Question

If $b\ne 0,$ then every vector $a$ can be written in a unique manner as the sum of a vector $a_{||}$ parallel to b and a vector $a_{\perp }$ perpendicular to $b$. If $a$ is parallel to $b,$ then $a_{||}=a$ and $a_{\perp }=0$. If $a$ is perpendicular to $b$, then $a_{||}=0$ and $a_{\perp }=a$. The vector $a_{||}$ is called the projection of $a$ on $b$ and is denoted by $pro{j}_{b}a$. Since $pro{j}_{b}a$ is parallel to $b$, it is a scalar multiple of the unit vector in the direction of $b$, i.e., $pro{j}_{b}a=\lambda u_b$
The scalar $\lambda$ is called the component of $a$ in the direction of $b$ and is denoted by $com{p}_{b}a$. In fact, $pro{j}_{b}a=(a\cdot u_b)u_b$ and $com{p}_{b}a=a\cdot u_b$

If $a=-2i+j+k$ and $b=4i-3j+k$, then $pro{j}_{b}a$ is equal to

• A. $4i-3j+2k$
• B. $-\frac{5}{13}(4i-3j+k)$
• C. $\frac{5}{13}(4i-3j+k)$
• D. $-\frac{4}{11}(2i-j+2k)$

Answer 1

The correct option is B $-\frac{5}{13}(4i-3j+k)$

$\left|b\right|=\sqrt{26},u_b=\frac{b}{\left|b\right|}=\frac{1}{\sqrt{26}}(4i-3j+k)$
${comp}_{b}a=a.u_b=\frac{-10}{\sqrt{26}}$
${Proj}_{b}a=\left({comp}_{b}a\right)u_b=-\frac{5}{13}(4i-3j+k)$