A line having direction ratios 1, -4, 2 inter | vedclass

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(A) Let $\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1} = \lambda$. Then $x = 3\lambda + 7, y = 1 - \lambda, z = \lambda - 2$. Let $\frac{x}{2} = \fra

Vedclass · Accepted Answer

$A(-8, 6, -7), B(-6, -2, -3)$

Vedclass · Answer

(A) Let $\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1} = \lambda$. Then $x = 3\lambda + 7, y = 1 - \lambda, z = \lambda - 2$. Let $\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1} = \mu$. Then $x = 2\mu, y = 3\mu + 7, z = \mu$. Coordinates of point $A$ on the first line are $(3\lambda + 7, 1 - \lambda, \lambda - 2)$. Coordinates of point $B$ on the second line are $(2\mu, 3\mu + 7, \mu)$. The direction ratios of line $AB$ are $(3\lambda - 2\mu + 7, -\lambda - 3\mu - 6, \lambda - \mu - 2)$. Since the direction ratios of the line are $1, -4, 2$,we have: $\frac{3\lambda - 2\mu + 7}{1} = \frac{-\lambda - 3\mu - 6}{-4} = \frac{\lambda - \mu - 2}{2}$. From $\frac{3\lambda - 2\mu + 7}{1} = \frac{\lambda + 3\mu + 6}{4}$,we get $12\lambda - 8\mu + 28 = \lambda + 3\mu + 6$,which simplifies to $11\lambda - 11\mu + 22 = 0$,or $\lambda - \mu + 2 = 0$ $(i)$. From $\frac{\lambda + 3\mu + 6}{4} = \frac{\lambda - \mu - 2}{2}$,we get $\lambda + 3\mu + 6 = 2\lambda - 2\mu - 4$,which simplifies to $\lambda - 5\mu - 10 = 0$ $(ii)$. Solving $(i)$ and $(ii)$,we subtract $(ii)$ from $(i)$: $(\lambda - \mu + 2) - (\lambda - 5\mu - 10) = 0$,so $4\mu + 12 = 0$,which gives $\mu = -3$. Substituting $\mu = -3$ into $(i)$,$\lambda - (-3) + 2 = 0$,so $\lambda = -5$. Thus,$A = (3(-5) + 7, 1 - (-5), -5 - 2) = (-8, 6, -7)$ and $B = (2(-3), 3(-3) + 7, -3) = (-6, -2, -3)$.

$A$ line having direction ratios $1, -4, 2$ intersects the lines $\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1}$ and $\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1}$ at the points $A$ and $B$ respectively. Then,the coordinates of points $A$ and $B$ are:

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