# Dot products of vectors in three dimensions.

• Sep 23rd 2007, 09:55 AM
niyati
Dot products of vectors in three dimensions.
Consider three vectors:

A = 4 i - 9 j + 6 k
B = -1 i + 2 j
C = (6cos60) i + (6sin60) j

Evaluate the magnitude of the quantity A + (B x C )

I cannot have a graphing calculator on the test (only a scientific one), so matrices are not possible.

I tried taking the dot product of B and C and got about 53.65. But then finding the quadrature of the x components, y components, and z components throws me off.

Help?
• Sep 23rd 2007, 10:05 AM
Jhevon
Quote:

Originally Posted by niyati
Consider three vectors:

A = 4 i - 9 j + 6 k
B = -1 i + 2 j
C = (6cos60) i + (6sin60) j

Evaluate the magnitude of the quantity A + (B x C )

I cannot have a graphing calculator on the test (only a scientific one), so matrices are not possible.

I tried taking the dot product of B and C and got about 53.65. But then finding the quadrature of the x components, y components, and z components throws me off.

Help?

there is no dot product here. you have an addition and a cross-product

here's what you need to know:

Let $\bold {A}$ be $\left< a_1,a_2,a_3 \right>$ and,
let $\bold {B}$ be $\left< b_1,b_2,b_3 \right>$

then, $\bold {A + B} = \left< a_1 + b_1, a_2 + b_2, a_3 + b_3 \right>$

and $\bold {A \times B} = \left| \begin {array}{ccc} \bold {i} & \bold {j} & \bold {k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end {array} \right| = (a_2b_3 - a_3b_2) \bold {i} + (a_3b_1 - a_1b_3) \bold {j} + (a_1b_2 - a_2b_1) \bold {k}$

the magnitude of a vector $\bold {A}$ is given by:

$| \bold {A} | = \sqrt {a_1^2 + a_2^2 + a_3^2}$
• Sep 23rd 2007, 10:27 AM
niyati
I was never taught how to do matrices by hand. :S What is the general rule?
• Sep 23rd 2007, 10:29 AM
Jhevon
Quote:

Originally Posted by niyati
I was never taught how to do matrices by hand. :S What is the general rule?

i gave you the expanded formula as well. it would be a pain to explain here how i did it, you should try looking it up on google and/or wikipedia
• Sep 23rd 2007, 11:02 AM
niyati
Instead of using $< \ \ \ >$ use $\left< \ \ \ \right>$ to make it look nicer.