(3, 0, 2) and (0, 2, k) are the direction rat | vedclass

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(B) Let the direction ratios of the two lines be $\vec{a} = (3, 0, 2)$ and $\vec{b} = (0, 2, k)$.The formula for the cosine of the angle $	heta$ betw

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$\pm 3$

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(B) Let the direction ratios of the two lines be $\vec{a} = (3, 0, 2)$ and $\vec{b} = (0, 2, k)$.The formula for the cosine of 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 $|\cos 	heta| = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.Substituting the given values: $|\cos 	heta| = \frac{|(3)(0) + (0)(2) + (2)(k)|}{\sqrt{3^2 + 0^2 + 2^2} \sqrt{0^2 + 2^2 + k^2}} = \frac{|2k|}{\sqrt{13} \sqrt{4 + k^2}}$.Given $|\cos 	heta| = \frac{6}{13}$,we have $\frac{6}{13} = \frac{|2k|}{\sqrt{13} \sqrt{4 + k^2}}$.Squaring both sides: $\frac{36}{169} = \frac{4k^2}{13(4 + k^2)}$.Simplifying: $\frac{9}{13} = \frac{k^2}{4 + k^2}$.$9(4 + k^2) = 13k^2 \Rightarrow 36 + 9k^2 = 13k^2 \Rightarrow 4k^2 = 36 \Rightarrow k^2 = 9$.Thus,$k = \pm 3$.

$(3, 0, 2)$ and $(0, 2, k)$ are the direction ratios of two lines and $\theta$ is the angle between them. If $|\cos \theta| = \frac{6}{13}$,then $k =$

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