Question

18. Consider the two planes given by

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)

$v = n_{1} \times n_{2}$

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

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

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

$\vec n_1 \times \vec n_2 = \boxed{\langle 14, 2, 15\rangle}$

(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

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

Then the intersection can be parameterized by equations

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

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.