Math / Geometry

QuestionFind an equation of the plane. the plane that passes through the point $(3,6,-2)$ and contains the line $x=4-t, y=2 t-1, z=-3 t$

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a plane that goes through a specific point and also contains a given line. Watch out! Make sure you understand how to use the point-normal form of a plane equation and how to find a normal vector using the cross product!

STEP 2

1. Find two points on the line.
2. Create two vectors in the plane.
3. Find the normal vector.
4. Write the equation of the plane.

STEP 3

We're given the parametric equations of a line inside the plane: $x = 4 - t$ , $y = 2t - 1$ , and $z = -3t$ .
Let's find two points on this line by picking two different values for $t$ .
Let's choose $t = \mathbf{0}$ and $t = \mathbf{1}$ because they're easy to work with!

STEP 4

When $t = \mathbf{0}$ , we have $x = 4 - 0 = \mathbf{4}$ , $y = 2(0) - 1 = \mathbf{-1}$ , and $z = -3(0) = \mathbf{0}$ .
This gives us the point $(\mathbf{4}, \mathbf{-1}, \mathbf{0})$ .

STEP 5

When $t = \mathbf{1}$ , we have $x = 4 - 1 = \mathbf{3}$ , $y = 2(1) - 1 = \mathbf{1}$ , and $z = -3(1) = \mathbf{-3}$ .
This gives us the point $(\mathbf{3}, \mathbf{1}, \mathbf{-3})$ .

STEP 6

We're also given that the plane passes through the point $(\mathbf{3}, \mathbf{6}, \mathbf{-2})$ .
Let's call this point $P$ .
We can use this point and the two points we found on the line to create two vectors that lie in the plane.

STEP 7

Let's use the points $(\mathbf{4}, \mathbf{-1}, \mathbf{0})$ (which we'll call $A$ ) and $(\mathbf{3}, \mathbf{6}, \mathbf{-2})$ (point $P$ ).
A vector from $A$ to $P$ is given by $\langle 3 - 4, 6 - (-1), -2 - 0 \rangle = \langle \mathbf{-1}, \mathbf{7}, \mathbf{-2} \rangle$ .
Let's call this vector $\mathbf{v}$ .

STEP 8

Now, let's use the points $(\mathbf{3}, \mathbf{1}, \mathbf{-3})$ (which we'll call $B$ ) and $(\mathbf{3}, \mathbf{6}, \mathbf{-2})$ (point $P$ ).
A vector from $B$ to $P$ is given by $\langle 3 - 3, 6 - 1, -2 - (-3) \rangle = \langle \mathbf{0}, \mathbf{5}, \mathbf{1} \rangle$ .
Let's call this vector $\mathbf{w}$ .

STEP 9

The normal vector to the plane is a vector that is perpendicular to every vector in the plane.
We can find a normal vector by taking the cross product of the two vectors $\mathbf{v}$ and $\mathbf{w}$ that we found in the plane.

STEP 10

$\mathbf{n} = \mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 7 & -2 \\ 0 & 5 & 1 \end{vmatrix} = (7 + 10)\mathbf{i} - (-1)\mathbf{j} + (-5)\mathbf{k} = 17\mathbf{i} + \mathbf{j} - 5\mathbf{k} = \langle 17, 1, -5 \rangle$

STEP 11

Now that we have a normal vector $\langle \mathbf{17}, \mathbf{1}, \mathbf{-5} \rangle$ and a point $(\mathbf{3}, \mathbf{6}, \mathbf{-2})$ in the plane, we can use the point-normal form of the equation of a plane: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$ , where $\langle a, b, c \rangle$ is the normal vector and $(x_0, y_0, z_0)$ is the point.

STEP 12

Plugging in our values, we get $17(x - 3) + 1(y - 6) - 5(z - (-2)) = 0$ .
Simplifying, we get $17x - 51 + y - 6 - 5z - 10 = 0$ , which simplifies to $17x + y - 5z - 67 = 0$ , or $17x + y - 5z = 67$ .

STEP 13

The equation of the plane is $17x + y - 5z = 67$ .

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