Question
Find the direction cosines and direction angles of the vector.
$2 \hat{i}+\hat{j}+2 \hat{k}$
$2 \hat{i}+\hat{j}+2 \hat{k}$
$|\bar{a}|=\sqrt{2^2+1^2+2^2}=\sqrt{4+1+4}=\sqrt{9}=3$
$\therefore$ unit vector along $\bar{a}$
$=\hat{a}=\frac{\bar{a}}{|\bar{a}|}=\frac{2 \hat{i}+\hat{j}+2 \hat{k}}{3}=\frac{2}{3} \hat{i}+\frac{1}{3} \hat{j}+\frac{2}{3} \hat{k}$
$\therefore$ its direction cosines are $\frac{2}{3}, \frac{1}{3}, \frac{2}{3}$.
If $\alpha, \beta, y$ are the direction angles, then $\cos \alpha=\frac{2}{3}, \cos \beta=\frac{1}{3}$
$\cos \gamma=\frac{2}{3}$
$\therefore \alpha=\cos ^{-1}\left(\frac{2}{3}\right), \beta=\cos ^{-1}\left(\frac{1}{3}\right), \gamma=\cos ^{-1}\left(\frac{2}{3}\right)$
Hence, direction cosines are $\frac{2}{3}, \frac{1}{3}, \frac{2}{3}$ and direction angles
are $\cos ^{-1}\left(\frac{2}{3}\right), \cos ^{-1}\left(\frac{1}{3}\right), \cos ^{-1}\left(\frac{2}{3}\right)$
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is $\frac{1}{2}[\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}]$
Question is modified.
Show that the area of a triangle $A B C$, the position vectors of whose vertices are $\bar{a}, \bar{b}$ and $\bar{c}$
is $\frac{1}{2}[\bar{a} \times \bar{b}+\bar{b} \times \bar{c}+\bar{c} \times \bar{a}]$.
$\cot ^{-1}(0.999)$