Writing a vector as the sum of two other vectors.

Question

Suppose you have 2 vectors $\vec a = (1,1,2)$ and $\vec b = (3,4,-2)$, how would you write $\vec a$ as the sum of 2 vectors $\vec c$ and $\vec d$ where $\vec c$ is in the direction of $\vec b$ and $\vec d$ is perpendicular to $\vec b$.

I think $\vec c$ may be the dot product of $\vec a$ and $\vec b$ multiplied by $\vec b$'s unit vector and $\vec d$ may be the cross-product of $\vec a$ and $\vec b$ as that is orthogonal to $\vec b$?

Thanks!

Answer 1

What you want looks like this $$ \vec a = \alpha \vec b + \vec d, \text{ where } \vec d\perp \vec b $$ Hence, $$ \langle \vec a,\vec b\rangle = \alpha \langle \vec b,\vec b\rangle $$ This helps you find $\alpha$. Now just take $\vec d := \vec a-\alpha \vec b$ and you are done!

Answer 2

You want to find the scalar $k$ and the vector $d=(d_1,d_2,d_3)$ such that $a=kb+d$ and $b\perp d\iff b\cdot d=0$.

$ a=kb+d\implies (1,1,2)=(3k,4k,-2k)+(d_1,d_2,d_3)=(3kd_1,4k+d_2,-2k+d_3) $

so

\begin{align} 3k+d_1&=1\tag{1}\\ 4k+d_2&=1\tag{2}\\ -2k+d_3&=2\tag{3} \end{align}

On the other hand, $$b\cdot d=0\implies 3d_1+4d_2-2d_3=0\tag{4}$$

Solve these four equations in the four unknowns $k,d_1,d_2,d_3$ to get $$ k={3\over 29},\ d_1={20\over 29},\ d_2={17\over 29},\ d_3={64\over 29}.$$

Thus, $a={3\over 29}b+({20/29},{17/29},{64/29})$ is the decomposition you seek.