# Help me with cross product question!

Find two vectors which are perpendicular, and one of them is twice the magnitude of the other. Also their sum should be (6,8).

First of all let the coords be (x,y) and (a,b)

then the magnitude of one of them is twice the other i .e . $$2 \sqrt{x^2+y^2} = \sqrt{a^2+b^2}$$
also their dot product is zero

$$(x,y) \cdot (a,b) = 0 -> xa + yb = 0$$

also the cross product is just the magnitudes of the vectors multiplied

$$(x,y) \times (a,b) = 2 (x^2 + y^2)$$

But somehow i cannot get the fact that they add up to 6,8 to work into this framework i have setup

Is it simpler than i think it is??

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Hurkyl
Staff Emeritus
Gold Member
Well, first off, you haven't written equations describing all of the conditions of the problem...

ok then one more equation i missed is $$(x,y) + (a,b) = (6,8)$$

Fredrik
Staff Emeritus
Gold Member
The vectors are perpendicular: xa+yb=0
The magnitude of one of the vectors is twice the magnitude of the other: 4(x²+y²)=a²+b²
The sum is (6,8): x+a=6; y+b=8

I don't know what you're doing with that cross product. The cross product of two vectors is perpendicular to both vectors.

Fredrik said:
The vectors are perpendicular: xa+yb=0
The magnitude of one of the vectors is twice the magnitude of the other: 4(x²+y²)=a²+b²
The sum is (6,8): x+a=6; y+b=8

I don't know what you're doing with that cross product. The cross product of two vectors is perpendicular to both vectors.
ok what am i supposed to do after writing the equations down (sorry i dont mean to be sarcastic, but i'm not getting a reasonable reply here!)

vsage
What you ultimately want is from the four unknowns, a relationship of three to the other one. For example, a = 3x, b = 4x, y = 5x. It's not quite that simple in this case but that is about the best scenario you could get. I'll try and type out what I found. OK here is goes:

1. The sum of the vectors must be (6,8)
(a,b)+(x,y) = (6,8)
a+x = 6
b+y = 8
a+x+b+y=14

2. The two vectors must be perpendicular
$$(a,b) \cdot (x, y) = 0$$
$$ax + by = 0$$

3. The magnitude of one is twice the magnitude of the other
$$\sqrt{x^2+y^2} = 2 \sqrt{a^2+b^2}$$

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vsage
Hrm I wasn't that keen on what I said above so here's a little shove in the direction I would go to attack the problem. Use equations 1, 2 and 3 and solve for x:

1. x = 14-a-b-y
2. $$x=\frac {-by}{a}$$
3. $$x = \sqrt{4(a^2+b^2)-y^2}$$

Set equation 1 = equation 2 and solve for another variable. Set equation 2 equal to equation 3 and solve for that same variable. You will now have two equations with two unknowns relating to another variable. Set those two equal and solve for another variable. Simple substitution back into your many derived equations will give you a relationship of three variables to the other one.

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Fredrik
Staff Emeritus