambda\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}

for a different lambda of couse, and sin(2θ)0\sin(2\theta)\not=0

Note - this is the direction vector


I see - the position vector is still confusing me a bit though; what were you implying in the previous post?
Original post by r3l3ntl3ss
I see - the position vector is still confusing me a bit though; what were you implying in the previous post?


Substitute your value of x,y into the equation of the plane, what do you get for z?
Original post by ghostwalker
Substitute your value of x,y into the equation of the plane, what do you get for z?


ah I see, z=dz = d haha. Thank you!
Original post by r3l3ntl3ss
ah I see, z=dz = d haha. Thank you!


Seriously?!
Original post by ghostwalker
Seriously?!


sorry - I'm really out of practice
(edited 8 years ago)
Reply 14
Original post by r3l3ntl3ss
sorry - I'm really out of practice


Please explain how you got z=dz=d from your equation.

You have:

ysinθ+zcosθ=dysin\theta + zcos\theta = d

Correct?

You've already said y=0y=0
Original post by Phichi
Please explain how you got z=dz=d from your equation.

You have:

ysinθ+zcosθ=dysin\theta + zcos\theta = d

Correct?

You've already said y=0y=0


yeah sorry I meant to say z=dcos(θ)z = \frac{d}{\cos(\theta)} - don't think I was thinking when I posted that

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