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# Vectors watch

1. I don't have a clue about how to even start this question, would anyone be able to help? Thanks
2. (Original post by r3l3ntl3ss)

I don't have a clue about how to even start this question, would anyone be able to help? Thanks
If a point lies on the intersection of both planes then it must satisfy both equations. Sub in (x,y,z) and solve.
3. (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: , and after subbing in I got to .

Am I correct in saying the answer is ? It just seemed a bit odd.
4. (Original post by r3l3ntl3ss)
I did the cross product of both of the direction vectors to get the direction vector of the line: ,
OK, that's useful - forgot about that

and after subbing in I got to .
Not sure how you got that

Am I correct in saying the answer is ? 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.
5. (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 was the same in both planes, I just equated one to the other and it all cancelled out leaving ; i.e. . I just thought it may be more difficult than that since it was part of a question which was allocated 45% of the marks.
6. (Original post by r3l3ntl3ss)
Since was the same in both planes, I just equated one to the other and it all cancelled out leaving ; i.e. . 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, , so for , we have y =0; but what about z? That's not zero.
7. (Original post by ghostwalker)
Yep, sorry, my brain is starting to go.

OK, , so for , we have y =0; but what about z? That's not zero.
can be any value can't it, since LHS cancels out RHS?
8. (Original post by r3l3ntl3ss)
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.
9. (Original post by r3l3ntl3ss)
Am I correct in saying the answer is ? It just seemed a bit odd.
Since theta is fixed, that can be reduced to

for a different lambda of couse, and

Note - this is the direction vector
10. (Original post by ghostwalker)
Since theta is fixed, that can be reduced to

for a different lambda of couse, and

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?
11. (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?
12. (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, haha. Thank you!
13. (Original post by r3l3ntl3ss)
ah I see, haha. Thank you!
Seriously?!
14. (Original post by ghostwalker)
Seriously?!
sorry - I'm really out of practice
15. (Original post by r3l3ntl3ss)
sorry - I'm really out of practice

You have:

Correct?

16. (Original post by Phichi)

You have:

Correct?

yeah sorry I meant to say - don't think I was thinking when I posted that

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