A line with direction ratios 2, 1, 2 intersec | vedclass

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Question

(A) Let the two given lines be $L_1: \frac{x}{1} = \frac{y+a}{1} = \frac{z}{1} = \lambda$ and $L_2: \frac{x+a}{2} = \frac{y}{1} = \frac{z}{1} = \mu$.A

Vedclass · Accepted Answer

$(3a, 2a, 3a), (a, a, a)$

Vedclass · Answer

(A) Let the two given lines be $L_1: \frac{x}{1} = \frac{y+a}{1} = \frac{z}{1} = \lambda$ and $L_2: \frac{x+a}{2} = \frac{y}{1} = \frac{z}{1} = \mu$.Any point on $L_1$ is $P = (\lambda, \lambda - a, \lambda)$ and any point on $L_2$ is $Q = (2\mu - a, \mu, \mu)$.The direction ratios of the line $PQ$ are $(2\mu - a - \lambda, \mu - (\lambda - a), \mu - \lambda)$,which simplifies to $(2\mu - a - \lambda, \mu - \lambda + a, \mu - \lambda)$.Since the line $PQ$ has direction ratios proportional to $2, 1, 2$,we have:$\frac{2\mu - a - \lambda}{2} = \frac{\mu - \lambda + a}{1} = \frac{\mu - \lambda}{2}$.From $\frac{\mu - \lambda + a}{1} = \frac{\mu - \lambda}{2}$,we get $2\mu - 2\lambda + 2a = \mu - \lambda$,which implies $\mu - \lambda = -2a$.Substituting this into $\frac{2\mu - a - \lambda}{2} = \frac{\mu - \lambda}{2}$,we get $2\mu - a - \lambda = \mu - \lambda$,so $\mu = a$.Then,$a - \lambda = -2a$,which gives $\lambda = 3a$.Substituting $\lambda = 3a$ into $P$,we get $P = (3a, 3a - a, 3a) = (3a, 2a, 3a)$.Substituting $\mu = a$ into $Q$,we get $Q = (2a - a, a, a) = (a, a, a)$.Thus,the points of intersection are $(3a, 2a, 3a)$ and $(a, a, a)$.

$A$ line with direction ratios $2, 1, 2$ intersects the lines $x = y + a = z$ and $x + a = 2y = 2z$. Find the coordinates of the points of intersection.

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