Finding Equation of plane: (x-1)/-3 = (y-3)/2 = (z+2)/1, w/ point (0,7,-7)

[Devi415]

The equation of a line is \(\displaystyle \dfrac{x\, -\, 1}{-3}\, =\, \dfrac{y\, -\, 3}{2}\, =\, \dfrac{z\, +\, 2}{1}.\) Also, the coordinates of a point is (0, 7, -7). If both these line and point lie in the same plane, then the equation of the plane is

 

[Deleted member 4993]

The equation of a line is \(\displaystyle \dfrac{x\, -\, 1}{-3}\, =\, \dfrac{y\, -\, 3}{2}\, =\, \dfrac{z\, +\, 2}{1}.\) Also, the coordinates of a point is (0, 7, -7). If both these line and point lie in the same plane, then the equation of the plane is

There are multiple ways to do perform this task. I would do the following way:

Find a point (P) on the given line.

Find the equation of the line (l) through the point (P) and the given line.

Calculate the cross-product between the given line and (l). That will be the equation of the normal of the plane that you need to determine.

 

[HallsofIvy]

One method: taking x= 1, y= 3 in the equation of the line, z+ 2= 0 so (1, 3, -2) is a point on that line and, since the entire line is in the plane, a point in the plane. Taking x= 1, y= 4, z+ 2= 5 so (1, 4, 3) is also a point in that plane.

Since we are given that (0, 7, -7) we have three points in the plane. The vectors <0- 1, 7- 3, -7+ 2>= <-1, 4, -5> and <0- 1, 7- 4, -7- 3>= <-1, 3, -10> are vectors in the plane so their cross product is a vector perpendicular to the plane.

The equation of a plane with <A, B, C> perpendicular to the plane and point \(\displaystyle (x_0, y_0, z_0)\) in the plane is \(\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0\).

 

[Devi415]

Hi HallsofIvy
Can you please provide more explanation..? I couldn't get your instructions clearly.

 

[stapel]

Hi HallsofIvy
Can you please provide more explanation..? I couldn't get your instructions clearly.

Please reply with specification regarding where you are getting lost. You followed the step-by-step instructions until... what point?

Thank you!

 

[HallsofIvy]

Three points, or two intersecting lines (or two parallel lines, but that's not relevant here) determine a plane. If you have three points use them, as I did above, to find two vectors lying in the plane. Two intersecting lines will immediately give you two vectors.

As I said, the cross product of two vectors will determine a vector perpendicular to the plane. Here, those two vectors are <-1, 4, -5> and <-1, 3, -10>. Do you know how to find the cross product of two vectors? For these two vectors that would be the determinant \(\displaystyle \left| \begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ -1 & 4 & -5 \\ -1 & 3 & -10\end{array} \right|\). Again, with vector <A, B, C> perpendicular to the plane and point \(\displaystyle (x_0, y_0, z_0)\) in the plane, the equation of the plane is \(\displaystyle A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0\).