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

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(C) Let the direction ratios of the two lines be $\vec{a} = (2, -1, 2)$ and $\vec{b} = (K, 3, 5)$.Given the angle between the lines is $	heta = 45^{\

Vedclass · Accepted Answer

$4$

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(C) Let the direction ratios of the two lines be $\vec{a} = (2, -1, 2)$ and $\vec{b} = (K, 3, 5)$.Given the angle between the lines is $	heta = 45^{\circ}$.The formula for the cosine of the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos 	heta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.Substituting the values:$\cos 45^{\circ} = \frac{|(2)(K) + (-1)(3) + (2)(5)|}{\sqrt{2^2 + (-1)^2 + 2^2} \sqrt{K^2 + 3^2 + 5^2}}$$\frac{1}{\sqrt{2}} = \frac{|2K - 3 + 10|}{\sqrt{4 + 1 + 4} \sqrt{K^2 + 9 + 25}}$$\frac{1}{\sqrt{2}} = \frac{|2K + 7|}{3 \sqrt{K^2 + 34}}$$3 \sqrt{K^2 + 34} = \sqrt{2} |2K + 7|$Squaring both sides:$9(K^2 + 34) = 2(2K + 7)^2$$9K^2 + 306 = 2(4K^2 + 28K + 49)$$9K^2 + 306 = 8K^2 + 56K + 98$$K^2 - 56K + 208 = 0$Factoring the quadratic equation:$(K - 4)(K - 52) = 0$Thus,$K = 4$ or $K = 52$.Since $4$ is one of the given options,the correct value is $K = 4$.

If $(2, -1, 2)$ and $(K, 3, 5)$ are the triads of direction ratios of two lines and the angle between them is $45^{\circ}$,then the value of $K$ is

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