WEBVTT
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Given vector π equals six, one, four and vector π equals three, one, two, find ππ.
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We recall that to find ππ, we subtract vector π from vector π.
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The reason for this can be shown on a two-dimensional grid.
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This can then be transferred to three-dimensional vectors.
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Letβs consider two points on the grid, A and B.
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The vector π is given from its displacement from the origin O.
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The same is true of vector π.
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We wish to travel from point A to point B.
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This is the same as traveling from point A to point O and then point O to point B.
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Going from point A to point O would be equivalent to negative vector π.
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And going from point O to point B is equivalent to vector π.
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ππ is equal to negative vector π plus vector π.
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This can be rewritten as vector π minus vector π.
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Weβre told in this question that vector π is equal to six, one, four.
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This could also be written as a column vector as shown.
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It is also sometimes written in terms of vectors π’, π£, and π€.
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In this case, six π’ plus π£ plus four π€, where six, one, and four are the coefficients of π’, π£, and π€.
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For the purposes of this question, we will stick with the notation given.
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And weβre also told that vector π is equal to three, one, two.
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In order to calculate π minus π, we simply subtract the individual components.
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Three minus six is equal to negative three.
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One minus one is equal to zero.
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Finally, two minus four is equal to negative two.
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ππ is therefore equal to negative three, zero, negative two.