Find the equation for the plane containing the point

In summary: Otherwise, you may need to double check your calculations. In summary, the equation for the plane containing the given point and line is -x+y+z=-4.
  • #1
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Homework Statement



Find the equation for the plane containing the point (3,-8,7) and the line

l(t)=<1,-8,5>+t<3,-2,5>

Homework Equations



Cross product

Equation of a plane: A(x-x0)+B(y-y0)+C(z-z0)+D=0

The Attempt at a Solution



First I set t=0 in l(t)=<1,-8,5>+t<3,-2,5> to obtain the point (1,-8,5)

I then proceed to to subtract the given point (3,-8,7) by (1,-8,5) to get a vector <2,0,2>.

Since I have two vectors, that is, <3,-2,5> and <2,0,2>, I can use the cross product on

the two vectors to obtain the vector for the plane. By <3,-2,5> x <2,0,2>, I obtained the

vector <-4,4,4>. Simplifying it would result in <-1,1,1>.

Thus, using the equation for the plane:

A(x-x0)+B(y-y0)+C(z-z0)+D=0

where <A,B,C> = <-1,1,1> and (x0,y0,z0) = (3,-8,7)

So now, the equation looks like this:

-(x-3)+(y+8)+(z-7)=0

-x+3+y+8+z-7=0

-x+y+z=-4I am not even sure if I did it right. This question was a question on my last exam (and the

only one that I missed). Sadly, my professor did not even post up the solutions. Can anyone

please see if my results and methods were correct? Thanks! :)
 
Last edited:
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  • #2
You can check your answer yourself.
Does your plane contain the point (3, -8, 7)?
Is every point in the given line also on your plane?

If the answers to both questions are yes, then your work is correct.
 

1. What is the equation for a plane containing a given point?

The equation for a plane containing a given point (x0, y0, z0) is Ax + By + Cz = D, where A, B, and C are the coefficients of the x, y, and z variables and D is a constant determined by the given point.

2. How do you find the equation for a plane containing a point and a given vector?

To find the equation for a plane containing a point (x0, y0, z0) and a given vector v = ai + bj + ck, you can use the point-normal form of the equation, which is (x - x0, y - y0, z - z0) · v = 0. This can be simplified to Ax + By + Cz = D, where A, B, and C are the coefficients of v and D is a constant determined by the given point.

3. Can the equation for a plane containing a point have negative coefficients?

Yes, the equation for a plane containing a point can have negative coefficients. The important thing is that the coefficients are not all equal to zero, as this would result in an equation for a line instead of a plane.

4. How do you determine if a point lies on a plane defined by an equation?

To determine if a point (x0, y0, z0) lies on a plane defined by the equation Ax + By + Cz = D, you can simply plug in the values of x0, y0, and z0 into the equation. If the resulting expression is equal to D, then the point lies on the plane. Otherwise, it does not.

5. What is the normal vector of a plane defined by an equation?

The normal vector of a plane defined by an equation Ax + By + Cz = D is simply the vector n = Ai + Bj + Ck. This vector is perpendicular to the plane and can be used to find the angle between the plane and other surfaces or to determine the orientation of the plane in space.

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