Find the angle between the lines whose direction ratios are a, b,… - Vidyadip

Question

Find the angle between the lines whose direction ratios are a, b, c and b - c, c - a, a - b.

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Answer

Direction ratios of one line are a, b, c
$\Rightarrow $ A vector along this line is ${\vec b_1} = a\hat i + b\hat j + c\hat k$
Direction ratios of second line are $b - c,c - a,a - b$
$\Rightarrow$ A vector along second line is ${\vec b_2} = \left( {b - c} \right)\hat i + \left( {c - a} \right)\hat j + \left( {a - b} \right)\hat k$
Let $\theta $ be the angle between the two lines, then
$\cos \theta = \frac{{\left| {{{\vec b}_1}.{{\vec b}_2}} \right|}}{{\left| {{{\vec b}_1}} \right|.\left| {{{\vec b}_2}} \right|}} = \frac{{a\left( {b - c} \right) + b\left( {c - a} \right) + c\left( {a - b} \right)}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2} + {{\left( {a - b} \right)}^2}} }}$
$= \frac{{ab - ac + bc - ab + ac - bc}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2} + {{\left( {a - b} \right)}^2}} }} = 0 = \cos {90^0}$
$\Rightarrow \theta = {90^0}$

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