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find the components of the vector 3i+2j+8k in the direction of the vector 2i+2j+2k

Mathematics
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3i+2j+8k +2i+2j+2k ______________ 5i +4j +10k Check if its right. Been a looong time since ive done this.
im guessing find magnitude of 1st vector --> sqrt(3^2 + 2^2+8^2) = sqrt(77) and then find components of directional vector with same magnitude unit vector of 2nd vector = 1/sqrt3 , 1/sqrt3 , 1/sqrt3 multiply by magnitude --> sqrt(77/3) i + sqrt(77/3) j + sqrt(77/3) k
r u sure about the answer

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Other answers:

.....NO!!! Im not confident thats right. I suggest asking this question in the physics study group. And look at t his site: http://www.ehow.com/how_8396057_determine-resultant-vector.html
im pretty sure i did it right as long as i understood the question correctly. you taking 1st vector and placing it so its going same direction as 2nd vector but has same magnitude...correct?
am not sure...am still trying to solve it
R u doing college physics?? Did u read anything about "dot product"??
dumb cow is correct ,
wait, ik the right answer.
we take the dot product of first vector with unit vector of second
Instead of adding, do what I showed you in the SAME set up, except u multiply straight down: (3*2)+(2*2)+(8*2) = 26
Yeaaah!! I dont remember how to get angle..if its needed
zara my answer is the right one, cause i did this question last day
thankss
\(<3,2,8> . <2,2,2> = 6 + 4+16 = 26\) \[|<2,2,2>|= \sqrt{12} \] Vector component of \(\vec{a} \)in the direction of \(\vec{b}\) is given by \( \frac {\mathbf{a} \cdot \mathbf{b}} {|\mathbf{b}| } \frac {\mathbf{b}} {|\mathbf{b}|} \) Hence, \(\frac {26}{12} <2,2,2> = \frac {13}6<2,2,2> \)
I have assumed that you meant vector component and not the scalar one.
@dumbcow: I am not sure from where you came up with that definition, can you please explain? http://en.wikipedia.org/wiki/Vector_projection
foolformath's right
@foolFormath, oops did not recognize this as vector projection...for some reason i thought the magnitude of 1st vector should remain the same

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