Question

Find the equation of the plane through (2,3,-4) and (1,-1,3) and parallel to x-axis?

Answer 1

Answer on Question #83677 – Math – Analytic Geometry

Question

Find the equation of the plane through (2,3,-4) and (1,-1,3) and parallel to x-axis?

Solution

Assume that $Ax + By + Cz + D = 0$ is an equation of the plane.

The plane is parallel to x-axis, hence

$$A = 0 \quad (1)$$

The points (2,3,-4), (1,-1,3) lie on the plane, hence

$$\left\{ \begin{array}{l} 3B - 4C + D = 0, \\ -B + 3C + D = 0. \end{array} \right.$$ $$\left\{ \begin{array}{l} D = 4C - 3B, \\ D = B - 3C. \end{array} \right.$$ $$4C - 3B = B - 3C,$$ $$-4B = -7C,$$ $$C = \frac{4}{7}B, \quad (2)$$ $$D = 4C - 3B = 4 * \frac{4}{7}B - 3B = \frac{16}{7}B - 3B = \frac{16 - 21}{7}B = -\frac{5}{7}B,$$ $$D = -\frac{5}{7}B, \quad (3)$$ $$By + cz + d = 0,$$ $$By + \frac{4}{7}Bz - \frac{5}{7}B = 0, \quad (4)$$

If

$$B = 0, \quad (5)$$

then it follows from (2), (3), (5) that

$$C = 0, \quad D = 0 \quad (6)$$

In case of (5) taking (1), (5), (6) into account one gets that all coefficients in the equation

$$Ax + By + Cz + D = 0$$

are zero, which is impossible, therefore

$$B \neq 0 \quad (7)$$

Using (7) divide the equation (4) through by $B \neq 0$.

Then

$$7y + 4z - 5 = 0$$

is the equation of the plane through $(2,3,-4)$, $(1,-1,3)$ and parallel to x-axis.

Answer: $7y + 4z - 5 = 0$.

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