Finding an othogonal vector (Elimiante x and y)

I'm stuck on this format of question:
For what values of x and y is vector v = [6, 2, x]T orthogonal to the plane containing points
A(1, 8, 2), B(1, −1, −4) and C(5, y, 15)? Record x then y on the same line.

I'm not sure how to solve for x and y. With out them there I'd form two vectors AB and AC then use the cross product to get an equation for the plane and compare the co-efficients with the vector but I honestly have no idea how to get rid of the x and y values... would I set them to 0 and solve, or do I need simultaneous equations?

Reply 1

ghostwalker

Original post by matte30

Proceed as normal(!). What do you get? It does work out.

(edited 11 years ago)

Reply 2

matte30

Original post by ghostwalker

Proceed as normal(!). What do you get? It does work out.

Unparseable latex formula:

AB = (1-1, -1-8, -4-2)[br] = (0, -9, -6)[br]AC = (5-1, y-8, -4-15)[br] = (4, y-8, -19)[br]\\[br]\\[br]

Unparseable latex formula:

[br]AB\times AC[br]\[ \left| \begin{array}{ccccc} i & j & k & i & j \\ 0 & -9 & -6 & 0 & -9 \\ 4 & y-8 & -19 & 4 & y-8 \end{array} \right|\] [br]

Unparseable latex formula:

[br]\\[br]\\[br] [br] = i(-9)(-19) + j(-6)(-4) + k(0)(y-8) - j(0)(-19) -i(-6)(y-8) - k(-9)(4)[br] = -219i + 6yi + 24j + 36k[br]

(I think that's the best I can do for formatting; I hope it helps)

(edited 11 years ago)

Reply 3

Smaug123

Original post by ghostwalker

Proceed as normal(!).

PRSOM
matte30, it would be really helpful if you could LaTeX your working; you can literally just write [ latex] around it (without the spaces), and use \times for a cross if you want it.

Reply 4

ghostwalker

Original post by matte30

AB = (1-1, -1-8, -4-2)[br] = (0, -9, -6)[br]AC = (5-1, y-8, -4-15)[br] = (4, y-8, -19)[br]

(Matrix removed because I'm not sure how to format it; but am working on it now)

[br] AB\times AC[br] = i(-9)(-19) + j(-6)(-4) + k(0)(y-8) - j(0)(-19) -i(-6)(y-8) - k(-9)(4)[br] = -219i + 6yi + 24j + 36k[br]

Agree with your k component, but not the i, or j. j should be the negative of what you have.

AC = (4,y-8,15-2)

Then just compare the 2nd coordinate of your result with that of your vector v, to get the scaling factor, and ....

Edit: Corrected.

(edited 11 years ago)

Reply 5

matte30

Original post by ghostwalker

Agree with your k component, but not the i, or j. j should be the negative of what you have.

AC = (4,y-8,15-2)

Then just compare the 1st coordinate of your result with that of your vector v, to get the scaling factor, and ....

Edit: Corrected.

AC = (4, y-8, 13)

[br]AB \times AC = -144i + 6yi -48i -24j + 36k[br]= -192i +6yi -24j +36k

I found why the j should be negative; thank you.
To get the first co-ordinate do I set y to 0?

Reply 6

ghostwalker

Original post by matte30

AC = (4, y-8, 13)

[br]AB \times AC = -144i + 6yi -48i -24j + 36k[br]= -192i +6yi -24j +36k

I found why the j should be negative; thank you.
To get the first co-ordinate do I set y to 0?

Don't agree with the "-192i", I make it "-165i", otherwise we are in agreement.

No, you don't set y to 0.

This (corrected) vector is orthogonal to the plane, as is v.

So, one must be a scalar multiple of the other.

You want to compare the coefficients of the coordinates to work out the scaling factor. You need to check j - not i as I previously said (sorry about that) - to get the scaling factor.

(edited 11 years ago)

Reply 7

matte30

Original post by ghostwalker

Don't agree with the "-192i", I make it "-156i", otherwise we are in agreement.

No, you don't set y to 0.

This (corrected) vector is orthogonal to the plane, as is v.

So, one must be a scalar multiple of the other.

You want to compare the coefficients of the coordinates to work out the scaling factor. You need to check j - not i as I previously said (sorry about that) - to get the scaling factor.

I cannot get -156i; the closet I have is -165i.
The corrected k column finishes with +13.

-9(13)i = -117i
6(y-8)i = 6yi -48i
-117 - 48 = 165

----------------------------------------------

For the actual question I've been asked I've ended up comparing my j value with a 0 in the vector... once again, I'm not sure what to do now since there is no scale factor...

(edited 11 years ago)

Reply 8

ghostwalker

Original post by matte30

I cannot get -156i; the closet I have is -165i.
The corrected k column finishes with +13.

-9(13)i = -117i
6(y-8)i = 6yi -48i
-117 - 48 = 165

Yes, you're right -165i. Clearly, I can't multiply.

So, (-165+6y)i-24j+36k

I presume you can now finish off.

Reply 9

matte30

Original post by ghostwalker

Yes, you're right -165i. Clearly, I can't multiply.

So, (-165+6y)i-24j+36k

I presume you can now finish off.

For this question; yes. But what if the unit you're comparing with a 0? example v is changed from [6, 2, x]^T to [6, 0, x]^T?

Reply 10

ghostwalker

Original post by matte30

For this question; yes. But what if the unit you're comparing with a 0? example v is changed from [6, 2, x]^T to [6, 0, x]^T?

Then it's not solveable. Your scalar multiple would have to be zero. And with that x would be 0, and there's no way you can get 6 by multiplying something by zero.

Reply 11

matte30

Original post by ghostwalker

Then it's not solveable. Your scalar multiple would have to be zero. And with that x would be 0, and there's no way you can get 6 by multiplying something by zero.