Ytukyg

Find the scalar equation of the plane that passes through the line of intersection of two planes: $x+y+z-4 = 0$ and $y+z-2=0$ and the plane is 3 units from the point (5,-3,7).

I am able to form a plane that goes through the line of intersection of two planes by using a linear combination:

$$x+y+z-4 + k(y+z-2)=0$$

where $k$ is real number.

How do I find the value of k so that the plane is 3 units from the point?

You’ve got a good start, although you really need to use a more general linear combination of the two planes instead: $$lambda(x+y+z-4)+mu(y+z-2)=0. tag 1$$ The problem with your family of planes is that they exclude $y+z-2=0$ itself. That would leave you stuck with no solution if that plane happened to be the correct one. You can easily verify that the distance from $(5,-3,7)$ to this plane is not $3$, though, so in this case you can proceed with your original formulation.

Now you can simply use the formula for the distance of a point $P=(x_p,y_p,z_p)$ from the plane $ax+by+cz+d=0$: $${|ax_p+by_p+cz_p+d| over sqrt{a^2+b^2+c^2}}.$$ Plug in the coordinates of the point from the problem and use (1) for the plane, and set this equal to the required distance from the point. Squaring both sides and rearranging will give you a quadratic equation in $k$ to solve.

Since the plane passes through the intersection of 2 other planes, lets start by finding the intersection line:

$ x+y+z-4 = 0$ and $ y + z - 2 = 0$

Finding two points on both planes to get the line equation. Points (2,0,2) and (2,2,0) are on the both plane, hence is enough to construct the line equation.

$$L1: (2,0,2) + k(0,1,-1) = 0$$

Since the plane has to contain L1, it's normal has to be normal to L1 as well. Let $n = (a,b,c)$, if n is normal to L1, the inner product has to be 0.

$$0*a + b*1 + c*-1 = 0$$

$$b=c$$

$$n = (a,b,b)$$

Writing the plane equation using the normal vector:

$$P1: ax + by + bz - d = 0$$

we know that 2,0,2 is on the plane, $2a+2b-d = 0$ Substituting d in the equation:

$$P1: ax + by + bz - 2a - 2b = 0$$

The vector from origin to the closest point has to be perpendicular to the plane. We can deduct that the closest point has the form (0,0,0) + k(n), the closest point to plane is, therefore k(a,b,b) is the closest point. Set the normal vector to have length 2 for ease of calculation.

$$a^2 + b^2 + b^2 = 2^2$$
Since (a,b,b) is on the plane and closest to origin
$$a^2 + b^2 + b^2 - 2a - 2b = 0 $$
$$4 - 2a - 2b = 0 $$
$$a = 2 - b$$
$$(2 - b)^2 + b^2 + b^2 = 2^2$$
$$4 -4.b +b^2 + 2.b^2 = 4$$
$$3.b^2=4.b$$
$$b_1 = 0, b_2 = frac {4} 3, a_1 = 2, a_2 = frac 2 3$$

Hence the two solutions are:

$P_1: x - 2 = 0$ with closest point (2,0,0)

$P_2: frac 2 3 x + frac 4 3 y + frac 4 3 z - 4 = 0$ with closest point $(frac 2 3,frac 4 3,frac 4 3)$