Question

A third plane can be found that passes through the line of intersection of two existing planes.?
a. Two planes are given by the equations 2x − 3y + z − 6 = 0 and -3x + 5y + 4z + 6 = 0. Find the scalar equation of the plane that passes through the line of intersection of these two planes, and also passes through the point (1, 3, -1).

b. Give the equations of two planes. Create a third plane that passes through the line of intersection of the original two and which is parallel to the z-axis. Explain your reasoning

Answer 1

A third plane can be found that passes through the line of intersection of two existing planes.?

- Two planes are given by the equations $2x-3y+z-6=0$ and $-3x+5y+4z+6=0$. Find the scalar equation of the plane that passes through the line of intersection of these two planes, and also passes through the point $A=(1,3,-1)$.

- Give the equations of two planes. Create a third plane that passes through the line of intersection of the original two and which is parallel to the $z$-axis. Explain your reasoning

Solution

a) At first, find equation of the line of intersection of two planes.

$2x-3y+z-6=-3x+5y+4z+6=0$

$6x-9y+3z-18=6x-10y-8z-12$

$y+11z-6=0$

Let $z=\lambda$. Then the general solution of equation

$\begin{cases}2x-3y+z-6=0\cr y=6-11z\end{cases}$

is

$\begin{cases}x=12-17\lambda\cr y=6-11\lambda\cr z=\lambda\end{cases}$

Then find two points on this line, e.g. $B=(12,6,0)$ and

$C=(-5,-5,1)$. Let

$\vec{AB}\,=(11,3,1)$

$$\vec {A C} = (- 6, - 8, 2)$$

. The normal of the two vectors is:

$$n = \vec {A B} \times \vec {A C} = \left| \begin{array}{c c c} \mathbf {i} & \mathbf {j} & \mathbf {k} \\ 1 1 & 3 & 1 \\ - 6 & - 8 & 2 \end{array} \right| =$$ $$\mathbf {i} \left| \begin{array}{l l} 3 & 1 \\ - 8 & 2 \end{array} \right| - \mathbf {j} \left| \begin{array}{l l} 1 1 & 1 \\ - 6 & 2 \end{array} \right| + \mathbf {k} \left| \begin{array}{l l} 1 1 & 3 \\ - 6 & - 8 \end{array} \right| = 1 4 \mathbf {i} - 2 8 \mathbf {j} - 7 0 \mathbf {k}$$

The general equation of a plane is:

$$n \cdot (x - x _ {0}, y - y _ {0}, z - z _ {0}) = 0$$ $$(1 4, - 2 8, - 7 0) \cdot (x - 1, y - 3, z + 1) = 0$$ $$1 4 (x - 1) - 2 8 (y - 3) - 7 0 (z + 1) = 0$$ $$1 4 x - 2 8 y - 7 0 z = 0$$

b) Equation of the plane which is parallel to the $z$-axis looks like $ax + by + d = 0$. The equation of the line of intersection and points $B$ and $C$ are the same as in previous task. So far as points $B = (12,6,0)$ and $C = (-5, -5,1)$ belong to this plane:

$$\left\{ \begin{array}{l} 1 2 a + 6 b + d = 0 \\ - 5 a - 5 b + d = 0 \end{array} \right.$$ $$\left\{ \begin{array}{l} 1 7 a + 1 1 b = 0 \\ - 5 a - 5 b + d = 0 \end{array} \right.$$

Let $a = t$. The general solution of this equation is:

$$\left\{ \begin{array}{l} a = t \\ b = - \frac {1 7 t}{1 1} \\ d = - \frac {3 0 t}{1 1} \end{array} \right.$$

The general equation of this plane is

$$t x - \frac {1 7 t y}{1 1} - \frac {3 0 t}{1 1} = 0 \Rightarrow 1 1 x - 1 7 y - 3 0 = 0$$

Answer

a) $14x - 28y - 70z = 0$

b) $11x - 17y - 30 = 0$

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