# Question Video: Comparing the Magnitude of Subtracting Two 3D Vectors with the Difference between Their Magnitudes Mathematics

𝐕 and 𝐖 are two vectors, where 𝐕 = 〈−1, 5, −2〉 and 𝐖 = 〈3, 1, 1〉. Comparing |𝐕 − 𝐖| and |𝐕| − |𝐖|, which quantity is larger?

03:31

### Video Transcript

𝐕 and 𝐖 are two vectors, where vector 𝐕 is equal to negative one, five, negative two and vector 𝐖 is equal to three, one, one. Comparing the magnitude of vector 𝐕 minus vector 𝐖 and the magnitude of vector 𝐕 minus the magnitude of vector 𝐖, which quantity is larger?

We recall that the magnitude of any vector 𝐀 with components 𝑥, 𝑦, and 𝑧 is equal to the square root of 𝑥 squared plus 𝑦 squared plus 𝑧 squared. We find the sum of the squares of the individual components and then square root the answer. This means that the magnitude of vector 𝐕 is equal to the square root of negative one squared plus five squared plus negative two squared. This is equal to the square root of one plus 25 plus four. The magnitude of vector 𝐕 is equal to the square root of 30. We can repeat this process for vector 𝐖. This is equal to the square root of three squared plus one squared plus one squared, which simplifies to the square root of nine plus one plus one, which is the square root of 11. The magnitude of vector 𝐖 is the square root of 11.

Before calculating a numerical value for the magnitude of vector 𝐕 minus the magnitude of vector 𝐖, we will firstly work out vector 𝐕 minus vector 𝐖. We know that to subtract two vectors, we subtract the corresponding components. Negative one minus three is equal to negative four. Five minus one is equal to positive four. And negative two minus one is equal to negative three. Vector 𝐕 minus vector 𝐖 is equal to negative four, four, negative three. The magnitude of this is equal to the square root of negative four squared plus four squared plus negative three squared. This simplifies to the square root of 16 plus 16 plus nine. The magnitude of vector 𝐕 minus vector 𝐖 is therefore equal to the square root of 41.

We now need to calculate the decimal values of the magnitude of vector 𝐕 minus vector 𝐖 and the magnitude of vector 𝐕 minus the magnitude of vector 𝐖. The square root of 41 is equal to 6.4031 and so on. The square root of 30 minus the square root of 11 is equal to 2.1606 and so on. This means that the square root of 41 is greater than the square root of 30 minus the square root of 11. We can therefore conclude that the magnitude of vector 𝐕 minus vector 𝐖 is greater than the magnitude of vector 𝐕 minus the magnitude of vector 𝐖. The larger quantity is the magnitude of vector 𝐕 minus vector 𝐖.