Find the angles at which the following vectors are inclined to each of the coordinate axes:
(i) $\hat{i} - \hat{j} + \hat{k}$

(ii) $\hat{j} - \hat{k}$

(iii) $4 \hat{i} + 8 \hat{j} + \hat{k}$

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Solution

(i) Let $\vec{r}$ be the given vector, and let it make an angle $α, β, γ$ with $O X, O Y, O Z$ respectively. Then, its direction cosines are $\cos α, \cos β, \cos γ$ . So, direction ratios of $\vec{r}$ = $\hat{i} - \hat{j} + \hat{k}$ are proportional to $1, - 1, 1$ . Therefore,
Direction cosine of $\vec{r}$ are $\frac{1}{\sqrt{1^{2} + {(- 1)}^{2} + 1^{2}}}, \frac{- 1}{\sqrt{1^{2} + {(- 1)}^{2} + 1^{2}}}, \frac{1}{\sqrt{1^{2} + {(- 1)}^{2} + 1^{2}}}$
or, $\frac{1}{\sqrt{3}}, \frac{- 1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ .
∴ $\cos α = \frac{1}{\sqrt{3}}, \cos β = \frac{- 1}{\sqrt{3}}, \cos γ = \frac{1}{\sqrt{3}}$
$α = \cos^{- 1} (\frac{1}{\sqrt{3}}), β = \cos^{- 1} (\frac{- 1}{\sqrt{3}}), γ = \cos^{- 1} (\frac{1}{\sqrt{3}})$

(ii) Let $\vec{r}$ be the given vector, and let it make an angles $α, β, γ$ with $O X, O Y, O Z$ respectively. Then, its direction cosines are $\cos α, \cos β, \cos γ$ . So, direction ratios of $\vec{r}$ $= \hat{j} - \hat{k}$ are proportional to $0, 1, - 1$ . Therefore, direction cosines of $\vec{r}$ are
$\frac{0}{\sqrt{0 + 1^{2} + {(- 1)}^{2}}}, \frac{1}{\sqrt{0 + 1^{2} + {(- 1)}^{2}}}, \frac{- 1}{\sqrt{0 + 1^{2} + {(- 1)}^{2}}}$ or, $0, \frac{1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}}$

∴ $\cos α = 0, \cos β = \frac{1}{\sqrt{2}}, \cos γ = \frac{- 1}{\sqrt{2}}$
$\Rightarrow α = \cos^{- 1} (0), β = \cos^{- 1} (\frac{1}{\sqrt{2}}), γ = \cos^{- 1} (\frac{- 1}{\sqrt{2}}) \Rightarrow α = \frac{π}{2}, β = \frac{π}{4}, γ = \frac{3 π}{4}$

(iii) Let $\vec{r}$ be the given vector, and let it make an angle $α, β, γ$ with $O X, O Y, O Z$ respectively. Then, its direction cosines are $\cos α, \cos β, \cos γ$ . So, direction ratio of $\vec{r}$ $= 4 \hat{i} + 8 \hat{j} + \hat{k}$ are proportional to $4, 8, 1$
Therefore, direction ratio of $\vec{r}$ are
$\frac{4}{\sqrt{4^{2} + 8^{2} + 1^{2}}}, \frac{8}{\sqrt{4^{2} + 8^{2} + 1^{2}}}, \frac{1}{\sqrt{4^{2} + 8^{2} + 1^{2}}}$ or, $\frac{4}{9}, \frac{8}{9}, \frac{1}{9}$ .

∴ $α = \cos^{- 1} (\frac{4}{9}), β = \cos^{- 1} (\frac{8}{9}), γ = \cos^{- 1} (\frac{1}{9}) .$

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