Video: Finding the Norm of a Given Vector

Given that vector 𝐀 = βˆ’5𝐒 βˆ’ 3𝐣, where 𝐒 and 𝐣 are perpendicular unit vectors, find |𝐀|.

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Video Transcript

Given that vector 𝐀 is equal to negative five 𝐒 minus three 𝐣, where 𝐒 and 𝐣 are perpendicular unit vectors, find the magnitude of vector 𝐀.

We can begin by drawing this on a grid where 𝐒 and 𝐣 are perpendicular unit vectors. Our vector 𝐀 moves a distance of negative five in the 𝐒-direction and a distance of negative three in the 𝐣-direction. Vector 𝐀 can therefore be drawn as shown. As the magnitude of any vector is its length, we can calculate this by drawing a right triangle as shown. The magnitude of any vector 𝐯 can therefore be calculated using Pythagoras’s theorem, where the magnitude equals the square root of π‘Ž squared plus 𝑏 squared.

Lower case π‘Ž and 𝑏 are the 𝐒- and 𝐣-components, respectively. Therefore, the magnitude of vector 𝐀 is equal to the square root of negative five squared plus negative three squared. Squaring a negative number gives a positive answer. So, the square root of negative five is 25, and the square root of negative three is nine. This means that the magnitude of vector 𝐀 is equal to the square root of 34. As 34 is not a square number, we can leave our answer in surd or radical form. If vector 𝐀 is equal to negative five 𝐒 minus three 𝐣, then its magnitude is equal to the square root of 34.

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