Answer to Question #209781 in Analytic Geometry for Jaguar

Question

(7.1) Find a point-normal form of the equation of the plane passing through P = (1, 2, −3) and

having ~n =< 2, −1, 2 > as a normal.

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

(a) x + y + 3z + 10 = 0 and x + 2y − z = 1 ,

(b) 3x − 2y + z − 6 = 0 and 4x + 2y − 4z = 0 ,

(c) 3x + y + z − 1 = 0 and −x + 2y + z + 3 = 0,

(d) x − 3y + z + 1 = 0 and 3x − 4y + z − 1 = 0.

Answer 1

(7.1)

Point-normal form of the plane is

$$2(x − 1) -(y − 2) + 2(z +3) = 0.$$

We can also write this as

$$2x-y+2z=-6$$

(7.2)

(a). 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.

(b). 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.

(c). 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.

(d). 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.