# Thread: Vectors - cross and dot products

1. ## Vectors - cross and dot products

The unknown vector v satisfies $b \cdot v = \lambda$ and $b \times v = c$ where lambda, b and c are fixed and known.
Find v in terms of lambda, b and c

At first I was thinking of expanding everything out, and so I'd get:

$b_1 v_1 + b_2 v_2 + b_3 v_3 = \lambda$
$b_2 v_3 - v_2 b_3, b_3 v_1 - b_1 v_3, b_1 v_2 - b_2 v_1) = (c_1, c_2, c_3)$

But then that didn't seem to help. Any ideas?

2. ## Re: Vectors - cross and dot products

Hey Educated.

Hint: You have three unknowns (v1,v2,v3) and four equations that are linear. Try setting up a linear system and reducing to get v1,v2,v3 in terms of your other known parameters. Also if there is any chance for in-consistency, then deal with that as well (there shouldn't but you need to check since you have four equations in three unknowns).

3. ## Re: Vectors - cross and dot products

Sorry, I'm not quite sure how to do that. I'll give it a go though:

$b_1 v_1 + b_2 v_2 + b_3 v_3 = \lambda$
$b_2 v_3 - b_3 v_2 = c_1$
$b_3 v_1 - b_1 v_3 = c_2$
$b_1 v_2 - b_2 v_1 = c_3$

$\begin{pmatrix} b_1 & b_2 & b_3 & \lambda \\ 0 & -b_3 & b_2 & c_1 \\ b_3 & 0 & -b_1 & c_2\\ -b_2 & b_1 & 0 & c_3 \end{pmatrix}$

I don't know how to reduce it down though. Is there another way of doing this?

4. ## Re: Vectors - cross and dot products

Personally, I don't like to use matrices to solve systems of equations.
you have
$b_1v_1+ b_2v_2+ b_3v_3= \lambda$
$b_2v_3- b_3v_2= c_1$
$b_3v_1- b_1v_3= c_2$
$b_1v_2- b_2v_1= c_3$

Multiply the first equation by $b_2$ to get $b_1b_2v_1+ b_2^2v_2+ b_2b_3v_3= b_2\lambda$ and the second equation by $b_3$ to get $b_2b_3v_3- b_3^2v_2= b_3c_1$. Those equations both have " $b_2b_3v_3$ so subtracting the second fro the first eliminates $v_3$ giving $b_1b_2v_1+(b_2^2+ b_3^2)v_2= b_2\lambda- b_3c_1$. Now the fourth equation $b_1v_2- b_2v_1= c_3$ has only $v_1$ and $v_2$ so we can eliminate either of those.

For example, if we multiply $b_1v_2- b_2v_1= c_3$ by $b_1$ we get $b_1^2v_2- b_1b_2v_1= b_1c_3$ and adding that to $b_1b_2v_1+ (b_2^2+ b_3^2)v_2= b_2\lambda- b_2c_1$ gives (b_1^2+ b_2^2+ b_3^2)v_2= b_2\lambda+ b_1c_3- b_2c_1[/tex] and can soplve for $v_2$.

5. ## Re: Vectors - cross and dot products

b.v=λ
bxv=c

bx(bxv)=b(b.v)-v(b.b)=bxc
v=(λb+bxc)(1/b2)

6. ## Re: Vectors - cross and dot products

If
v = 2i + 4j
and
w = i + 5j
the
v . w = (2)(1) + (4)(5) = 22

7. ## Re: Vectors - cross and dot products

Originally Posted by marybalogh
If
v = 2i + 4j
and
w = i + 5j
the
v . w = (2)(1) + (4)(5) = 22
and 2 + 2 = 4

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