Question

18. Consider the two planes given by

3x - 6y - 2z = 3, 2x + y - 2z = 2

(a) The point (x, 0, 0) is on both planes. Find x.
(b) Find a vector n normal to the first plane.
(c) Find a vector n normal to the second plane.
(d) Find a vector normal to both n₁ and n₂.
(e) Find parametric equations for the line in which the two planes intersect. ​

Answer 1

a)

`3x - 6(0) - 2(0) = 3 \\ 3x = 3 \\ x = 1 \\ \\ or \\ \\ 2x + 0 - 2(0) = 2 \\ 2x = 2 \\ x = 1`

b)

`n_(1) \: (3, - 6, - 2)`

c)

`n_(2) \: \: (2,1, - 2)`

d)

`v = n_(1) * n_(2)`

`i \: \: \: \: \: \: \: \: \: \: \: j \: \: \: \: \: \: \: \: \: k \\ 3 \: \: \: \: \: - 6 \: \: \: \: - 2 \\ 2 \: \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: - 2`

`v \:(12 + 2, - ( - 6 + 4),3 + 12) \\ v \: (14,2,15)`

e)

`x = 14m + 1 \\ y = 2m \\ z = 15m`

Answer 2

(a) If
`(x,0,0)` lies on both planes, then

`y=z=0 \implies 3x = 3 \implies \boxed{x=1}`

and at the same time

`2x = 2 \implies x=1`

(b) A plane with normal vector
`\vec n` containing the point
`(a,b,c)` can be written in the form

`\vec n \cdot \langle x - a, y - b, z - c \rangle = 0`

Expanding the left side, we see that the components of
`\vec n` correspond to the coefficients of
`x,y,z`. So the normal vector to
`3x-6y-2z` is
`\vec n_1 = \boxed{\langle3,-6,-2\rangle}`.

(c) Similarly, the normal to
`2x+y-2z=2` is
`\vec n_2 = \boxed{\langle2,1,-2\rangle}`.

(d) The cross product of any two vectors
`\vec x` and
`\vec y` is perpendicular to both of the vectors. So we have

`\vec n_1 * \vec n_2 = \boxed{\langle 14, 2, 15\rangle}`

(e) Solve the two plane equations for
`z`.

`3x - 6y - 2z = 3 \implies 2z = 3x - 6y - 3`

`2x + y - 2z = 2 \implies 2z = 2x + y - 2`

By substitution,

`3x - 6y - 3 = 2x + y - 2 \implies x = 7y + 1`

Let
`y=t\in\Bbb R`. Then
`x=7t+1` and

`2z = 3(7t+1) - 6t - 3 \implies 2z = 15t \implies z = \frac{15}2t`

Then the intersection can be parameterized by equations

`\begin{cases} x(t) = 7t + 1 \\\\ y(t) = t \\\\ z(t) = \frac{15}2 t \end{cases}`

for
`t\in\Bbb R`.

We can also set
`x=t` or
`z=t` first, then solve for the other variables in terms of the parameter
`t`, so this is by no means a unique parameterization.