# Video: Finding the Norm of a Given Vector

Given that vector π = β5π’ β 3π£, where π’ and π£ are perpendicular unit vectors, find |π|.

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### Video Transcript

Given that vector π is equal to negative five π’ minus three π£, where π’ and π£ are perpendicular unit vectors, find the magnitude of vector π.

We can begin by drawing this on a grid where π’ and π£ are perpendicular unit vectors. Our vector π moves a distance of negative five in the π’-direction and a distance of negative three in the π£-direction. Vector π 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 π― can therefore be calculated using Pythagorasβs theorem, where the magnitude equals the square root of π squared plus π squared.

Lower case π and π are the π’- and π£-components, respectively. Therefore, the magnitude of vector π 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 π 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 π is equal to negative five π’ minus three π£, then its magnitude is equal to the square root of 34.