How do I work out the calculation to find the unknown vector Can't quite figure out how the answer is obtained here and if someone can help me solve it it'd be much appreciated
"Find components of vector 'v' in direction of and perpendicular to a vector 'd'"
I'm trying to find the last vector, v2
v = (3i + 3j + 5k)
v1 = -1/9(2i + j - 2k)
v2 = v - v1
= (3i + 3j + 5k) - -1/9(2i + j - 2k)
Answer is: 1/9(29i + 28j + 43k)
-
v = 2i + 3j - 2k
v1 = 24/49(3i + 2j - 6k)
v2 = v - v1
= (2i + 3j - 2k) - 24/49(3i + 2j - 6k)
Answer is: 1/49(26i + 99j + 46k)
Could someone please show me the working? Thanks.
 A: $\vec{v}=(3,3,5)$ and $\vec{v_1}=-\frac{1}{9}(2,1,-2)$. So, $\vec{v_2}=(3,3,5)+\frac{1}{9}(2,1,-2)=\frac{1}{9}((27,27,45)+(2,1,-2))=\frac{1}{9}(29,28,43)$.
A: Since you have been given an answer for the first one, I'll answer the second one.
Write it out in column vector notation and factor out the $\frac{1}{49}$ so that you can subtract the vectors easily: $$\vec v_2=\begin{bmatrix}
 2   \\
 3   \\
 -2   \\
\end{bmatrix}-\cfrac{24}{49}\begin{bmatrix}
 3   \\
 2   \\
 -6   \\
\end{bmatrix}=\begin{bmatrix}
 \frac{98}{49}   \\
 \frac{147}{49}   \\
 -\frac{98}{49}   \\
\end{bmatrix}-\cfrac{1}{49}\begin{bmatrix}
 72   \\
 48   \\
 -144   \\
\end{bmatrix}=\cfrac{1}{49}\left(\begin{bmatrix}
 {98}   \\
 {147}   \\
 -{98}   \\
\end{bmatrix}-\begin{bmatrix}
 72   \\
 48   \\
 -144   \\
\end{bmatrix}\right)=\cfrac{1}{49}\begin{bmatrix}
 {98}-72   \\
 {147}-48   \\
 -{98}--144   \\
\end{bmatrix}=\cfrac{1}{49}\begin{bmatrix}
 26   \\
 99   \\
 46   \\
\end{bmatrix}=\color{blue}{\cfrac{1}{49}\left(26 \hat i +99 \hat j +46 \hat k\right)}$$
