Analytic Geometry question, planes and lines

Question

Let there be a plane going through three points $(0,2,-9), (0,-1,0), (-\frac{3}{m},1,-3)$.

For which value of $m$ is the line $l: (3,0,-9)+t(2m,-5,7)$ onto (or 'inside') the plane?

Not sure how to do this. Thanks in advance!

Answer 1

We show m = 2. Let A = (0,2,-9), B = (0,-1,0), and C = (-3/m,1, -3). Then vectors AB = (0,-3,9), and AC = (-3/m,-1,6). The normal to the plane is N = AB * AC = (-9,-27/m,-9/m). So the equation of the plane is: -9(x-0) -27/m*(y-2) -9/m(z+9) = 0. The line l is on the plane if two of its points are on it. So let t = 0 we get the point (3,0,-9) on it and this means its coordinates satisfies the equation of the plane and this gives: m = 2. Put t = 1 get the second point (3+2m,-5,-2) and plug the coordinates of this point into the plane equation yields: 3 + 2m = 14/m. So m = 2.

Answer 2

Here is how you advance.

1) Find the parametric equation of the plane which should be in the form $(x,y,z)=(g(\tau,s),\tau,s)$.

2) Solve the two equations, the parametric line equation $(x,y,z)=(3+2mt,-5t,-9+7t)$ and the plane equation

Answer 3

Note: This answer is not as general as the ones provided by E-theory and Mhenni Benghorbal. It is, however, a bit easier to do as it involves no use of the variable $m$ until the very end. So it is a little less complicated computationally.

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We are asked to find the value of $m$ so that the line lies in the plane. The standard approach would be to find the plane and then make sure the line lies in the plane. In this problem though, the plane is not fixed as the last given point in the plane contains the variable $m$. This will make the normal vector a variable expression. We can, instead, take advantage of the fact that we must have the line in the plane at the end. That means that the point $(3,0,-9)$ from the line also needs to be on the plane. So we can use the points $A=(0,2,-9)$, $B=(0,-1,0)$ and $C=(3,0,-9)$ to find the normal vector. Taking $\vec{BA} = \langle 0,3,-9\rangle$ and $\vec{BC} = \langle3,1,-9\rangle$, we find that $\vec{BA} \times \vec{BC} = \langle -18, -27, -9\rangle=-9\langle 2, 3, 1\rangle$. So we can take $\vec{n} = \langle 2,3,1\rangle$. We now have two choices for obtaining $m$: write the equation of the plane and substitute the point we didn't use to find or use the fact that the normal vector to the plane is perpendicular to the direction vector of all lines in the plane.

Equation of plane: Use $\vec{n}$ and the point $B=(0,-1,0)$, we get the equation of the plane is $2(x-0)+3(y+1)+(z-0)=0 \iff 2x+3y+z=-3$. Substitute $(-\frac{3}{m},1,-3)$ and we get $2(-\frac{3}{m})+3(1)+1(-3) = -3 \iff -\frac{6}{m} = -3 \iff m =2$.

Normal vector: Recall two vectors are perpendicular iff their dot product is $0$. So $\langle 2,3,1\rangle \cdot \langle 2m,-5,7\rangle = 0 \iff 4m-15+7 =0 \iff 4m =8 \iff m =2$.