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Find the angle between the following pairs of lines: (i) r vector = 2 i caret - 5 j caret + k caret + lambda 3 i caret + 2 j caret + 6 k caret

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

      (i)     \overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda (3\widehat{i}+2\widehat{j}+6\widehat{k}) and \overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu (\widehat{i}+2\widehat{j}+2\widehat{k})

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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}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda (3\widehat{i}+2\widehat{j}+6\widehat{k}) and

\overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu (\widehat{i}+2\widehat{j}+2\widehat{k})

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

where \vec{b_{1}}= 3\widehat{i}+2\widehat{j}+6\widehat{k}   and   \vec{b_{2}}= \widehat{i}+2\widehat{j}+2\widehat{k} respectively,

Then we have

\vec{b_{1}}.\vec{b_{2}} =(3\widehat{i}+2\widehat{j}+6\widehat{k}).(\widehat{i}+2\widehat{j}+2\widehat{k})

=3+4+12 = 19

and |\vec{b_{1}}| = \sqrt{3^2+2^2+6^2} = 7

|\vec{b_{2}}| = \sqrt{1^2+2^2+2^2} = 3

Therefore we have;

\cos A = \left | \frac{19}{7\times3} \right | = \frac{19}{21}

or A = \cos^{-1} \left ( \frac{19}{21} \right )

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