Question Video: Finding the Norms of Vectors Mathematics

If |𝐀 Γ— 𝐁|Β² + |𝐀 β‹… 𝐁|Β² = 17,424 and |𝐀| = 12, find |𝐁|.

07:25

Video Transcript

If the magnitude of 𝐀 cross 𝐁 squared plus the absolute value of 𝐀 dot 𝐁 squared equals 17,424 and the magnitude of 𝐀 is 12, find the magnitude of 𝐁.

First things first, let’s understand why we read the vertical bars as denoting a magnitude in this term but an absolute value in this term. This is because the cross product of two vectors is another vector, but the dot product of two vectors is a scalar. When we put vertical bars around a vector, we mean find its length or magnitude. When we put vertical bars around a scalar, we mean find its absolute value.

The reason we use the same notation with vertical bars to represent both quantities is that the magnitude of a vector plays a very similar role to the absolute value of a scalar. Nevertheless, because finding the magnitude of a vector is really a different operation from finding the absolute value of a scalar, it’s important to understand what kind of quantity is inside the vertical bars to correctly understand what operation the vertical bars are meant to represent.

So likewise, the vertical bars around 𝐀 and the vertical bars around 𝐁 both represent vector magnitudes because the quantities inside the bars are vectors. Anyway, turning back to our goal, it’s not obvious from this equation how to come up with an expression for the magnitude of 𝐁. However, if we focus on the quantities inside the vertical bars, we see that we have a vector cross product and a vector dot product, both of which can be expressed in terms of the magnitudes of the two vectors. The dot product of two vectors 𝐀 and 𝐁 has a value given by the magnitude of 𝐀 times the magnitude of 𝐁 times the cosine of the angle between them represented with the Greek letter πœƒ.

Note that the magnitude of 𝐀, the magnitude of 𝐁 , and the cos of πœƒ are all scalars. So the right-hand side of this expression is a product of three scalar quantities, which means it itself is a scalar, which is good because we know that 𝐀 dot 𝐁 should be a scalar.

There is a similar formula for 𝐀 cross 𝐁 in terms of the magnitude of 𝐀, the magnitude of 𝐁, and the angle between them. But remember, 𝐀 cross 𝐁 is a vector, so this formula will actually give us the magnitude of 𝐀 cross 𝐁. The magnitude of the vector 𝐀 cross 𝐁 is equal to the magnitude of 𝐀 times the magnitude of 𝐁 times the sine of the angle between them. Now, since 𝐀 cross 𝐁 is a vector, it also has a direction. And it turns out that this direction is actually given by what’s known as the right-hand rule. But in our particular statement, we only care about the magnitude of 𝐀 cross 𝐁, so we don’t need to worry about its direction. Also, just like the dot product, on the right-hand side, we have a product of three scalars, which gives us another scalar, which is exactly what we should have because the magnitude of a vector is in fact a scalar.

Finally, we also note that the angle between two vectors, πœƒ, is between zero and 180 degrees. The sine of any angle between zero and 180 degrees is positive or zero. Also, vector magnitudes are always positive or zero. So this expression is the product of three terms that are positive or zero, so it itself is positive or zero. And this is consistent because this formula gives us the magnitude of a vector, which, as we just said, is positive or zero. The dot product, however, can be positive, zero, or negative because the magnitudes of 𝐀 and 𝐁 are always positive. But the cosine of angles between zero and 180 degrees can be positive, negative, or zero. So in our equation, we take the absolute value of 𝐀 dot 𝐁 because 𝐀 dot 𝐁 can be positive or negative.

Okay, so now that we’ve expressed both the dot product and the magnitude of the cross product in terms of the magnitudes of 𝐀 and 𝐁, let’s plug these formulas into our equation. So we have that the magnitude of 𝐀 times the magnitude of 𝐁 times the sin of πœƒ all squared plus the absolute value of the magnitude of 𝐀 times the magnitude of 𝐁 times the cos of πœƒ squared is equal to 17,424. Note that the vertical bars around this first term disappeared because what we replaced was the quantity the magnitude of 𝐀 cross 𝐁. However, we still have the absolute value bars in the second term because what we replaced was 𝐀 dot 𝐁 not the absolute value of 𝐀 dot 𝐁.

However, 𝐀 dot 𝐁 is a real number. And for any real number, its square is the same as the square of its absolute value. For example, negative two squared is negative two times negative two, which is positive four, which is the same thing as two squared, and two is the absolute value of negative two. So, going back to our equation, we can remove the vertical bars from around this term because the square of a real number is always positive or zero. We are actually quite close to our solution. The next step is to replace these two squares of products with the products of squares. That is, the magnitude of 𝐀 times the magnitude of 𝐁 times the sin of πœƒ all squared is the magnitude of 𝐀 squared times the magnitude of 𝐁 squared times the sin of πœƒ squared.

Here, we use the convention where we represent the sin of πœƒ squared and the cos of πœƒ squared with the two coming before the πœƒ. This way, we don’t accidentally get confused and square the angle before applying the trigonometric function. Now, observe that we have a sin πœƒ squared and a cos πœƒ squared in two separate terms. And both of these terms have the same coefficient, the magnitude of 𝐀 squared times the magnitude of 𝐁 squared. This should remind us of the very important identity that for any angle πœƒ the cos of πœƒ squared plus the sin of πœƒ squared is always equal to one. If we factor out the common factor from these two terms, we get the magnitude of 𝐀 squared times the magnitude of 𝐁 squared times the quantity cos πœƒ squared plus sin πœƒ squared.

But according to our trigonometric identity, this term in the parentheses is exactly one. So we have the magnitude of 𝐀 squared times the magnitude of 𝐁 squared times one, which is just equal to the magnitude of 𝐀 squared times the magnitude of 𝐁 squared. Now, since this is equivalent to the left-hand side of our previous equation, it must be equal to the right-hand side of our previous equation 17,424. And now we have exactly what we need. The magnitude of 𝐀 squared times the magnitude of 𝐁 squared is 17,424, which is a number on one side of an equation, and the magnitude of 𝐀 and the magnitude of 𝐁 on the other side of the equation. But the magnitude of 𝐁 is what we’re looking for, and we’re given a value for the magnitude of 𝐀. So there’s actually only one unknown quantity in this equation. And we can directly solve for the magnitude of 𝐁.

We could start by dividing both sides by 𝐀 squared or taking the square root of both sides. But let’s instead start by substituting in 12 for the magnitude of 𝐀. If the magnitude of 𝐀 is 12, then the magnitude of 𝐀 squared is 12 squared, which is 144. So 144 times the magnitude of 𝐁 squared is 17,424. And dividing both sides by 144, we find that the magnitude of 𝐁 squared is 121. Now, 121, we may recognize, is exactly 11 squared. If we didn’t recognize this, we could just go ahead and take the square root of 121 and find that it is 11. Now, as usual, we have to be careful when taking the square root of 121 because 121 has two square roots: positive 11 and negative 11. However, recall that we are solving for the magnitude of a vector, and the magnitude of a vector is always positive or zero. So we want the positive square root of 121, which is 11.

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