Video Transcript
Which of these vectors has the greatest magnitude? Option (A) the vector 𝐢 plus 𝐣. Option (B) the vector 𝐢 plus 𝐣 minus 𝐤. Option (C) the vector three 𝐢 minus 𝐤. Option (D) the vector two 𝐢 plus three 𝐣 minus 𝐤. Or option (E) the vector three 𝐢 minus two 𝐤.
In this question, we need to determine which of five vectors has the greatest magnitude. And we can see that our five vectors are given in terms of the unit directional vectors 𝐢, 𝐣, and 𝐤. To do this, let’s start by recalling what we mean by the magnitude of a vector. In fact, there’s two ways of thinking about this. First, recall when we’re talking about vectors, all vectors have two properties. They have a magnitude, and they have a direction. So for example, if we were to draw our vector 𝐯 graphically, the direction that the vector points is represented by the arrow and its magnitude is represented by the length.
So one method we could use to find the magnitude of all five of the given vectors would be to draw them and then find the length of the given vectors. We would do this by using something like the Pythagorean theorem. And this would work. And we could find the magnitude of all of our vectors using this method. However, it’s actually easier to do this by using a formula.
We already know how to find the magnitude of any vector in general. The magnitude of a vector will be the square root of the sum of the squares at its components. So for example, the magnitude of the vector 𝑎𝐢 plus 𝑏𝐣 plus 𝑐𝐤 will be equal to the square root of 𝑎 squared plus 𝑏 squared plus 𝑐 squared. All we need to do is apply this formula to all five of the vectors given to us in the question.
Let’s start by finding the magnitude of option (A). That’s the magnitude of the vector 𝐢 plus 𝐣. To find the magnitude of this vector, it might be easier to rewrite our vector as one 𝐢 plus one 𝐣 plus zero 𝐤. Then we can see the components of this vector are one, one, and zero. So to find the magnitude of this vector, we need to take the square root of the sum of the squares of these components. This gives us the square root of one squared plus one squared plus zero squared, which we can calculate is equal to the square root of two. And because we need to find which of these vectors has the greatest magnitude, it would be easier to write this in decimal form. So we’ll write this to two decimal places. It’s 1.41.
We’ll do exactly the same to find the magnitude of option (B). That’s the magnitude of the vector 𝐢 plus 𝐣 minus 𝐤. This time, we can see the components of our vector, the coefficients of our unit directional vectors, will be one, one, and negative one. So to find the magnitude of this vector, we need to take the square root of the sum of the squares of these components. That’s the square root of one squared plus one squared plus negative one squared, which we can calculate is equal to the square root of three. And once again, we’ll write this to two decimal places is 1.73.
Let’s now calculate the magnitude of the vector in option (C). That’s the magnitude of three 𝐢 minus 𝐤. We need to find the components of our vector. That’s the coefficients of the unit directional vectors 𝐢, 𝐣 and 𝐤. We can see this will be three, zero, negative one. And we know the coefficient of 𝐣 is zero because it doesn’t appear in our vector. So we must have zero 𝐣. We can then find the magnitude of this vector by taking the square root of the sum of the squares of the components. That’s the square root of three squared plus zero squared plus negative one all squared. And we can calculate this is equal to root 10. And to two decimal places, this is 3.16.
Let’s now do the same for the vector in option (D). We can see its components will be two, three, and negative one. We can calculate the magnitude of this vector. It’s going to be the square root of two squared plus three squared plus negative one all squared, which we can calculate is the square root of 14. And to two decimal places, this is 3.74.
And we can do exactly the same for the vector in option (E). Its components will be three, zero, and negative two. And the magnitude of this vector will be the square root of three squared plus zero squared plus negative two all squared, which we can calculate is the square root of 13. And to two decimal places, we can calculate this is 3.61.
And now we can see it wasn’t actually necessary to write these to two decimal places since we know the largest of these will be the square root of the biggest number. It will be the square root of 14. However, we were still able to show of the five given vectors, the vector with the greatest magnitude was option (D), the vector two 𝐢 plus three 𝐣 minus 𝐤, which had a magnitude of the square root of 14.
