Find the distance between the point negative five, negative eight, negative six and the plane negative two 𝑥 plus 𝑦 plus two 𝑧 equals seven.
In this question, we are asked to find the distance between a point and a plane. We can recall that the distance between a point and a plane means the perpendicular distance, since this is the shortest distance between the two objects. And there is a formula to help us do this. This formula tells us that the perpendicular distance, denoted uppercase 𝐷, between a point 𝑥 sub one, 𝑦 sub one, 𝑧 sub one and a 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.
But before we can use this formula, we need to change the equation of the plane so that it is in the same form as we have in the formula. By subtracting seven from both sides of the equation, we would get negative two 𝑥 plus 𝑦 plus two 𝑧 minus seven equals zero. This simply means that we have the 𝑑-value of negative seven along with the other terms on the same side of the equation. And it’s also a good idea to note down the values that we’re using. 𝑥 sub one will be negative five, 𝑦 sub one will be negative eight, and 𝑧 sub one will be negative six.
In the equation of the plane, we have 𝑎 is equal to negative two, 𝑏 is equal to one because the coefficient of 𝑦 is one, 𝑐 is equal to two, and 𝑑 is equal to negative seven. We can now fill in the values into the formula for the perpendicular distance. And so we have 𝐷 is equal to the magnitude of negative two times negative five plus one times negative eight plus two times negative six plus negative seven over the square root of negative two squared plus one squared plus two squared.
At this point, it is worth noting that in the denominator of this formula, sometimes it’s easy to think that we might add on another 𝑑 squared within the square root. However, we do not. The denominator within this formula is simply the square root of 𝑎 squared plus 𝑏 squared plus 𝑐 squared. When we simplify the right-hand side of our equation, we get the magnitude of 10 minus eight minus 12 minus seven over the square root of four plus one plus four. And we can further simplify this by recognizing that the denominator of the square root of nine will be equal to three. And so we have the magnitude of negative 17 over three.
of course, the magnitude of negative 17 is 17. And so we have worked out the value of 𝐷. Since 𝐷 is a distance, we can add the units of length units to the given value. We can therefore give the answer that the distance between the given point and the plane is 17 over three length units.