Lines frac{1-x}{3}=frac{y-2}{1}=frac{z-1}{2} | vedclass

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(C) The given lines are $\frac{x-1}{-3} = \frac{y-2}{1} = \frac{z-1}{2}$ and $\frac{x-2}{p} = \frac{y-1}{2} = \frac{z-2}{1}$.The direction ratios of t

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$\frac{4}{3}$

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(C) The given lines are $\frac{x-1}{-3} = \frac{y-2}{1} = \frac{z-1}{2}$ and $\frac{x-2}{p} = \frac{y-1}{2} = \frac{z-2}{1}$.The direction ratios of the first line are $\vec{a_1} = (-3, 1, 2)$.The direction ratios of the second line are $\vec{a_2} = (p, 2, 1)$.Since the lines are mutually perpendicular,the dot product of their direction ratios must be zero:$\vec{a_1} \cdot \vec{a_2} = 0$$(-3)(p) + (1)(2) + (2)(1) = 0$$-3p + 2 + 2 = 0$$-3p + 4 = 0$$3p = 4$$p = \frac{4}{3}$.

Lines $\frac{1-x}{3}=\frac{y-2}{1}=\frac{z-1}{2}$ and $\frac{x-2}{p}=\frac{y-1}{2}=\frac{z-2}{1}$ are mutually perpendicular to each other,then $p=$ . . . . . . .

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