The angle between the lines with direction ra | vedclass

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

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the direction ratios of the two lines be $a_1, b_1, c_1 = (3, 4, 5)$ and $a_2, b_2, c_2 = (4, -3, 5)$.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 the formula:$\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$Substituting the values:$a_1 a_2 + b_1 b_2 + c_1 c_2 = (3)(4) + (4)(-3) + (5)(5) = 12 - 12 + 25 = 25$$\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$$\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{4^2 + (-3)^2 + 5^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$Therefore,$\cos 	heta = \frac{25}{(5\sqrt{2})(5\sqrt{2})} = \frac{25}{25 	imes 2} = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,we have $	heta = 60^o$.

The angle between the lines with direction ratios $3, 4, 5$ and $4, -3, 5$ is equal to ......... $^o$.

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