Question Video: The Magnitude of a Vector | Nagwa Question Video: The Magnitude of a Vector | Nagwa

# Question Video: The Magnitude of a Vector Mathematics • First Year of Secondary School

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What is the magnitude of the vector π΄π΅, where π΄ = (5, β9) and π΅ = (9, 1)?

02:18

### Video Transcript

What is the magnitude of the vector π΄π΅, where π΄ is the point with coordinates five, negative nine and π΅ is the point with coordinates nine, one?

First, letβs recall how to calculate the magnitude of a vector. If a vector π£ is in component form with components π₯ and π¦, then the magnitude of this vector, which is denoted using the vertical bars on either side, is found by calculating the square root of π₯ squared plus π¦ squared.

This is just an application of Pythagorean theorem as the vector forms a right-angled triangle with its π₯- and π¦-components. So the first step to answering this problem is we need to write the vector π΄π΅ in component form.

The π₯-component of the vector π΄π΅ is found by subtracting five from nine. The π¦-component is found by subtracting negative nine from one. Therefore, in component form, the vector π΄π΅ is four, 10.

So now we can substitute into our formula for calculating the magnitude of a vector. The magnitude of π΄π΅ is the the square root of four squared plus 10 squared. Four squared is 16 and 10 squared is 100, so the magnitude of π΄π΅ is the square root of 116.

Now this surd can be simplified if we recall that 116 is equal to four multiplied by 29. The laws of surds tell us that we can separate this out into the the square root of four multiplied by the square root of 29. Four is a square number, so we can find its square root exactly; itβs just two.

29 is not a square number and, in fact, itβs a prime number, so we canβt simplify this surd anymore. Therefore the magnitude of the vector π΄π΅ is two root 29.

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