Question

Obtain the equation of the line passing through (1,-1,2) having direction ratios (2,0,1)

Answer 1

Answer on Question #75326 – Math – Analytic Geometry

Question

1. Obtain the equation of the line passing through $(1,-1,2)$ having direction ratios $(2,0,1)$.

Solution

The symmetric form of the equation of a line passing through a point $M(x_0, y_0, z_0)$ and having direction ratios $a$, $b$ and $c$ is:

$$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$$

The parametric form of the equation of a line passing through a point $M(x_0, y_0, z_0)$ and having direction ratios $a$, $b$ and $c$ is:

$$\left\{ \begin{array}{l} x = x_0 + a t \\ y = y_0 + b t \\ z = z_0 + c t, \quad t \in R \end{array} \right.$$

Putting the values of $x_0, y_0, z_0$ and $a, b, c$ in (1) and (2) we get the equation of the line passing through $(1,-1,2)$ having direction ratios $(2,0,1)$:

$$\frac{x - 1}{2} = \frac{y + 1}{0} = \frac{z - 2}{1} \leftrightarrow \left\{ \begin{array}{l} x = 1 + 2 t \\ y = -1 \\ z = 2 + t \end{array} \right., \qquad t \in R$$

Answer:

$$\frac{x - 1}{2} = \frac{y + 1}{0} = \frac{z - 2}{1} \leftrightarrow \left\{ \begin{array}{l} x = 1 + 2 t \\ y = -1 \\ z = 2 + t \end{array} \right., \qquad t \in R$$

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