Question

Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes OX, OY and OZ.

Answer 1

Let $a=\hat{i}+\hat{j}+\hat{k}$
Then, magnitude of vector $a=|a|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$
Therefore, the direction cosines of a are $\frac{x}{|a|},\frac{y}{|a|},\frac{z}{|a|}i.e.,\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$
If a makes angles $\alpha ,\beta$ and $\gamma$ respectively with (positive) OX,OY and OZ.
Then, we have cos $\alpha =1/\sqrt{3}$
$(\because$Direction cosines are the cosines of the angles made by $\hat{i},\hat{j},\hat{k}$ components of the vector with X,Y and Z axes)
$cos\beta =\frac{1}{\sqrt{3}}andcos\gamma =\frac{1}{\sqrt{3}},\alpha =\beta =\gamma =co{s}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
Hence, the given vector a is equally inclined with OX, OY and OZ.