If the direction ratios of two lines are 5, - | vedclass

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(A) Let the direction ratios of the two lines be $a_1, b_1, c_1 = 5, -12, 13$ and $a_2, b_2, c_2 = -3, 4, 5$.The angle $	heta$ between two lines with

Vedclass · Accepted Answer

$\cos^{-1}(1/65)$

Vedclass · Answer

(A) Let the direction ratios of the two lines be $a_1, b_1, c_1 = 5, -12, 13$ and $a_2, b_2, c_2 = -3, 4, 5$.The angle $	heta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by:$\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$Calculating the numerator:$a_1 a_2 + b_1 b_2 + c_1 c_2 = (5)(-3) + (-12)(4) + (13)(5) = -15 - 48 + 65 = 2$Calculating the denominators:$\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{5^2 + (-12)^2 + 13^2} = \sqrt{25 + 144 + 169} = \sqrt{338} = 13\sqrt{2}$$\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{(-3)^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$Therefore,$\cos 	heta = \frac{2}{(13\sqrt{2})(5\sqrt{2})} = \frac{2}{65 	imes 2} = \frac{1}{65}$Thus,$	heta = \cos^{-1}(1/65)$.

If the direction ratios of two lines are $5, -12, 13$ and $-3, 4, 5$,then the angle between them is:

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