Let v1=(3/√14, 1/√14, -2/√14), v2= (1/√3,-1/√3,1/√3), v3=(1/√42,5/√42,4/√42) and its also known that the set {v1,v2,v3} is an orthonormal basis of R3. Write the vector u=(100,-200,300) as a linear combination of v1,v2,v3.
I am so lost in changing basis relatively for a vector, and i would appreciate any sort of tip or instructions on about doing this sort of question.
Use the change of basis formula: $u^*_i = \alpha_{ij} u_i$, where $\alpha_{ij} = v_i \cdot \hat{e}_j$, where the $\hat{e}_j$ are the standard basis vectors of $R^3$.
So $\alpha_{ij} u_i$ in this case is just $$ \begin{bmatrix} \dfrac{3}{\sqrt{14}} & \dfrac{1}{\sqrt{14}} & \dfrac{-2}{\sqrt{14}} \\ \dfrac{1}{\sqrt{3}} & \dfrac{-1}{\sqrt{3}} & \dfrac{1}{\sqrt{3}} \\ \dfrac{1}{\sqrt{42}} & \dfrac{5}{\sqrt{42}} & \dfrac{4}{\sqrt{42}} \end{bmatrix} \begin{bmatrix} 100 \\ -200 \\ 300 \end{bmatrix} $$ which equals $$ \begin{bmatrix} \dfrac{-500}{\sqrt{14}} \\ \dfrac{600}{\sqrt{3}} \\ \dfrac{300}{\sqrt{42}} \\ \end{bmatrix} $$
The ith row of this resulting column vector is $u^*_i$, the ith component of $u$ after the change of basis, so the vector $u$ would be the linear combination $u^*_1 (\dfrac{3}{\sqrt{14}} , \dfrac{1}{\sqrt{14}} , \dfrac{-2}{\sqrt{14}}) + \ldots $ etc.