Question Video: Finding the Distance between a Point and a Plane | Nagwa Question Video: Finding the Distance between a Point and a Plane | Nagwa

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

Which of the following is the distance of the point (3, βˆ’4, 5) from the plane (4π‘₯/21) + (8𝑦/21) + (5𝑧/21) = 1 rounded to two decimal places? [A] 0.64 [B] 0.39 [C] 1.56 [D] 6.10 [E] 3.88

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Video Transcript

Which of the following is the distance of the point three, negative four, five from the plane four π‘₯ over 21 plus eight 𝑦 over 21 plus five 𝑧 over 21 equals one rounded to two decimal places? Option (A) 0.64, option (B) 0.39, option (C) 1.56, option (D) 6.10, or option (E) 3.88.

In this question, we want to find the distance between a point and a plane. We should recall that the distance between a point and a plane will be given by the perpendicular distance, since this is the shortest distance between the two objects. There is a formula which we can use to help us work out this perpendicular distance. For the point π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and the plane π‘Žπ‘₯ plus 𝑏𝑦 plus 𝑐𝑧 plus 𝑑 equals zero, then we can say that the perpendicular distance 𝐷 is equal to 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.

Before we use this formula, however, we might notice that the equation of the plane is not in the same form that we need it to be in. Now, we could subtract one from both sides, but it might be easier to begin rearranging by multiplying through by 21. So let’s rearrange this. And the first thing we can do is multiply throughout by 21. This will give us four π‘₯ plus eight 𝑦 plus five 𝑧 equals 21. Then we need to have an equation equal to zero. So let’s subtract 21 from both sides. And that gives us the equation of the plane in a way in which we can extract the values of π‘Ž, 𝑏, 𝑐, and 𝑑.

We can also list down the values that we’ll use. π‘Ž is equal to four, 𝑏 will be equal to eight, 𝑐 is equal to five, and 𝑑 is equal to negative 21. The coordinates of the point will give us the next values. So π‘₯ sub one is three, 𝑦 sub one is negative four, and 𝑧 sub one is five. Now we are ready to substitute into the formula. This gives us that 𝐷 is equal to the magnitude of four times three plus eight times negative four plus five times five plus negative 21 over the square root of four squared plus eight squared plus five squared.

Simplifying this, the right-hand side will be equal to the magnitude of 12 minus 32 plus 25 minus 21 over the square root of 16 plus 64 plus 25. This is equal to the magnitude of negative 16 over the square root of 105. The magnitude of negative 16 is 16, and so we have 16 over the square root of 105. Since we are asked for an answer to two decimal places, then we can use our calculator to find a decimal equivalent for this value. This is 1.5614 and so on.

Rounding to two decimal places will give us the value of 1.56. And since the perpendicular distance is a length, we can add the units of length units. We can therefore give the answer that option (C) 1.56 is the distance between the given point and the plane.

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