# Question Video: Finding the Distance between Two Planes Mathematics

Find, to the nearest two decimal places, the distance between the two planes 2(π₯ β 2) + (π¦ β 3) + 3(π§ β 1) = 0 and π« β β¨4, 2, 6β© = 12.

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### Video Transcript

Find, to the nearest two decimal places, the distance between the two planes two times π₯ minus two plus π¦ minus three plus three times π§ minus one equals zero and vector π« dot four, two, six is equal to 12.

In this question, we are asked to find the distance between two planes. The first thing we need to do is to check if the planes are parallel. If they are, we can then apply the formula for perpendicular distance. We can remember that two planes are parallel if the normal vectors to each plane are parallel. The first plane is given in a Cartesian form. When we expand the parentheses, we can find the coefficients of π₯, π¦, and π§. Using these coefficients, we can give the normal vector to this plane as the vector two, one, three. The normal vector to the second plane can be given as the vector four, two, six.

We can then observe that these two normal vectors are scalar multiples of each other. And that means that the two planes are parallel. Now that we know that these two planes are parallel, we can find the distance between them by finding a point which lies on one of the planes and working out the distance between this point and the other plane. Letβs find a point which lies on the first plane. One way to do this is by setting π₯ is equal to zero and π¦ is equal to zero. When we do this, we get the equation two times negative two plus negative three plus three π§ minus three is equal to zero. Simplifying this, we get three π§ minus 10 is equal to zero, and so π§ is equal to 10 over three. As we set π₯ and π¦ equal to zero, we can say that the point zero, zero, 10 over three lies on this plane.

We can then apply the formula that the perpendicular distance, denoted uppercase π·, between the point π₯ sub one, π¦ sub one, π§ sub one and the plane vector π« dot π, π, π equals negative π is given by π· equals the magnitude of ππ₯ sub one plus ππ¦ sub one plus ππ§ sub one plus π over the square root of π squared plus π squared plus π squared. The values that we substitute into the formula can be given as π₯ sub one equals zero, π¦ sub one equals zero, and π§ sub one equals 10 over three. Then, from the plane, we have π is equal to four, π is equal to two, and π is equal to six. Since the value of negative π is equal to 12, then we know that π must be equal to negative 12.

We are now ready to substitute these values into the formula. This will give us π· is equal to the magnitude of four times zero plus two times zero plus six times 10 over three plus negative 12 over the square root of four squared plus two squared plus six squared. We can simplify this to the magnitude of 20 minus 12 over the square root of 16 plus four plus 36. On the numerator, the magnitude of eight is eight. And on the denominator, we have the square root of 56. Since weβre asked to find the answer to two decimal places, letβs use our calculators to work out a decimal equivalent. And so we have 1.0690 and so on. Rounded to two decimal places, we can give the answer that the distance between the two given planes is 1.07 length units.

Notice that, in this example, we found a point on the first given plane and worked out the distance between this point and the second plane. However, we couldβve done it in the other direction. For example, we couldβve found a point on the second plane and worked out the distance between this and the first plane. Either method would produce the same perpendicular distance of 1.07 length units to two decimal places.