Question Video: Finding the Distance between Two Planes | Nagwa Question Video: Finding the Distance between Two Planes | Nagwa

# Question Video: Finding the Distance between Two Planes Mathematics • Third Year of Secondary School

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Find, to the nearest hundredth, the distance between the two planes π₯ + 2π¦ + 4π§ = 4 and (2π₯/13) + (4π¦/13) + (8π§/13) = 1.

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### Video Transcript

Find, to the nearest hundredth, the distance between the two planes π₯ plus two π¦ plus four π§ equals four and two π₯ over 13 plus four π¦ over 13 plus eight π§ over 13 is equal to one.

When we are finding the distance between two planes, we must first check if the two planes are parallel. We can recall that two planes are parallel if the normal vectors to each plane are parallel. The normal vector to each plane is given by the coefficients of π₯, π¦, and π§. And so the normal vector to the first given plane is the vector one, two, four. The normal vector to the second plane can be given as two thirteenths, four thirteenths, eight thirteenths.

Since we can multiply each component of the first plane by two over 13 to give the corresponding component of the second plane, then that means that the normal vectors are scalar multiples of each other, and so they are parallel. Hence, the planes themselves are also parallel. And so, in order to find the distance between two planes, we can find a point on one of the planes and then work out the perpendicular distance from that point to the other plane.

It doesnβt matter which plane we find a point on. So letβs take the first plane of π₯ plus two π¦ plus four π§ equals four and see if we can find a point on this plane. We can substitute the values of π₯ equals zero and π¦ equals zero into the equation of this plane. This would give zero plus two times zero plus four π§ is equal to four. Since four π§ is equal to four, then we know that π§ is equal to one. We can therefore say that the point zero, zero, one lies on this first plane.

We can now recall the formula that the perpendicular distance, denoted uppercase π·, between the point π₯ sub one, π¦ sub one, π§ sub one and the plane ππ₯ plus ππ¦ plus ππ§ plus π equals zero 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 coordinates of the point zero, zero, one gives us the π₯ sub one, π¦ sub one, and π§ sub one values.

For the coefficients π, π, π, and π from the plane, we may find it useful to rearrange this equation by multiplying through by 13. So we have the equation two π₯ plus four π¦ plus eight π§ minus 13 equals zero. Notice how we have also rearranged the constant term so that π is on the same side as the equation in the form that we use in the formula. So weβll be using the values π is equal to two, π is equal to four, π is equal to eight, and π is equal to negative 13. Now we can substitute these into the formula.

And so we have π· is equal to the magnitude of two times zero plus four times zero plus eight times one plus negative 13 over the square root of two squared plus four squared plus eight squared. Simplifying, we have the magnitude of eight minus 13 over the square root of four plus 16 plus 64. On the numerator, the magnitude of negative five is five. And on the denominator, we have the square root of 84.

Because we are asked for a value to the nearest hundredth, then we need to find a decimal equivalent. And this will be 0.5455 and so on length units. We can therefore give the answer that the distance between the two given planes is 0.55 length units to the nearest hundredth.

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