Question

Let R be the point which divides the line segment joining P(2,1,0) and Q(-1,3,4) in the ratio 1:2 such that PR < PQ.
Find the equation of the line passing through R and parallel to the line x/2=y/1=z/3 .

Answer 1

Answer on Question #45092 – Math - Analytic Geometry

Problem.

Let $R$ be the point which divides the line segment joining $P(2,1,0)$ and $Q(-1,3,4)$ in the ratio 1:2 such that $PR < PQ$.

Find the equation of the line passing through $R$ and parallel to the line $x/2 = y/1 = z/3$.

Solution.

Suppose that $R$ has coordinates $(a, b, c)$. Then $\overrightarrow{PQ}$ has coordinates $(-3,2,4)$ and $\overrightarrow{PR}$ has coordinates $(a - 2, b - 1, c)$. $\overrightarrow{PR} = \frac{1}{3}\overrightarrow{PQ}$, so $\frac{1}{3}(-3,2,4) = (a - 2, b - 1, c)$. Therefore $a = 1$, $b = \frac{5}{3}$, $c = \frac{4}{3}$.

The line parallel to $\frac{x}{2} = \frac{y}{1} = \frac{z}{3}$ has direction vector $(2,1,3)$. Hence the equation of the line passing through $R$ and parallel to the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{3}$ is $\frac{x - 1}{2} = \frac{y - \frac{5}{3}}{1} = \frac{z - \frac{4}{3}}{3}$, i.e. $\frac{x - 1}{2} = \frac{3y - 5}{3} = \frac{3z - 4}{9}$.

Answer: $\frac{x - 1}{2} = \frac{3y - 5}{3} = \frac{3z - 4}{9}$.

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