# Question Video: Calculating the Magnitude of 3D Vectors Mathematics

Which of these vectors has the least magnitude? [A] <−1, 7, −2> [B] <−3, −1, −7> [C] <3, 5, 6> [D] <−2, −4, 5> [E] <4, 6, 7>

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### Video Transcript

Which of these vectors has the least magnitude? Is it (A) negative one, seven, negative two? (B) Negative three, negative one, negative seven. (C) Three, five, six. (D) Negative two, negative four, five. Or (E) four, six, seven.

We see we’ve been given five three-dimensional vectors and asked to describe them in relation to their magnitude. So, let’s think about this in general terms. Suppose we have a three-dimensional vector 𝐚 given by 𝑥𝐢 plus 𝑦𝐣 plus 𝑧𝐤. We can alternatively represent this to match the vectors in the question as shown. The magnitude of our vector is denoted using these vertical bars. Then this is given by the square root of 𝑥 squared plus 𝑦 squared plus 𝑧 squared. This essentially tells us the length of the vector 𝐚. So, with this in mind, we’re able to find the magnitude of each of our vectors.

For the vector in part (A), it’s the square root of negative one squared plus seven squared plus negative two squared. That’s the square root of 54. Then, for the second vector, it’s the square root of negative three squared plus negative one squared plus negative seven squared, which is the square root of 59. For our third vector, it’s root three squared plus five squared plus six squared, which is the square root of 70. The magnitude of the vector in part (D) is root 45. And in the same way, the magnitude of the vector in part (E) is the square root of 101.

Now we notice at this stage that we could have simplified each of our radicals or our surds. But in fact, not doing so allows us to more easily compare each of the values. Suppose we have two numbers 𝑝 and 𝑞 which are greater than or equal to one. If 𝑞 is greater than 𝑝, then the square root of 𝑞 is also greater than the square root of 𝑝. This essentially means that the vector which has the least magnitude is the one in which we’re finding the square root of the smallest number. Well, we notice that’s in part (D): the square root of 45. Root 45 is smaller than all of the remaining numbers.

And so, the answer is (D). The vector that has the least magnitude is the vector negative two, negative four, five.