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All of the following are unit vectors except blank.
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Option (A) the vector 0.6, negative 0.8; option (B) the vector one, negative one; option (C) the vector one, zero; or option (D) the vector zero, one.
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In this question, we’re given a list of four vectors.
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We need to determine which of these four vectors is not a unit vector.
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So, to answer to this question, let’s start by recalling what we mean by unit vector.
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We call any vector with magnitude one a unit vector.
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So we need to determine which of these four vectors does not have magnitude one.
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There’s a few different ways we could do this.
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One way of doing this would be to draw all four of our vectors onto a diagram, starting at the origin.
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Then a vector having magnitude one means it would end on the unit circle.
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So, we’d just need to find a vector which does not end on the unit circle.
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This would work and will give us the correct answer.
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However, there’s another method.
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We can use the following formula: the magnitude of the vector 𝐚, 𝐛 will be equal to the square root of 𝑎 squared plus 𝑏 squared.
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And it’s worth pointing out here that this is exactly the same as saying the magnitude of a vector is its length sketched graphically, since this formula is just applying the Pythagorean theorem to a right-angled triangle with width modulus 𝑎 and height modulus of 𝑏.
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We’ll use this to find the magnitude of all four of the vectors given to us in the question.
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Let’s start with the vector in option (A).
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That’s the magnitude of the vector 0.6, negative 0.8.
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To find the magnitude of this vector, we need to take the square root of the sum of the squares of its components.
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This gives us the square root of 0.6 squared plus negative 0.8 squared.
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And if we evaluate this, we get the square root of one, which is equal to one.
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So, the vector in option (A) is a unit vector.
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Let’s now do the same for the vector in option (B).
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We want to find the magnitude of the vector one, negative one.
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To find the magnitude of this vector, we need to find the square root of the sum of the square of its components.
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That gives us the square root of one squared plus negative one all squared.
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And we can calculate this; it’s equal to the square root of two.
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Since this is not equal to one, this vector is not a unit vector.
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Therefore, we’ve shown the vector in option (B) is not a unit vector.
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We could stop here; however, let’s also check the magnitude of our other two vectors.
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And instead of doing this by using our formula, let’s instead do this by using a sketch.
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Remember, with any vectors, we can represent these graphically by using the first component to be the horizontal component of our vector and the second component to be the vertical component of our vector.
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So, for the vector in option (A), it will have horizontal component 0.6 and vertical component negative 0.8.
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So, we can represent the vector in option (A) on our diagram as shown.
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And it’s worth pointing out we don’t need our vector to start at the origin.
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However, this does give us a useful property.
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The magnitude of our vector will be the distance our vector lies from the origin.
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In other words, because this vector lies on the unit circle centered at the origin, we can see its magnitude is one.
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We can then do exactly the same for the vectors in options (B), (C), and (D).
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Let’s do the vector in option (C).
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The horizontal component of this vector is one, and the vertical component is zero.
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So, we can add this vector onto our sketch starting at the origin.
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And we can see that the magnitude of this vector or the length of the vector graphically is equal to one.
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We can do exactly the same for the vector in option (D).
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Its horizontal component is zero and its vertical component is one.
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And we can also add this vector to our sketch.
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It will start at the origin and go up to the point zero, one.
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And we can also see the magnitude of this vector is one.
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And for completeness sake, let’s add the vector in option (B) to our sketch.
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Its horizontal component is one and its vertical component is negative one.
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And starting this vector at the origin, we can see its magnitude is greater than one.
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Therefore, we were able to show of the four given vectors, only the vector in option (B) — that’s one, negative one — was not a unit vector.