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Express the vector π with components negative five over two and negative 19 using the unit vectors π’ and π£.
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So weβre given these two unit vectors π’ and π£.
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These are sometimes written with circumflex accents on top which look like hats.
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This emphasises the fact they are unit vectors, that is vectors whose magnitude is one.
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Of course, there are many unit vectors.
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There are unit vectors pointing in any direction you like.
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π’ and π£ are particular unit vectors; π’ is the unit vector pointing in the π₯-direction, and π£ is the unit vector pointing in the π¦-direction.
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π’ has components one, zero and π£ has components zero, one.
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Any two-dimensional vector π can be written in terms of π’ and π£.
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The vector π is equal to the π₯-component of π£ times π’ plus the π¦-component of π£ times π£.
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So the vector π is equal to the π₯-component of π, negative five over two, times π’ plus the π¦-component of π, negative 19, times π£.
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Writing the plus negative 19π£ as minus 19π£, we get our final answer: π is equal to negative five over two π’ minus 19π£.
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We can check this using the components of π’ and π£ and what we know about the scale of multiplication and subtraction of vectors.
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Negative five over two π’ minus 19π£ is equal to negative five over two times the vector with components one, zero minus 19 times the vector with components zero, one.
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Here weβre using the component forms of π’ and π£.
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And of course, when multiplying a vector by a scaler, we just multiply the components of that vector by the scaler.
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So we get the vector with components negative five over two, zero minus the vector with components zero, 19.
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And subtracting one vector from another means subtracting the components of that vector from the other.
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So we get the vector with components negative five over two, negative 19.
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And as desired, this is the vector π.