Question

Find the angle between the pair of lines given below.

(x + 3)/3 = (y -1)/5 = (z + 3)/4

(x + 1)/1 = (y – 4)/1 = (z – 5)/2

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Answer 1

Solution

$\bf \: Given$

$\tt \to \: \dfrac{(x + 3)}{3} = \dfrac{(y - 1)}{5} = \dfrac{(z + 3)}{4}$

$\tt \to \: \dfrac{(x + 1)}{1} = \dfrac{(y - 4)}{1} = \dfrac{(z - 5)}{2}$

$\bf \: Formula$

$\tt \to \: cos \theta \: = \bigg| \dfrac{a_{1} a_2 + b_1b_2 + c_1c_2}{ \sqrt{a_1 {}^{2} + b_1 {}^{2} + c_1 {}^{2} } \sqrt{a_2 {}^{2} + b_2 {}^{2} + c_2 {}^{2}}} \bigg|$

$\bf \: Now \: we \: get \: from \: above \: equation$

$\tt \to \: a_1 = 3 ,\: b_1 = 5, \: and \: c_1 = 4 \\ \tt \to \: a_2 = 1 ,\: b_2 = 1 ,\: and \: c_2 = 2$

$\bf \: Now \: put \: the \: value \: on \: formula$

$\tt \to \: cos \theta = \bigg| \dfrac{3 \times 1 + 5 \times 1 + 4 \times 2}{ \sqrt{(3) {}^{2} + (5) {}^{2} + (4) {}^{2} } \sqrt{(1) {}^{2} + (1) {}^{2} + (2) {}^{2} } } \bigg|$

$\tt \to \: cos \theta = \bigg |\dfrac{3 + 5 + 8}{ \sqrt{9 + 25 + 16} \sqrt{1 + 1 + 4} } \bigg |$

$\tt \to \: cos \theta \: = \bigg| \dfrac{16}{ \sqrt{50} \sqrt{6} } \bigg |$

$\tt \to \: cos \theta = \bigg| \dfrac{16}{ \sqrt{50 \times 6} } \bigg|$

$\tt \to \: cos \theta = \bigg| \dfrac{16}{ \sqrt{300} } \bigg|$

$\tt \to \: cos \theta = \dfrac{16}{ 10\sqrt{3} }$

$\tt \to \: cos \theta = \dfrac{8}{ 5\sqrt{3} }$

$\bf \: Answer$

$\tt \to \: \theta = cos {}^{ - 1} \bigg(\dfrac{8}{ 5\sqrt{3} } \bigg)$

Answer 2

Question :-

Find the angle between the pair of lines given below -

$\displaystyle \sf \: \frac{(x + 3) }{3} = \frac{(y - 1)}{5} = \frac{(z + 3)}{4} \\ \\ \sf \: \frac{(x + 1)}{1} = \frac{(y - 4)}{1} = \frac{(z - 5)}{2}$

Solution :-

Comparing both pair of lines with

$\displaystyle \sf \frac{( {x - x_{1}) } }{a} = \frac{(y - y_{1})}{b} = \frac{(z - z_{1})}{c}$

$\star \sf \: a_{1} = 3 \: \: \: b_{1} = 5 \: \: \: c_{1} = 4 \\ \\ \star \sf \:a_{2} = 1 \: \: \: b_{2} = 1 \: \: \:c_{2} = 2$

Now we have the formula to find the angle -

$\displaystyle \boxed{\star \: \sf \cos( \theta) = | \frac{ a_{1}^{2} a_{2}^{2} \: + \: b_{1}^{2} b_{2}^{2} \: + \: c_{1}^{2} c_{2}^{2} }{ \sqrt{a_{1}^{2} \: + \: b_{1}^{2} \: + \: c_{1}^{2} \: } \times \sqrt{a_{2}^{2} \: + \: b_{2}^{2} \: + \: c_{2}^{2}} } | }$

So,

$\mapsto \sf \: \cos( \theta) = \bigg| \frac{ {(3)} \times {(1)}+ {(5)}\times ( {1)} + ( {4)} \times ( {2)}}{ \sqrt{( {3)}^{2} + ( {5)}^{2} + ( {4)}^{2} } \times \sqrt{ {(1)}^{2} + ( {1)}^{2} + ( {2)}^{2} } } \bigg| \\ \\ \mapsto \sf \: \cos( \theta) = \bigg| \frac{3 \times 1 + 5 \times 1 + 4 \times 2}{ \sqrt{9+ 25 + 16} \times \sqrt{1 + 1 + 4} } \bigg| \\ \\ \mapsto \sf \: \cos( \theta) = \bigg| \frac{3+ 5 + 8}{ \sqrt{50} \times \sqrt{6} } \bigg| \\ \\ \mapsto \sf \: \cos( \theta) = \bigg| \frac{16}{ \sqrt{300} } \bigg| \\ \\ \mapsto \sf \: \cos( \theta) = \frac{16}{10 \sqrt{3} } \\ \\ \sf \theta = {cos}^{ - 1} \frac{16}{10 \sqrt{3} }$