Question

For given vectors a=(3/4, -5/3, 6) and
b=(-5/20, 2/3, -12/5) express b in terms of a (if possible, if not possible explain why).

Answer 1

$$
\vec a\cdot \vec b=ab\cos\theta\\
$$
Solve for \(\theta \).

If \(\theta \). is zero, then vectors are parallel an one can be expressed as a function of the other; otherwise, they are independent and one can not be expressed a function of the other.

see - http://en.wikipedia.org/wiki/Dot_product#Geometric_definition

Answer 2

so if i solve for theta and it's zero as in (axb)/abcos=thetha, then i what do i do afterwards.
i attempted a solution by
lal=\[\sqrt{3/4^{2}+-5/3^{2}+6^{2}}\]
and l bl=
\[\sqrt{-5/20^{2}+2/3^{2}+-12/5^{^{2}}}\]
but i've got no clue what to do with the results

Answer 3

i got theta equals zero but i don't know if i did that right

Answer 4

\[\cos^{-1} (ab/ab)=0\] umm not sure how to move past this or whether its right

Answer 5

okay wait i did that wrong
the second time i did this
\[\cos^{-1} ((a \times b)\div(\left| a \right|\times \left| b \right|)) =\]
a x b is (3/4 x -5/20) + (-5/3 x 2/3) + (6 x -12/5) =11303/720
\[\left| a \right| = \sqrt{5665}/12\]
\[\left| b \right|=\sqrt{22291}/60\]
and if i plugged this all into the rearranged equation then \[\theta=87.85\] rounded to two decimals so i can't actually express b in terms of a?

Answer 6

I get that they are almost parallel but not quite -

http://www.wolframalpha.com/input/?i=acos%28-11303%2F720%2F%28sqrt%285665%29%2F12*sqrt%2822561%29%2F60%29%29%29

Answer 7

http://www.wolframalpha.com/input/?i=vector%28%7B3%2F4%2C+-5%2F3%2C+6%7D+%2C%7B-5%2F20%2C+2%2F3%2C+-12%2F5%7D+%29

Answer 8

The last link shows that they are linearly independent and thus can not be expressed as functions of each other.

Answer 9

thank you so much

Answer 10

you're welcome