If (2, -1, 2) and (K, -3, -5) are the triads | vedclass

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(C) Given the direction ratios of two lines are $(a_1, b_1, c_1) = (2, -1, 2)$ and $(a_2, b_2, c_2) = (K, -3, -5)$, and the angle between them is $	h

Vedclass · Accepted Answer

$7K^2 - 112K - 110 = 0$

Vedclass · Answer

(C) Given the direction ratios of two lines are $(a_1, b_1, c_1) = (2, -1, 2)$ and $(a_2, b_2, c_2) = (K, -3, -5)$, and the angle between them is $	heta = 60^{\circ}$.Using the formula for the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$:$\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$Substituting the given values:$\cos 60^{\circ} = \frac{|2K + (-1)(-3) + 2(-5)|}{\sqrt{2^2 + (-1)^2 + 2^2} \sqrt{K^2 + (-3)^2 + (-5)^2}}$$\frac{1}{2} = \frac{|2K + 3 - 10|}{\sqrt{4 + 1 + 4} \sqrt{K^2 + 9 + 25}}$$\frac{1}{2} = \frac{|2K - 7|}{3 \sqrt{K^2 + 34}}$$3 \sqrt{K^2 + 34} = 2 |2K - 7|$Squaring both sides:$9(K^2 + 34) = 4(2K - 7)^2$$9K^2 + 306 = 4(4K^2 - 28K + 49)$$9K^2 + 306 = 16K^2 - 112K + 196$$7K^2 - 112K - 110 = 0$

If $(2, -1, 2)$ and $(K, -3, -5)$ are the triads of direction ratios of two lines and the angle between the lines is $60^{\circ}$, then

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