grishmabeli6 wrote:Find the scalar equation for the plane passing through the points P1=(−5, −4, −5), P2=(−5, −2, −7), and P3=(0, −4, −7).

Find x so that the triangle with vertices A=(1, −7, 1), B=(−5, −16, 11), and C=(x, −3, −6) has a right angle at A.

how can one go about solving these questions?

You need one point in the plane and a normal vector. So first calculate vectors P1P2 and P2P3. (Let's call them u and v)

Now we need to find a vector n that is perpendicular to both u and v

That means that u*n=0 and v*n=0 (dot product)

https://9apps.ooo/Now say n=(n_x,n_y,n_z)

n*u = n_x * u_x + n_y * u_y + n_z * u_z

Do the same for n*v

You have the coordinates of u and v, now you have 2 equations containing the coordinates of n. Two equations with three variables => assume one of them (for example say n_x = 1 and solve for n_y and n_z)

Now that you have the normal vector, you easily get the equation of the plane.

It's gonna be: ax+by+cz+d=0 where a,b,c are n_x,n_y and n_z