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In this video, we will learn how to find the coordinates of a point in three dimensions.
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We will also calculate the distance between two points in 3D and then midpoint.
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We will begin by recalling what we know about points, midpoints, and distances in two dimensions.
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The two-dimensional π₯π¦-coordinate plane is drawn below.
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Any point on this coordinate plane will have an π₯- and π¦-coordinates.
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Letβs consider the two points π΄ and π΅ with coordinates π₯ one, π¦ one and π₯ two, π¦ two, respectively.
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In order to find the midpoint of π΄ and π΅, we find the average of the π₯- and π¦-coordinates.
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The π₯-coordinate of the midpoint will be equal to π₯ one plus π₯ two divided by two.
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And the π¦-coordinate will be equal to π¦ one plus π¦ two over two.
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In order to calculate the distance between two points on the π₯π¦-plane, we use an adaption of the Pythagorean theorem.
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The distance between point π΄ and point π΅ is the square root of π₯ two minus π₯ one squared plus π¦ two minus π¦ one squared.
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We find the difference between the π₯-coordinates and square the answer.
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We then find the difference between the π¦-coordinates and square this answer.
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The sum of these square rooted is the distance between the two points on the π₯π¦-plane.
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We will now look at how we can adapt these two formulas when dealing in three dimensions.
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The three-dimensional π₯π¦π§-plane could be drawn in many ways on a two-dimensional surface.
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We know that any point will have an π₯-, π¦-, and π§-coordinates.
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For example, the two points shown have coordinates π₯ one, π¦ one, π§ one and π₯ two, π¦ two, π§ two.
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We can find the midpoint of π΄ and π΅ by finding the average of the π₯-, π¦-, and π§-coordinates.
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The π₯-coordinate of the midpoint will be equal to π₯ one plus π₯ two divided by two.
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The π¦-coordinate will be π¦ one plus π¦ two divided by two.
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And the π§-coordinate will be π§ one plus π§ two divided by two.
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We can extend the distance formula in the same way.
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The distance between two points in three dimensions is equal to the square root of π₯ two minus π₯ one squared plus π¦ two minus π¦ one squared plus π§ two minus π§ one squared.
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We simply repeat the process used with the π₯- and π¦-coordinates with the π§-coordinate.
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We will now look at some questions where we need to identify points in three dimensions.
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In which of the following coordinate planes does the point negative seven, negative eight, zero lie?
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Is it (A) the π₯π¦-plane, (B) the π₯π§-plane, or (C) the π¦π§-plane?
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We know that any point in three dimensions has an π₯-, π¦-, and π§-coordinate.
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In this question, the π₯-coordinate is negative seven, the π¦-coordinate is negative eight, and the π§-coordinate is zero.
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As π§ is equal to zero, the point will not move in the direction of the π§-axis.
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We can therefore conclude that as π§ is equal to zero, the point will lie on the π₯π¦-plane.
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If our π¦-coordinate was equal to zero but π₯ and π§ had a positive or negative value, the point would lie in the π₯π§-plane.
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In a similar way, a point would lie in the π¦π§-plane if it had coordinate zero, π¦, π§, where π¦ and π§ are positive or negative values.
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In our next question, we need to find the coordinates of a point graphically.
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Determine the coordinates of point π΄.
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Any point on the 3D plane will have an π₯-, π¦-, and π§-coordinate.
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We can see from our diagram that point π΄ has an π₯-coordinate of three.
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It has a π¦-coordinate of negative three.
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Finally, it has a π§-coordinate of three.
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We can therefore conclude that the coordinates of point π΄ are three, negative three, three.
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If we werenβt able to spot this immediately on our diagram, we could begin by considering the point π΅ in the two-dimensional π₯π¦-plane.
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Point π΅ has an π₯-coordinate equal to three and a π¦-coordinate equal to negative three.
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As it lies on the π₯π¦-plane, it will have a π§-coordinate equal to zero.
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Point π΄ lies directly above point π΅.
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This means its π₯- and π¦-coordinates will be the same.
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All we now need to work out is the distance traveled along the π§-axis to get from point π΅ to point π΄.
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As this is equal to three, the π§-coordinate of point π΄ is three.
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This confirms that point π΄ has coordinates three, negative three, three.
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In our next question, we need to work out the midpoint of a line segment.
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Points π΄ and π΅ have coordinates eight, negative eight, negative 12 and negative eight, five, negative eight, respectively.
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Determine the coordinates of the midpoint of line segment π΄π΅.
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We recall that in order to find the midpoint of two points in three dimensions, we find the average of the π₯-, π¦-, and π§-coordinates.
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We can begin by letting point π΄ have coordinates π₯ one, π¦ one, π§ one and point π΅: π₯ two, π¦ two, π§ two.
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The π₯-coordinate of our midpoint will be equal to eight plus negative eight divided by two.
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Eight plus negative eight is equal to zero and zero divided by two is equal to zero.
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The π¦-coordinates of π΄ and π΅ are negative eight and five.
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This means that the π¦-coordinate of the midpoint will be equal to negative eight plus five divided by two.
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This is equal to negative three over two, which we could write as negative one and a half or negative 1.5.
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We will leave the answer as a top heavy or improper fraction.
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The π§-coordinate of our midpoint is equal to negative 12 plus negative eight divided by two.
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Negative 12 plus negative eight is equal to negative 20.
