Question Video: Finding the Distance between a Point and a Plane Mathematics

Find, to the nearest two decimal places, the distance between the point (3, 2, 3) and the plane 2(π‘₯ βˆ’ 2) + 3(𝑦 βˆ’ 3) + (𝑧 βˆ’ 1) = 0.

03:19

Video Transcript

Find, to the nearest two decimal places, the distance between the point three, two, three and the plane two times π‘₯ minus two plus three times 𝑦 minus three plus 𝑧 minus one equals zero.

Here we are asked to find the distance between a point and a plane. We should remember that the distance between a point and a plane means the perpendicular distance since this is the shortest distance between these two objects. There is a formula which allows us to calculate the perpendicular distance. This formula tells us that the distance uppercase 𝐷 between the point π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and the plane π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero is given by 𝐷 equals the magnitude of π‘Žπ‘₯ sub one plus 𝑏𝑦 sub one plus 𝑐𝑧 sub one plus 𝑑 over the square root of π‘Ž squared plus 𝑏 squared plus 𝑐 squared.

But before we go to use this formula, we might notice that the equation of the plane that we are given does not match the form within the formula. We will need to change it into the form π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero. We can simplify this equation by distributing the values across the parentheses, which gives two π‘₯ minus four plus three 𝑦 minus nine plus 𝑧 minus one equals zero. We then collect together the numerical values negative four, negative nine, and negative one to give the equation two π‘₯ plus three 𝑦 plus 𝑐 minus 14 equal zero.

Before we use the formula, we can note down the values of the variables that we’ll be using. π‘₯ sub one is three, 𝑦 sub one is two, and 𝑧 sub one is three. In the equation of the plane, we have π‘Ž is equal to two, 𝑏 is equal to three, 𝑐 is equal to one β€” since that’s the coefficient of 𝑧 β€” and the constant term 𝑑 is equal to negative 14.

Filling these values into the formula, on the numerator, we have the magnitude of two times three plus three times two plus one times three plus negative 14. On the denominator, we have the square root of two squared plus three squared plus one squared. We simplify this to the magnitude of six plus six plus three minus 14 over the square root of four plus nine plus one. And so we have one over the square root of 14.

We then need to convert this value into a decimal. So we have 0.2672 and so on. Rounding to the nearest two decimal places, the third decimal digit is seven. And so this value rounds up to 0.27. We can therefore give the answer that to the nearest two decimal places, the distance between the given point and plane is 0.27 length units.

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