Answer
$\frac{2}{\sqrt {14}}$, $\frac{2}{14}i+ \frac{4}{14}j+\frac{6}{14}k$
Work Step by Step
Given: $a=i+2j+3k$ , $b=5i-k$
Change the form of $a$ and $ b$.
$a=\lt1,2,3\gt$ , $b=\lt5,0,-1\gt$
Scalar Projection $b$ onto $a$ can be calculated as follows:
$\frac{a \times b }{|a|}=\frac{(1 \times 5)+( 2 \times 0)+(3 \times -1)}{\sqrt {{(1)^{2}+(2)^{2}}+(3)^{2}}}$
$=\frac{5+0-3}{\sqrt {14}}$
$=\frac{2}{\sqrt {14}}$
Vector Projection $b$ onto $a$ can be calculated as follows:
$\frac{a \times b }{|a|^{2}}\times a=\frac{2}{14}\lt1,2,3\gt$
$=\lt\frac{2}{14}, \frac{4}{14},\frac{6}{14}\gt$
Change vector projection back into $i+j+k$ form.
$=\frac{2}{14}i+ \frac{4}{14}j+\frac{6}{14}k$
Hence,
Scalar Projection $b$ onto $a$ = $\frac{2}{\sqrt {14}}$,
Vector Projection $b$ onto $a$=$\frac{2}{14}i+ \frac{4}{14}j+\frac{6}{14}k$
