Vector Projections/Dot Product properties

Question

Suppose that $\boldsymbol{u}=\begin{bmatrix}-3\\-4\\-1\\-8\end{bmatrix}$, $\boldsymbol{v}=\begin{bmatrix}1\\5\\-6\\-7\end{bmatrix}$, $\boldsymbol{w}=\begin{bmatrix}-8\\-5\\-4\\-2\end{bmatrix}$ and $\boldsymbol{b}$ is another non-zero vector in $\Bbb{R}^4$. You are given $3\operatorname{proj}_{\boldsymbol{b}}\boldsymbol{u}+2\operatorname{proj}_{\boldsymbol{b}}\boldsymbol{v}+5\operatorname{proj}_{\boldsymbol{b}}\boldsymbol{w}=0$. Explain why the vector $\boldsymbol{c}=3\boldsymbol{u}+2\boldsymbol{v}+5\boldsymbol{w}=\begin{bmatrix}-47\\-27\\-35\\-48\end{bmatrix}$ is perpendicular to $\boldsymbol{b}$.

What I tried so far is expand out the dot-product of the vector projections like so: $$5\Big(\frac{\boldsymbol{u}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\Big)+2\Big(\frac{\boldsymbol{v}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\Big)+5\Big(\frac{\boldsymbol{w}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\Big)=0\\(5\boldsymbol{u}+2\boldsymbol{v}+5\boldsymbol{w})\cdot\Big(\frac{\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\Big)=0$$

However, I don't know if you're actually allowed to distribute the dot product like that and where to go from there.

Answer 1

Since $b\neq 0$ and $\left(3\frac{u\cdot b}{|b|^2}+2\frac{v\cdot b}{|b|^2}+5\frac{w\cdot b}{|b|^2}\right)b=0$, we have $$3\frac{u\cdot b}{|b|^2}+2\frac{v\cdot b}{|b|^2}+5\frac{w\cdot b}{|b|^2}=0$$ so, multiplying by $|b|^2$, and using the fact that the scalar product is distributive, we obtain $$(3u+2v+5w)\cdot b=0\,,$$ hence $c$ is perpendicular to $b$.

Answer 2

On the other way

1. by hypothesis

$$\boldsymbol{c}\cdot \boldsymbol{b}=(3\boldsymbol{u}+2\boldsymbol{v}+5\boldsymbol{w})\cdot \boldsymbol{b}=0$$

2. by distributive law of dot product

$$\iff 3\boldsymbol{u}\cdot \boldsymbol{b}+2\boldsymbol{v}\cdot \boldsymbol{b}+5\boldsymbol{w}\cdot \boldsymbol{b}=0$$

3. dividing both side by $|b|^2\neq 0$

$$\iff 3\frac{\boldsymbol{u}\cdot \boldsymbol{b}}{|b|^2}+2\frac{\boldsymbol{v}\cdot \boldsymbol{b}}{|b|^2}+5\frac{\boldsymbol{w}\cdot \boldsymbol{b}}{|b|^2}=0$$

4. using that $\lambda=0 \iff \lambda\,\boldsymbol{b}=\boldsymbol{0} $

$$\iff \left(3\frac{\boldsymbol{u}\cdot \boldsymbol{b}}{|b|^2}+2\frac{\boldsymbol{v}\cdot \boldsymbol{b}}{|b|^2}+5\frac{\boldsymbol{w}\cdot \boldsymbol{b}}{|b|^2}\right)\boldsymbol{b}=\boldsymbol{0}$$

5. by distributive law (i.e. linearity)

$$\iff 3\frac{\boldsymbol{u}\cdot \boldsymbol{b}}{|b|^2}\boldsymbol{b}+2\frac{\boldsymbol{v}\cdot \boldsymbol{b}}{|b|^2}\boldsymbol{b}+5\frac{\boldsymbol{w}\cdot \boldsymbol{b}}{|b|^2}\boldsymbol{b}=\boldsymbol{0}$$