### 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 Pythagorasβs 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.