MHB S6.12.3.17 Find the angle between the vectors

[karush]

$\tiny{s6.12.3.17}\\$
$\textsf{ Find the angle between the vectors $a$ and $b$}\\$
\begin{align}
\displaystyle
a&=\langle 3,-1,5\rangle &b&=\langle -2,4,3\rangle\\
\\
\cos\left({\theta}\right)&=\frac{(3\cdot -2)+(-1\cdot4)+(5\cdot3)}
{|\sqrt{35}|\cdot|\sqrt{29}|}\\
&\approx 0.1569 \\
\arccos(0.1569)&\approx80.97^o\approx81^o
\end{align}
tried W|A but dot product ?

 

[Joppy]

This looks correct. What's the problem exactly?

 

[Greg]

What about the dot product are you referring to?

The working, at least up to the expression for cosine of theta, is correct.