Question

Find the equation of the plane through the line x-2÷2=y-3÷3=z-4÷5 and parallel to x-axis?

Answer 1

Answer on Question #83693 – Math – Analytic Geometry

Question

Find the equation of the plane through the line $\frac{x - 2}{2} = \frac{y - 3}{3} = \frac{z - 4}{5}$ and parallel to x-axis?

Solution

From the equation of line

$$\frac {x - x _ {1}}{x _ {2} - x _ {1}} = \frac {y - y _ {1}}{y _ {2} - y _ {1}} = \frac {z - z _ {1}}{z _ {2} - z _ {1}},$$

we have

$$x _ {2} - 2 = 2;$$ $$y _ {2} - 3 = 3;$$ $$z _ {2} - 4 = 5;$$ $$x _ {2} = 4.$$ $$y _ {2} = 6.$$ $$z _ {2} = 9.$$

Because the line lies on the plane, the plane also contains points (2, 3, 4) and (4, 6, 9).

Consider the plane equation

$$A x + B y + C z + D = 0$$

or

$$\frac {A}{D} x + \frac {B}{D} y + \frac {C}{D} z + 1 = 0 \text{ if } D \neq 0$$

$(A = 0$ because the plane parallel to x-axis) make such a system:

$$\left\{ \begin{array}{l} 3 \frac {B}{D} + 4 \frac {C}{D} + 1 = 0; \\ 6 \frac {B}{D} + 9 \frac {C}{D} + 1 = 0. \end{array} \right.$$ $$A = 0, \frac {B}{D} = - \frac {5}{3} \text{ and } \frac {C}{D} = 1$$

From (3), (4) it follows that

$$- \frac {5}{3} y + z + 1 = 0 \text{ if } D \neq 0, \text{ hence } 5 y - 3 z - 3 = 0.$$

Answer: $5y - 3z - 3 = 0$ is the equation of the plane.

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