Find the perpendicular distance between the line vector 𝐫 equals one, two, four plus 𝑡 times negative two, one, four and the plane vector 𝐫 dot two, zero, one equals one.
This question involves finding the perpendicular distance between a line and a plane. When considering a line and a plane, there are really two general possibilities. Either the line is parallel to the plane or the line intersects the plane. If the line does intersect the plane, then the shortest distance at any point between the line and the plane will be zero. We will therefore need to determine if the line and the plane intersect.
We can take the equation of the line and rearrange it so that we have vector 𝐫 is equal to one minus two 𝑡, two plus 𝑡, four plus four 𝑡. We can now substitute this vector into the equation of the plane. Instead of vector 𝐫, we will have one minus two 𝑡, two plus 𝑡, four plus four 𝑡, which we can then take the dot product of with two, zero, one. Finding the sum of the products of the corresponding components, we have two times one minus two 𝑡 plus zero times two plus 𝑡 plus one times four plus four 𝑡 is equal to one. Simplifying, we have two minus four 𝑡 plus four plus four 𝑡 is equal to one. But when we further simplify this, we get six on the left-hand side and one on the right-hand side. This means that this will not be true for any value of 𝑡. And so the line and plane do not intersect.
Therefore, we can say that this line and plane must be parallel. The distance between a line and a plane can be found by taking a point on the line and finding the perpendicular distance from that point to the plane. Let’s see if we can find a point which lies on the line. We can do this by taking the equation of the line and substituting in any value of 𝑡. So let’s use 𝑡 is equal to zero. By doing this, we can work out that the point one, two, four must lie on the line.
We can now recall and use the formula to find the perpendicular distance between a point and a plane given in vector form. This formula tells us that the perpendicular distance denoted with uppercase 𝐷 between point 𝑥 sub one, 𝑦 sub one, 𝑧 sub one and the plane vector 𝐫 dot 𝑎, 𝑏, 𝑐 equals negative 𝑑 is given by 𝐷 equals the magnitude of 𝑎𝑥 sub one plus 𝑏𝑦 sub one plus 𝑐𝑧 sub one plus 𝑑 over the square root of 𝑎 squared plus 𝑏 squared plus 𝑐 squared.
When we are substituting in a lot of values into a formula such as this, it’s always worthwhile jotting down the values that we’ll be using. The values of 𝑥 sub one, 𝑦 sub one, and 𝑧 sub one come from the point that we’re using. They’re one, two, and four, respectively. The values of 𝑎, 𝑏, 𝑐, and 𝑑 come from the equation of the plane. So 𝑎 is two, 𝑏 is zero, and 𝑐 is one. We need to be more careful with the term of 𝑑, however. Notice how in this form that we use in the formula we have negative 𝑑 on the right-hand side. Since the value in the equation of the plane that we are given has one on the right-hand side, then that means that the value of 𝑑 must be negative one.
And so we are now ready to substitute these values into the formula to find the perpendicular distance. So we have 𝐷 is equal to the magnitude of two times one plus zero times two plus one times four plus negative one over the square root of two squared plus zero squared plus one squared. When we simplify, we have the magnitude of two plus four minus one over the square root of four plus one. The numerator will be equal to the magnitude of five, which is five, and the denominator is the square root of five. We can rationalize this denominator by multiplying the numerator and denominator by root five. We can then take out the common factor of five to give the answer that the perpendicular distance between the given line and plane is root five length units.