Question

Find the projection vector of $\overline{b}=\hat{i}+2\hat{j}+\hat{k}$ along the vector $\overline{a}=2\hat{i}+\hat{j}+2\hat{k}$.
Also write $\overline{b}$ as the sum of a vector along $\overline{a}$ and a vector perpendicular to $\overline{a}$.

Answer 1

$\overline{b}=\hat{i}+2\hat{j}+\hat{k}$ and $\overline{a}=2\hat{i}+\hat{j}+2\hat{k}$

Projection of $\overline{b}$ onto $\overline{a}$ is given by

$\frac{\overline{a}\cdot \overline{b}}{|\overline{a}{|}^2}\overline{a}$

$=\frac{(2,1,2)\cdot (1,2,1)}{(\sqrt{2^2+1^2+2^2}{)}^2}(2,1,2)$

$=\frac{6}{9}(2,1,2)$

$=\frac{2}{3}(2\hat{i}+\hat{j}+2\hat{k})$

$=\frac{4}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{4}{3}\hat{k}$

Thus, the projection vector is $\frac{4}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{4}{3}\hat{k}$

Let $\overline{b}=\overline{b_1}+\overline{b_2}$ such that $\overline{b_1}\parallel \overline{a}$ and $\overline{b_2}\perp \overline{a}$

$\overline{b_1}=m\overline{a}=m(2\hat{i}+\hat{j}+2\hat{k})$, where $m$ is a scalar multiple.

Let $\overline{b_2}=p\hat{i}+q\hat{j}+r\hat{k}$

Since, $\overline{b_2}\perp \overline{a},\overline{b_2}\cdot \overline{a}=0$

$\Rightarrow (p\hat{i}+q\hat{j}+r\hat{k})\cdot (2\hat{i}+\hat{j}+2\hat{k})=0$

$\Rightarrow 2p+q+2r=0$ ....$\left(1\right)$

Now, $\overline{b}=\overline{b_1}+\overline{b_2}$

$\Rightarrow \hat{i}+2\hat{j}+\hat{k}=(2m\hat{i}+m\hat{j}+2m\hat{k})+(p\hat{i}+q\hat{j}+r\hat{k})$

$\Rightarrow \hat{i}+2\hat{j}+\hat{k}=(2m+p)\hat{i}+(m+q)\hat{j}+(2m+r)\hat{k}$

Comparing the components, we get

$2m+p=1$ ....$\left(2\right)$

$m+q=2$ ....$\left(3\right)$

$2m+r=1$ ....$\left(4\right)$

Subtracting $\left(2\right)$ from $\left(4\right)$, we get $r-p=0$

$\Rightarrow p=r$ ....$\left(5\right)$

From $\left(3\right)$, we get $m=2-q$. Substitute this value of $m$ in $\left(4\right)$,

$2m+r=1$

$\Rightarrow 2(2-q)+r=1$

$\Rightarrow 4-2q+r=1$

$\Rightarrow q=\frac{r+3}{2}$ ....$\left(6\right)$

Substitute the values of $p$ and $q$ from $\left(5\right)$ and $\left(6\right)$ in $\left(1\right)$,

$2r+\frac{r+3}{2}+2r=0$

$\Rightarrow 4r+r+3+4r=0$

$\Rightarrow r=-\frac{1}{3}=p$

$q=\frac{r+3}{2}=\frac{3-\frac{1}{3}}{2}=\frac{4}{3}$

$m=2-q=2-\frac{4}{3}=\frac{2}{3}$

$\therefore \overline{b_1}=\frac{4}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{4}{3}\hat{k}$

and $\overline{b_2}=-\frac{1}{3}\hat{i}+\frac{4}{3}\hat{j}-\frac{1}{3}\hat{k}$