10. Find the angle between the following pairs of lines:

     (ii)      \overrightarrow{r}= 3\widehat{i}+\widehat{j}-2\widehat{k}+\lambda (\widehat{i}-\widehat{j}-2\widehat{k}) and \overrightarrow{r}= 2\widehat{i}-\widehat{j}-56\widehat{k}+\mu (3\widehat{i}-5\widehat{j}-4\widehat{k})

Answers (1)

To find the angle A between the pair of lines \vec{b_{1}}\ and\ \vec{b_{2}} we have the formula;

\cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |

We have two lines :

\overrightarrow{r}= 3\widehat{i}+\widehat{j}-2\widehat{k}+\lambda (\widehat{i}-\widehat{j}-2\widehat{k}) and

\overrightarrow{r}= 2\widehat{i}-\widehat{j}-56\widehat{k}+\mu (3\widehat{i}-5\widehat{j}-4\widehat{k})

The given lines are parallel to the vectors \vec{b_{1}}\ and\ \vec{b_{2}};

where \vec{b_{1}}= \widehat{i}-\widehat{j}-2\widehat{k}   and   \vec{b_{2}}= 3\widehat{i}-5\widehat{j}-4\widehat{k} respectively,

Then we have

\vec{b_{1}}.\vec{b_{2}} =(\widehat{i}-\widehat{j}-2\widehat{k}).(3\widehat{i}-5\widehat{j}-4\widehat{k})

=3+5+8 = 16

and |\vec{b_{1}}| = \sqrt{1^2+(-1)^2+(-2)^2} = \sqrt{6}

|\vec{b_{2}}| = \sqrt{3^2+(-5)^2+(-4)^2} = \sqrt{50} = 5\sqrt2

Therefore we have;

\cos A = \left | \frac{16}{\sqrt6 \times5\sqrt2} \right | = \frac{16}{10\sqrt3}

or A = \cos^{-1} \left ( \frac{8}{5\sqrt3} \right )

 

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