Question

Show that the plane \(2 x-y-3 z=4\) is parallel to the line \(x=-2+2 t, y=-1+4 t, z=4,\) and find the distance between them

Short answer

The line and the plane are parallel, and the distance between them is 1 unit.

Step-by-step solution

Extract the Normal and Direction Vectors

Extract the normal vector of the plane from the coefficients of the given plane equation \(2x - y - 3z = 4\). Thus, the normal vector \(\vec{n}\) is \(\vec{n} = (2,-1,-3)\). Also, extract the direction vector from the given line equations as \(\vec{d} = (2, 4, 0)\).

Examine Perpendicularity

To be parallel, the direction vector of the line and the normal vector of the plane must be orthogonal (perpendicular). Check this by calculating the dot product of the two vectors and verifying that it is zero. The dot product is calculated as \(\vec{n} . \vec{d} = 2*2 + (-1)*4 + (-3)*0 = 0\). Since the dot product is zero, the vectors and thus their corresponding plane and line are parallel.

Find a Point and Calculate Distance

To calculate the distance between the plane and the line, we need a point on the line. The line equations can be written in parametric form as \(x = -2 + 2t, y = -1 + t, z = t\). Take \(t = 0\), we will have point P(-2,-1,0). The distance \(D\) from point P to the plane is given by the formula \(D = |(\vec{n} . \vec{AP} + d)| / ||\vec{n}||\) where \(\vec{AP}\) is the vector from the reference point A(0,0,0) on the plane to point P, and \(d\) is the right side of the plane equation. Calculate \(D = |(2*(-2) + (-1)*(-1) + (-3)*0 +4)| / sqrt(2^2 + (-1)^2 + (-3)^2) = 1\). Thus, the distance from the line to the plane is 1 unit.