Answer to Question #209806 in Analytic Geometry for Faith

Question

Determine in each case whether the given planes are parallel or perpendicular:

1.x+y+3z+10=0 and x+2y-z=1,

2.3x-2y+z-6=0 and 4x+2y-4z=0,

3.3x+y+z-1=0 and -x+2y+z+3=0,

4.x-3y+z+1=0 and 3x-4y+z-1=0.

Answer 1

1.x+y+3z+10=0 and x+2y-z=1

$$\vec n_1=\langle1,1,3\rangle, \vec n_2=\langle1,2,-1\rangle$$ $$\dfrac{1}{1}\not=\dfrac{2}{1}$$

The vectors $\vec n_1$ and $\vec n_2$ are not collinear.

$$\vec n_1\cdot\vec n _2=1(1)+1(2)+3(-1)=0$$

The vectors $\vec n_1$ and $\vec n_2$ are orthogonal.

Hence, the given planes are perpendicular.

2.3x-2y+z-6=0 and 4x+2y-4z=0

$$\vec n_1=\langle3,-2,1\rangle, \vec n_2=\langle4,2,-4\rangle$$ $$\dfrac{4}{3}\not=\dfrac{2}{-2}$$

The vectors $\vec n_1$ and $\vec n_2$ are not collinear.

$$\vec n_1\cdot\vec n _2=3(4)+(-2)(2)+1(-4)=4\not=0$$

The vectors $\vec n_1$ and $\vec n_2$ are not orthogonal.

Hence, the given planes are neither parallel nor perpendicular.

3.3x+y+z-1=0 and -x+2y+z+3=0

$$\vec n_1=\langle3,1,1\rangle, \vec n_2=\langle-1,2,1\rangle$$ $$\dfrac{-1}{3}\not=\dfrac{2}{1}$$

The vectors $\vec n_1$ and $\vec n_2$ are not collinear.

$$\vec n_1\cdot\vec n _2=3(-1)+1(2)+1(1)=0$$

The vectors $\vec n_1$ and $\vec n_2$ are orthogonal.

Hence, the given planes are perpendicular.

4.x-3y+z+1=0 and 3x-4y+z-1=0

$$\vec n_1=\langle1,-3,1\rangle, \vec n_2=\langle3,-4,1\rangle$$ $$\dfrac{3}{1}\not=\dfrac{-4}{-3}$$

The vectors $\vec n_1$ and $\vec n_2$ are not collinear.

$$\vec n_1\cdot\vec n _2=1(3)+(-3)(-4)+1(1)=16\not=0$$

The vectors $\vec n_1$ and $\vec n_2$ are not orthogonal.

Hence, the given planes are neither parallel nor perpendicular.