Answer
$\frac{1}{\sqrt {3}}$, $\lt\frac{1}{3}, \frac{1}{3},\frac{1}{3}\gt$
Work Step by Step
Given: $a=\lt1,1,1\gt$ , $b=\lt1,-1,1\gt$
Scalar Projection $b$ onto $a$ can be calculated as follows:
$\frac{a \times b }{|a|}=\frac{(1 \times 1)+( 1 \times -1)+(1 \times 1)}{\sqrt {{(1)^{2}+(1)^{2}}+(1)^{2}}}$
$=\frac{1-1+1}{\sqrt {3}}$
$=\frac{1}{\sqrt {3}}$
Vector Projection $b$ onto $a$ can be calculated as follows:
$\frac{a \times b }{|a|^{2}}\times a=\frac{1}{3}\lt1,1,1\gt$
$=\lt\frac{1}{3}, \frac{1}{3},\frac{1}{3}\gt$
Hence,
Scalar Projection $b$ onto $a$ = $\frac{1}{\sqrt {3}}$,
Vector Projection $b$ onto $a$=$\lt\frac{1}{3}, \frac{1}{3},\frac{1}{3}\gt$
