Question

Find the direction cosines of the vector -2i + j -5k.

Answer 1

Ok, first find the magnetude of this verctor:
v = -2i + j - 5k
|v| = sqrt((-2)^2 + 1^2 + (-5)^2)
|v| = sqrt(4 + 1 + 25)
|v| = sqrt(30)

now, let's find the cosines, just make inner products:
v . i = |v||i|cos(o1)
cos(o1) = -2/sqrt(30)

v . j = |v||j|cos(o2)
cos(o2) = 1/sqrt(30)

v . k = |v||k|cos(o3)
cos(o3) = -5/sqrt(30)

Answer 2

Okay I got you till finding the magnitude of the vector, what does o1, o2, o3 mean? is it possible for you to create an equation for me?

Answer 3

o1, o2 and o3 are the angles that this given vector makes with the x, y and z axis respectively

Answer 4

So -2 being the cosine of i we multiply it with the |v|? that's it?

Answer 5

No, we must use the definition of inner product, given a vector v = ai + bj + ck, then the directors angles are defined as:
\[\vec{v} \cdot î =|{\vec{v}}||î|\cos(\theta_x)\]
\[\vec{v} \cdot j =|{\vec{v}}||j|\cos(\theta_y)\]
\[\vec{v} \cdot k =|{\vec{v}}||k|\cos(\theta_z)\]

\[|\vec{v}| = \sqrt{a^2 + b^2 + c^2}\]

Answer 6

for example:
\[\vec{v} \cdot i = (ai + bj + ck)\cdot 1i + 0j + 0k = 1\times a + b\times 0+ c\times 0 = a\]

Answer 7

that is a generalized way of solving this

Answer 8

just apply this theory to your particular case, as i did above, and you find this cosines

Answer 9

Alright hang on, I'll try some more examples, it'll help me understand. I'll write down what you said, gimme 5min

Answer 10

Ok

Answer 11

So if I have p = i + j + k

The direction cosine of p will be

|p| = sqrt/1^2 + 1^2 + 1^2

It comes to the form of 1/sqrt/3

Answer 12

exactly

Answer 13

Thank you @M4thM1nd your really sweet :)

Answer 14

:)