If lines frac{x-1}{-3}=frac{y-2}{2k}=frac{z-3 | vedclass

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(A) The given lines are $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$.The direction ratios of the fi

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$-\frac{10}{7}$

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(A) The given lines are $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$.The direction ratios of the first line are $\vec{b_1} = -3\hat{i} + 2k\hat{j} + 2\hat{k}$.The direction ratios of the second line are $\vec{b_2} = 3k\hat{i} + 1\hat{j} - 5\hat{k}$.Since the lines are mutually perpendicular,their dot product must be zero,i.e.,$\vec{b_1} \cdot \vec{b_2} = 0$.$(-3)(3k) + (2k)(1) + (2)(-5) = 0$.$-9k + 2k - 10 = 0$.$-7k - 10 = 0$.$-7k = 10$.$k = -\frac{10}{7}$.Thus,the value of $k$ is $-\frac{10}{7}$.

If lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are mutually perpendicular,then $k$ is equal to

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