WEBVTT
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What is the unit vector in the same direction as the vector ๐ which has components three and four?
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So weโve been asked to find a unit vector in the same direction as the given vector ๐.
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Now, this means that two things must be true.
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Firstly, our vector must be parallel to the vector ๐.
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But it must have a magnitude of one.
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Diagrammatically, that will look a little like this.
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The two vectors are parallel.
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Theyโre in the same direction.
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But the vector ๐ฎ will have a magnitude of one.
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Now, the way that we can achieve this is to take the vector ๐ and divide it by its own magnitude, which weโll first need to calculate.
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We recall that for the general two-dimensional vector with components ๐ and ๐, its magnitude is equal to the square root of ๐ squared plus ๐ squared.
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And this is simply an application of the Pythagorean theorem.
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The magnitude of our vector ๐ then is the square root of three squared plus four squared.
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Thatโs the square root of nine plus 16, which is the square root of 25, which is simply equal to five.
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In fact, we should recognise this as one of our Pythagorean triples.
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That is, right triangles in which all three sides have integer lengths.
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To find our unit vector ๐ฎ, which is in the same direction as ๐ then, we take the vector ๐ and we divide it by its magnitude.
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So thatโs the same as multiplying by one-fifth.
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We have that ๐ฎ is equal to one-fifth of the vector three, four.
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In order to multiply a vector by a scalar, we just multiply each component by that scalar.
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So the first component of our vector ๐ฎ will be one-fifth multiplied by three, which is three-fifths.
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And the second component will be one-fifth multiplied by four, which is four-fifths.
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Weโve, therefore, found a unit vector in the same direction as ๐.
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Itโs the vector with components three-fifths and four-fifths.
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We can, of course, check that the magnitude of this vector is equal to one.
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We have the square root of three-fifths squared plus four-fifths squared, which is the square root of nine over 25 plus 16 over 25.
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Thatโs the square root of 25 over 25 or the square root of one which is equal to one.
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So weโve confirmed that the magnitude of our vector is indeed equal to one.
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And so, it is a unit vector.