Question

Find the acute angle which the line with direction cosines $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{6}},n$ makes with positive direction of z-axis.

Answer 1

$di{r}^n$ cosines $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{6}},n):(l,m,n)$

as $l^2+m^2+n^2=1$

$\frac{1}{3}+\frac{1}{6}+n^2=1$

$n^2=1-\frac{1}{3}+\frac{1}{6}$

$=\frac{6-2-1}{6}=\frac{3}{6}=\frac{1}{2}$

$n=\pm \frac{1}{\sqrt{2}}$

$di{r}^n$ cosines makes with the $di{r}^n$ of z-axis

so $n=+\frac{1}{\sqrt{2}}$

as $cosr=n$

$cosr=\frac{1}{\sqrt{2}}$

$cos\frac{\pi }{4}=\frac{1}{\sqrt{2}}$

$\therefore$ acute angle = $\frac{\pi }{4}$

so, line make $\frac{\pi }{4}$ angle with ${+}^{ve}di{r}^n$ of z - axis