Solve 3D Vector Question: Find Plane Orthogonal to -7x+8y+5z=1

[BoundByAxioms]

I'm doing some multi-variable calculus review, and I had a question (my understanding of the class was not as good as I would have liked it to be).

<b> 1. Homework Statement </b>.
Find a plane containing the line r(t) = <6,-6,4> + t<-2,7,-4> and orthogonal to the plane -7x+8y+5z=1.

<b> 2. Homework Equations </b>.
I think I need to use a cross product. I cross <-2,7,-4> and <-7,8,5> to get a vector orthogonal to to the plane (and the line). Then, I use n (dot) (r-r_0), but I keep getting the wrong answer. I fear that my approach is wrong though.
answer is: 635/2

<b> 3. The Attempt at a Solution </b>.
So when I cross <-2,7,-4> and <-7,8,5> I get <67, 38, 33>, and my plane is 67x+38y+33z=46.

Help please.

 

[Defennder]

I think I need to use a cross product. I cross <-2,7,-4> and <-7,8,5> to get a vector orthogonal to to the plane (and the line). Then, I use n (dot) (r-r_0), but I keep getting the wrong answer. I fear that my approach is wrong though.
answer is: 635/2

That numerical answer is for which part of the question? You're asked to find the plane containing the line right? I got the same normal vector as you did <67,68,33>. But then you're told to find the equation of the plane, which you know has the general form \vec{n} \cdot (\vec{r}-\vec{r_0})

So what you're missing is a point which resides on the plane. Look at the question again, how can you get that missing point? How did you get 67x + 38y + 33z = 46?

 

[BoundByAxioms]

That numerical answer is for which part of the question? You're asked to find the plane containing the line right? I got the same normal vector as you did <67,68,33>. But then you're told to find the equation of the plane, which you know has the general form \vec{n} \cdot (\vec{r}-\vec{r_0})

So what you're missing is a point which resides on the plane. Look at the question again, how can you get that missing point? How did you get 67x + 38y + 33z = 46?

Yeah that fraction was part of what I copied and pasted, so ignore that. Ok, so all I need is a point, meaning I can use t=1 to get the point (4,1,0)? Then it would be:

67(x-4)+68(y-1)+33(z-0)=0
Then
67x+68y+33z=336?

I got 67x + 38y + 33z = 46 by doing <67,38,33> \cdot <x-6, y+6, z-4>. I used t=0 for my point. I used the wrong normal vector (just a writing mistake).

 

[Defennder]

Typo error on my part. The normal vector should be <67,38,33>, not <67,68,33>. Use that vector and just apply the formula for any value of t. After all, the line lies on the plane, does it not?

 

[BoundByAxioms]

Typo error on my part. The normal vector should be <67,38,33>, not <67,68,33>. Use that vector and just apply the formula for any value of t. After all, the line lies on the plane, does it not?

This is true. But I let t=0 and still got the wrong answer.

 

[Defennder]

What's the answer supposed to be?

 

[BoundByAxioms]

I got the correct answer. My arithmetic was wrong, as it usually is when doing cross-products. Thank you for your help!