Question

1. two planes intersect at a right angle and their line of intersection has the equations $x = t$, $y = 1 - t$ and $z = 2 + 3t$. if one of the planes passes through the origin, find the equations of the two planes in the form $ax + by + cz = d$.

Explanation:

Step1: Identify line direction vector

The line of intersection is $\vec{v} = \langle 1, -1, 3
angle$ (from $x=t, y=1-t, z=2+3t$).

Step2: Let plane 1 pass through origin

Let plane 1: $A_1x + B_1y + C_1z = 0$. Its normal $\vec{n_1} = \langle A_1, B_1, C_1
angle$ is perpendicular to $\vec{v}$, so:
$$\vec{n_1} \cdot \vec{v} = A_1 - B_1 + 3C_1 = 0$$

Step3: Use plane perpendicularity

Since planes are perpendicular, $\vec{n_1} \cdot \vec{n_2} = 0$. Plane 2 has normal $\vec{n_2} = \langle A_2, B_2, C_2
angle$, and $\vec{n_2} \cdot \vec{v} = A_2 - B_2 + 3C_2 = 0$.

Step4: Pick point on line for plane 2

Take $t=0$: point $(0,1,2)$ lies on plane 2, so $0 \cdot A_2 + 1 \cdot B_2 + 2 \cdot C_2 = D_2$.

Step5: Solve for $\vec{n_1}$

Choose $C_1=1$, then $A_1 - B_1 = -3$. Let $A_1=1$, so $B_1=4$. Thus $\vec{n_1} = \langle 1,4,1
angle$, plane 1: $x + 4y + z = 0$.

Step6: Solve for $\vec{n_2}$

Use $\vec{n_1} \cdot \vec{n_2} = A_2 + 4B_2 + C_2 = 0$ and $A_2 - B_2 + 3C_2 = 0$. Subtract equations:
$$(A_2 + 4B_2 + C_2) - (A_2 - B_2 + 3C_2) = 0 \implies 5B_2 - 2C_2 = 0 \implies B_2 = \frac{2}{5}C_2$$
Substitute $B_2$ into $A_2 - B_2 + 3C_2 = 0$:
$$A_2 = \frac{2}{5}C_2 - 3C_2 = -\frac{13}{5}C_2$$
Let $C_2=5$, then $A_2=-13, B_2=2$. Plane 2: $-13x + 2y + 5z = D_2$. Use point $(0,1,2)$:
$$D_2 = 2(1) + 5(2) = 12$$

Answer:

The equations of the two planes are $x + 4y + z = 0$ and $-13x + 2y + 5z = 12$ (or $13x - 2y - 5z = -12$).