Question

Find the angles made by the following vectors with the coordinate axes:
(i) $(\hat{i}-\hat{j}+\hat{k})$ (ii) $(\hat{j}-\hat{k})$ (iii) $(\hat{i}-4\hat{j}+5\hat{k})$

Answer 1

$\left(i\right)$ Vector is $(\hat{i}-\hat{j}+\hat{k})$

Here, direction of cosine are $=\frac{1}{\sqrt{(1{)}^2+(-1{)}^2+(1{)}^2}},$ $\frac{-1}{\sqrt{(1{)}^2+(-1{)}^2+(1{)}^2},}$ $\frac{1}{\sqrt{(1{)}^2+(-1{)}^2+(1{)}^2},}$

$=\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

$\therefore$ $\cos \alpha =\frac{1}{\sqrt{3}},\cos \beta =\frac{-1}{\sqrt{3}},\cos \gamma =\frac{1}{\sqrt{3}}$

$\therefore$ $\alpha ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right),\beta ={\cos }^{-1}\left(\frac{-1}{\sqrt{3}}\right),\gamma ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\left(ii\right)$ Vector is $(\hat{j}-\hat{k})$

Here, direction of cosine are $=\frac{0}{\sqrt{(0{)}^2+(1{)}^2+(-1{)}^2}},$ $\frac{1}{\sqrt{(0{)}^2+(1{)}^2+(-1{)}^2},}$ $\frac{-1}{\sqrt{(0{)}^2+(1{)}^2+(-1{)}^2},}$

$=0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}}$

$\therefore$ $\cos \alpha =,0\cos \beta =\frac{1}{\sqrt{2}},\cos \gamma =\frac{-1}{\sqrt{2}}$

$\alpha ={\cos }^{-1}\left(0\right)=\frac{\pi }{2}$

$\beta ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}$

$\gamma ={\cos }^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3\pi }{4}$

$\left(iii\right)$ $(\hat{i}-4\hat{j}+5\hat{k})$

Here, direction of cosine are $=\frac{1}{\sqrt{(1{)}^2+(-4{)}^2+(5{)}^2}},$ $\frac{-4}{\sqrt{(1{)}^2+(-4{)}^2+(5{)}^2},}$ $\frac{5}{\sqrt{(1{)}^2+(-4{)}^2+(5{)}^2},}$

$=\frac{1}{\sqrt{42}},\frac{-4}{\sqrt{42}},\frac{5}{\sqrt{42}}$

$\cos \alpha =\frac{1}{\sqrt{42}},\cos \beta =\frac{-4}{\sqrt{42}},\cos \gamma =\frac{5}{\sqrt{42}}$

$\alpha ={\cos }^{-1}\left(\frac{1}{\sqrt{42}}\right),\beta ={\cos }^{-1}\left(\frac{-4}{\sqrt{42}}\right),\gamma ={\cos }^{-1}\left(\frac{5}{\sqrt{42}}\right)$