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Let $v_1$, $v_2$, $v_3$ be mutually orthogonal non-zero vectors in $3$-space. So, any vector $v$ can be expressed as $v = c_1 v_1 + c_2 v_2 + c_3 v_3$.

(a)Show that the scalars $c_1$, $c_2$, $c_3$ are given by the formula $\displaystyle c_i=\frac{v\cdot v_i}{||v_i||^2}$, $i=1,2,3$

how to calculate the value of $v\cdot v_i$? what distributivity of scalar multiplication and addition?

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Given $v = c_1v_1 + c_2v_2 + c_3v_3$, take the dot product of both sides with $v_3$. The result is $v\cdot v_3 = c_3||v_3||^2$. Solve for $c_3$. Likewise for $i = 1, 2$.

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$c3={v.v3}/{||v3||}^2$ is this correct? – user136733 Mar 20 '14 at 21:30
Yes - and in the case i = 3, it is what was to be shown. – Jason Zimba Mar 20 '14 at 21:48
Show that v1 = 3i − j + 2k, v2 = i + j − k, v3 = i − 5j − 4k are mutually orthogonal. Now let v = i − j + k. Use the result from above to find scalars c1,c2 and c3 such that v = i − j + k = c1v1 + c2v2 + c3v3.can you plz show me how to do this?do I jsut simplely plug v=i-j+k and v3=i-5j-4k in c3? – user136733 Mar 20 '14 at 21:53
Compute $v\cdot v_1$. You should get 6 I think. Compute $||v_1||^2$. You should get 14 I think. Divide to get $c_1 = 6/14 = 3/7$. Do likewise for $c_2$ and $c_3$, and once you have all of the $c$'s as numbers, you can express $v$ as $v = c_1v_1 + c_2v_2 + c_3v_3$. Check the result by doing the right-hand side and showing that it comes to i - j + k. – Jason Zimba Mar 20 '14 at 22:06
I got c1=3/7,c2=-1/3,c3=1/21,dont know if it's right. – user136733 Mar 20 '14 at 22:15

For i=1

$c_1 =\frac{vv_i}{|v_i|^2} \Leftrightarrow c_1 =\frac{(c_1v_1 + c_2v_2 + c_3v_3)v_i}{|v_i|^2} \Leftrightarrow c_1 =\frac{c_1v_1^2}{|v_i|^2} $ which is true by the fact $v_1^2=|v_1|^2$

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