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Find the angle between the pairs of lines: $ \overrightarrow r = 3\hat i + \hat j -2\hat k + \lambda (\hat i - \hat j -2\hat k)\: and \: \overrightarrow r = 2\hat i - \hat j-56 \hat k+ \mu (3 \hat i -5\hat j - 4\hat k) $

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  • Angle between two line is given by
  • $\cos \theta=\bigg|\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt {{a_1}^2+{b_1}^2+{c_1}^2}\sqrt {{a_2}^2+{b_2}^2+{c_2}^2}}\bigg|$
Let $L_1$ be the line :$\overrightarrow r=3 \hat i+ \hat j-2\hat k+ \lambda ( \hat i- \hat j -2 \hat k)$
Let $L_2$ be the line :$\overrightarrow r=2 \hat i- \hat j-56 \hat k+\mu ( 3\hat i-5 \hat j -4 \hat k)$
Direction ratios of the vector $b_1$ in $L_1$ is $(1,1,-2)$
Let this be $(a_1,b_1,c_1)$
Direction ratios of the vector $b_2$ in $L_2$ is $(3,-5,-4)$
Let this be $(a_2,b_2,c_2)$
we know $\cos \theta=\bigg|\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt {{a_1}^2+{b_1}^2+{c_1}^2}\sqrt {{a_2}^2+{b_2}^2+{c_2}^2}}\bigg|$
Now substituting for $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$
$\cos \theta=\bigg|\large\frac{1 \times 3+-1 \times -5+-2 \times -4}{\sqrt {1^2+1^2+(-2)^2}\sqrt {3^2+(-5)^2+(-4)^2}}\bigg|$
$\cos \theta=\bigg|\large\frac{3+5+8}{\sqrt {6}\sqrt {50}}\bigg|$
$\cos \theta=\bigg|\large\frac{16}{\sqrt {10}\sqrt {3}}\bigg|$
$\cos \theta=\bigg|\large\frac{8}{5\sqrt {3}}\bigg|$
Therefore $\theta=\cos ^{-1} \bigg(\large\frac{8}{5 \sqrt 3}\bigg)$
answered Jun 5, 2013 by meena.p
 

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