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# Find the angle between the pairs of lines: $\overrightarrow r = 2\hat i - 5 \hat j + \hat k + \lambda (3\hat i + 2\hat j + 6\hat k)\: and \: \overrightarrow r = 7\hat i - 6 \hat k+ \mu (\hat i + 2\hat j + 2\hat k)$

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• Angle between two line is given by
• $\cos \theta=\bigg|\large\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt {{a_1}^2+{b_1}^2+{c_1}^2}\sqrt {{a_2}^2+{b_2}^2+{c_2}^2}}\bigg|$
Angle between the lines 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 :$\overrightarrow r=2 \hat i-5 \hat j+\hat k+ \lambda (3 \hat i+2 \hat j +6 \hat k)$
Let $L_2$ be the line :$\overrightarrow r=7 \hat i-6 \hat k+\mu ( \hat i+2 \hat j +2 \hat k)$
Direction ratios of the vector $b_1$ in $L_1$ is $(3,2,6)$
Let this be $(a_1,b_1,c_1)$
Direction ratios of the vector $b_2$ in $L_2$ is $(1,2,2)$
Let this be $(a_2,b_2,c_2)$
Now substituting for $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$
$\cos \theta=\bigg|\large\frac{3 \times 1+2 \times 2+6 \times 2}{\sqrt {3^2+2^2+6^2}\sqrt {1^2+2^2+2^2}}\bigg|$
$\cos \theta=\bigg|\large\frac{3 +4+12}{\sqrt {9+4+36}\sqrt {1+4+4}}\bigg|$
$\cos \theta=\bigg|\large\frac{19}{\sqrt {49}\sqrt {9}}\bigg|$
$\cos \theta=\bigg|\large\frac{19}{21}\bigg|$
Therefore $\theta=\cos ^{-1} \bigg(\large\frac{19}{21}\bigg)$
edited Jun 5, 2013 by meena.p

+1 vote