If $\mathbf{A}=\langle 2,1,2\rangle$ and $\mathbf{B}=\langle 1,0,2\rangle$, what is $\operatorname{proj}_{\mathbf{B}}\mathbf{A}$?
- $\left\langle \frac{6}{5},0,\frac{12}{5}\right\rangle$
- $\langle 6,0,12\rangle$
- $\left\langle \frac{5}{6},0,\frac{5}{3}\right\rangle$
- $\left\langle \frac{4}{5},\frac{1}{5},\frac{8}{5}\right\rangle$
Show answer and solution
Correct answer: A. $\left\langle \frac{6}{5},0,\frac{12}{5}\right\rangle$
First compute $\mathbf{A}\cdot\mathbf{B}=2\cdot 1+1\cdot 0+2\cdot 2=6$.
Also, $\|\mathbf{B}\|^2=1^2+0^2+2^2=5$.
Hence $\operatorname{proj}_{\mathbf{B}}\mathbf{A}=\dfrac{\mathbf{A}\cdot\mathbf{B}}{\|\mathbf{B}\|^2}\mathbf{B}=\dfrac{6}{5}\langle 1,0,2\rangle=\left\langle \frac{6}{5},0,\frac{12}{5}\right\rangle$.