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Dividing this by two gives us negative 10.
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The midpoint of the line segment π΄π΅ has coordinates zero, negative three over two, negative 10.
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We could check this answer by looking at the distances between these values and the corresponding values in points π΄ and π΅.
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Zero is eight away from both eight and negative eight.
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Negative three over two or negative 1.5 is 6.5 away from negative eight and also from five.
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Finally, negative 10 is two away from negative 12 and also two away from negative eight.
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This confirms that the midpoint of points π΄ and π΅ is zero, negative three over two, negative 10.
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In our next question, weβll need to find the distance between a point and one of the axes.
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What is the distance between the point 19, five, five and the π₯-axis?
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Any point that lies on the π₯-axis will have coordinates π₯, zero, zero.
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Both the π¦- and π§-coordinates must be equal to zero.
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Weβre given the coordinates of a point 19, five, five.
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The point on the π₯-axis that is closest to this will have coordinates 19, zero, zero.
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The shortest distance will be to the point where the π₯-coordinate is the same.
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We know that we can calculate the distance between two points in three dimensions using an adaption of the Pythagorean theorem.
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If we have two points with coordinates π₯ one, π¦ one, π§ one and π₯ two, π¦ two, π§ two, the distance between them is equal to the square root of π₯ two minus π₯ one squared plus π¦ two minus π¦ one squared plus π§ two minus π§ one squared.
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Substituting in our two coordinates gives us the square root of 19 minus 19 squared plus zero minus five squared plus zero minus five squared.
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19 minus 19 is equal to zero.
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Zero minus five is equal to negative five.
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So we are left with the square root of negative five squared plus negative five squared.
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Multiplying a negative number by a negative number gives us a positive answer.
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Therefore, negative five squared is equal to 25.
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This means that our answer simplifies to the square root of 50.
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It is worth pointing out that we could have subtracted the coordinates in the other order as five minus zero squared is also equal to 25.
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As squaring a number always gives a positive answer, it doesnβt matter which order we subtract our coordinates in.
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We can actually simplify our answer by using our laws of radicals or surds.
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The square root of 50 is equal to the square root of 25 multiplied by the square root of two.
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As the square root of 25 equals five, weβre left with five multiplied by the square root of two or five root two.
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The square root of 50 is equal to five root two.
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We can therefore conclude that the distance between the points 19, five, five and the π₯-axis is five root two length units.
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We might actually notice a shortcut here.
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To find the distance between any point and an axis, we simply find the sum of the squares of the other two coordinates and then square root the answer.
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As we want to calculate the distance to the π₯-axis, we square the π¦- and π§-coordinates, find their sum, and then square root our answer.
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If we needed to calculate the distance between a point and the π¦-axis, we would square the π₯- and π§-coordinates, find the sum of these, and then square root that answer.
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We would use the same method to find the distance between a point and the π§-axis, this time using the π₯- and π¦-coordinates.
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In our final question, we will find the distance between two points given their coordinates in three dimensions.
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Find the distance between the two points π΄: negative seven, 12, three and π΅: negative four, negative one, negative eight.
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We know that we can find the distance between two points in three-dimensional space using the following formula.
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The distance is equal to the square root of π₯ two minus π₯ one squared plus π¦ two minus π¦ one squared plus π§ two minus π§ one squared.
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In this question, we will let the coordinates of point π΄ be π₯ one, π¦ one, π§ one and the coordinates of point π΅ be π₯ two, π¦ two, π§ two.
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Substituting in these values gives us the square root of negative four minus negative seven squared plus negative one minus 12 squared plus negative eight minus three squared.
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Negative four minus negative seven is the same as negative four plus seven.
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This is equal to three.
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Negative one minus 12 is equal to negative 13.
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Finally, negative eight minus three is equal to negative 11.
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We know that squaring a negative number gives a positive answer.
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This means that three squared is equal to nine, negative 13 squared is 169, and negative 11 squared is 121.
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169 plus 121 is equal to 290.
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And adding nine to this gives us 299.
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We can therefore conclude that the distance between the two points negative seven, 12, three and negative four, negative one, negative eight is the square root of 299 length units.
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We will now summarize the key points from this video.
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In this video, we saw that any point in three dimensions has coordinates π₯, π¦, and π§.
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We saw that if our π§-coordinate is equal to zero, the point lies on the π₯π¦-plane.
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If the π¦-coordinate was equal to zero, it would lie on the π₯π§-plane.
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In the same way, if π₯ was equal to zero, the point would lie on the π¦π§-plane.
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We also saw that if a point has two coordinates that are equal to zero, for example, if π¦ equals zero and π§ equals zero, it will lie on one of the axes, in this case, the π₯-axis.
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If π₯ and π¦ were both equal to zero, the point would lie on the π§-axis.
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And in the same way, if π₯ and π§ were equal to zero, the point would lie on the π¦-axis.
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We saw that the midpoint of two points π΄ and π΅ has coordinates π₯ one plus π₯ two over two, π¦ one plus π¦ two over two, and π§ one plus π§ two over two.
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We find the average of the π₯-, π¦-, and π§-coordinates.
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We also saw that we can calculate the distance between the same two points by square rooting π₯ two minus π₯ one squared plus π¦ two minus π¦ one squared plus π§ two minus π§ one squared.
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These two formulas will allow us to solve practical problems involving coordinates in three dimensions.