Question Video: Calculating the Distance between a Line and a Plane Mathematics • Third Year of Secondary School

Video Transcript

Find the distance between the line 𝑥 minus one over two equals 𝑦 minus two over four equals 𝑧 plus five over two and the plane three 𝑥 minus two 𝑦 plus 𝑧 equals negative two. Give your answer to one decimal place.

In this question, we need to find the distance between a line and a plane. So, let’s firstly determine if these two objects intersect or not. We can first check if the line and plane are parallel.

For a line and a plane to be parallel, the direction vector of the line must be perpendicular to the normal vector of the plane. The direction vector of this line is given by the vector two, four, two. For the plane, its normal vector is given by the vector three, negative two, one. To find the normal vector of the plane, we recall that if a plane is in the form 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 equals 𝑑, then the normal vector is found by the coefficients 𝑎, 𝑏, 𝑐. The normal vector of this plane can be given as three, negative two, one.

And so to check if the direction vector of the line is perpendicular to the normal vector of the plane, we need to find their dot product. If the dot product is equal to zero, then they are perpendicular. The dot product of two vectors is found by the sum of the products of the corresponding components. So here, we have two times three plus four times negative two plus two times one. This is equal to six minus eight plus two. And this will simplify to give us zero. As already mentioned, since the dot product is equal to zero, then the direction vector of the line is perpendicular to the normal vector of the plane. So, the line and the plane are parallel.

In order to find the distance between a parallel line and plane, we can find a point that lies on the line and then find the perpendicular distance from this point to the plane. We can use the following formula to help us work out this distance. The perpendicular distance, denoted capital 𝐷, between the point 𝑥 sub one, 𝑦 sub one, 𝑧 sub one and the plane 𝑎𝑥 plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero 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.

In order to actually find a point which lies on the line, we can set each part of this Cartesian equation equal to zero. If 𝑥 minus one over two is equal to zero, then 𝑥 minus one is equal to zero. And so, 𝑥 must be equal to one. If 𝑦 minus two over four is equal to zero, then 𝑦 minus two is equal to zero, and 𝑦 must be equal to two. Finally, if 𝑧 plus five over two is equal to zero, then that means the value of 𝑧 must be equal to negative five. The point one, two, negative five lies on the line. And so, now, we have a point and a plane which we can use to find the values to substitute into this formula.

When we are using a formula like this which has a lot of values to substitute in, 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 are the values of the point, so they will be one, two, and negative five, respectively. The values of 𝑎, 𝑏, 𝑐, and 𝑑 come from the plane. So, 𝑎 is equal to three, 𝑏 is equal to negative two, and 𝑐 will be equal to one. Notice that we need to rearrange this so that the value of 𝑑 is given in the appropriate place in the equation. And so, the value of 𝑑 is equal to two.

We are now ready to substitute these values into the formula. And so, we have 𝐷 is equal to the magnitude of three times one plus negative two times two plus one times negative five plus two over the square root of three squared plus negative two squared plus one squared. When we simplify this, we get the magnitude of three minus four minus five plus two over the square root of nine plus four plus one. On the numerator, the magnitude of negative four is simply four. And on the denominator, we have the square root of 14.

We were asked to give the value to one decimal place. And so, we can use our calculators to find a decimal equivalent for four over root 14. When we’ve done this, we get a value of 1.0690 and so on. Rounding this to one decimal place, we can give the answer that the distance between the given line and plane is 1.1 length units.