Given that vector 𝐀 is equal to
negative five 𝐢 minus three 𝐣, where 𝐢 and 𝐣 are perpendicular unit vectors,
find the magnitude of vector 𝐀.
We can begin by drawing this on a
grid where 𝐢 and 𝐣 are perpendicular unit vectors. Our vector 𝐀 moves a distance of
negative five in the 𝐢-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 𝑎 squared plus 𝑏 squared.
Lower case 𝑎 and 𝑏 are the 𝐢-
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 𝐢 minus three 𝐣, then its magnitude is equal to the square root of 34.