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Find the angle 𝜃 between the vectors 𝐯 two, one, four and 𝐰 one, negative two, zero.
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We recall that the cos of angle 𝜃, the angle between two vectors, is equal to the dot product of the two vectors, 𝐚 dot 𝐛, divided by the product of the magnitude or modulus of the two vectors.
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In this question, vector 𝐯 is equal to two 𝐢 plus 𝐣 plus four 𝐤.
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Vector 𝐰 is equal to 𝐢 minus two 𝐣 plus zero 𝐤.
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This can be simplified to 𝐢 minus two 𝐣.
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The dot product of these two vectors can be found by multiplying the coefficients of 𝐢, the coefficients of 𝐣, and the coefficients of 𝐤.
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We then calculate the sum of these three answers.
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Two multiplied by one is equal to two.
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One multiplied by negative two is negative two.
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Finally, four multiplied by zero is equal to zero.
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Two minus two plus zero is equal to zero.
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Therefore, the dot product of vectors 𝐯 and 𝐰 is zero.
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The magnitude of vector 𝐚 is equal to the square root of 𝑥 squared plus 𝑦 squared plus 𝑧 squared, where 𝑥, 𝑦, and 𝑧 are the coefficients of 𝐢, 𝐣, and 𝐤, respectively.
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The magnitude of vector 𝐯 is therefore equal to the square root of two squared plus one squared plus four squared.
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This is equal to the square root of 21.
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The magnitude of 𝐰 can be calculated in the same way.
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One squared plus negative two squared plus zero squared.
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This is equal to the square root of five.
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We can now substitute these three values into the formula.
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The cos of 𝜃 is equal to zero divided by the square root of 21 multiplied by the square root of five.
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Zero divided by any number is equal to zero.
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Therefore, the cos of 𝜃 equals zero.
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Taking the inverse cos of both sides of this equation gives us 𝜃 is equal to cos to the minus one or inverse cos of zero.
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This is equal to 90 degrees.
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The angle between the two vectors 𝐯 and 𝐰 is 90 degrees.