Find a Perpendicular Plane to Line \vec{L}(t) and Passes Through Point (5,-5,0)

[JoshHolloway]

Homework Statement

Find a plane that is perpendicular to the line $$\vec{L}(t) = (5,0,2)t + (3,-1,1)$$ and passes through the point $$(5,-5,0)$$

Homework Equations

The equation of the plane that P through $$(x_{0},y_{0},z_{0})$$ that has a normal vector $$\vec{n} = A \vec{i} + B \vec{j} + C\vec{k}$$ is:
$$A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0}) = 0$$
that is, $$(x,y,z) \in P$$

The Attempt at a Solution

$$\vec{L}(t) = (5t + 3, -1, 2t + 1)$$
let $$t = 1 \Rightarrow \vec{L}(1) = (8,-1,3) = (A,B,C)$$
$$(x_{0},y_{0},z_{0}) = (5,-1,0)$$

$$A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0}) = 0$$
$$(8)(x - 5) + (-1)(y +1) + (3)(z - 0) = 0$$
$$8x - 40 - y -1 + 3z = 0;$$
$$8x - y + 3z = 41$$

 

[JoshHolloway]

The correct answer is $$5x +2z = 25$$

What am I doing wrong?

 

[Mathdope]

I have to assume that you haven't yet finished your corrections, because the L(t) functions seem different. Let us know when you've updated it correctly in your OP please.

OK, I'm assuming that your second post indicates that you are now thinking that you have the original problem correctly set up.

My first question is - Where are you getting the L(t) in your attempt at a solution? It looks to me as if it's different from the one in the problem.

 

[JoshHolloway]

Alright, the I am finished with the corrections.

 

[Mathdope]

From the problem, L(1) = (8,-1,3). Where are you getting the second L(t) that appears in your solution?

 

[JoshHolloway]

I don't see what I have written that is wrong.

 

[Mathdope]

Homework Statement

Find a plane that is perpendicular to the line $$\vec{L}(t) = (5,0,2)t + (3,-1,1)$$ and passes through the point $$(5,-5,0)$$

Homework Equations

The equation of the plane that P through $$(x_{0},y_{0},z_{0})$$ that has a normal vector $$\vec{n} = A \vec{i} + B \vec{j} + C\vec{k}$$ is:
$$A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0}) = 0$$
that is, $$(x,y,z) \in P$$

The Attempt at a Solution

$$\vec{L}(t) = (5t + 3, -1, 2t + 1)$$
let $$t = 1 \Rightarrow \vec{L}(1) = (8,-1,3) = (A,B,C)$$
$$(x_{0},y_{0},z_{0}) = (5,-1,0)$$

$$A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0}) = 0$$
$$(8)(x - 5) + (-1)(y +1) + (3)(z - 0) = 0$$
$$8x - 40 - y -1 + 3z = 0;$$
$$8x - y + 3z = 41$$

I don't see what I have written that is wrong.

Then there's something weird in the software or your formatting. I see in your OP a value of L(1) = (-1,5,4).

When I look at it in quote mode it is different than what I see when looking at the post itself.

It looks like you are assuming that the value L(1) =(8,-1,3) is the normal vector. It's not. You need to find a vector that is parallel to the line. That's not the same as a point on the line.

Think about what it means if (8,-1,3) and (3,-1,1) both lie on the line.

 

[JoshHolloway]

So do I take the cross product of (8,-1,3) and (3,-1,1) to find the normal vector?

 

[HallsofIvy]

You don't need to do that. In fact, you don't need to do any calculation at all. If you write a plane as A(x-x₀)+ B(y- y₀)+ C(z- z₀)= 0, then $A\vec{i}+ B\vec{j}+ C\vec{k}$ is perpendicular to the plane. You know that (5, -5, 0) is a point in the plane so you immediately know x₀, y₀, and z₀.

You know that the line $$\vec{L}(t) = (5,0,2)t + (3,-1,1)$$, which has "direction vector" $5\vec{i}+ 2\vec{k}$, is to be perpendicular to the plane so you immediately know A, B, and C.

 

[Mathdope]

So do I take the cross product of (8,-1,3) and (3,-1,1) to find the normal vector?

You need a vector that's parallel to the line. The direction vector from (3,-1,1) to (8,-1,3) fits. Notice what you get, and compare it to what HallsofIvy says.