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Question

(B) The formula for 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 = \lef

Vedclass · Accepted Answer

$\cos^{-1}\left(\frac{18\sqrt{2}}{35}\right)$

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(B) The formula for 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:$\cos 	heta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} ight|$Given direction ratios are $(2, 3, -6)$ and $(3, -4, 5)$.Substituting these values into the formula:$\cos 	heta = \left| \frac{(2)(3) + (3)(-4) + (-6)(5)}{\sqrt{2^2 + 3^2 + (-6)^2} \sqrt{3^2 + (-4)^2 + 5^2}} ight|$Calculating the numerator:$2(3) + 3(-4) + (-6)(5) = 6 - 12 - 30 = -36$Calculating the denominators:$\sqrt{4 + 9 + 36} = \sqrt{49} = 7$$\sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$Thus,$\cos 	heta = \left| \frac{-36}{7 	imes 5\sqrt{2}} ight| = \frac{36}{35\sqrt{2}} = \frac{18\sqrt{2}}{35}$Therefore,the acute angle $	heta = \cos^{-1}\left(\frac{18\sqrt{2}}{35}ight)$.

If the direction ratios of two lines are proportional to $(2, 3, -6)$ and $(3, -4, 5)$,then the acute angle between them is

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