Question Video: Using the Dot Product between Vectors to Find the Area of a Triangle Mathematics • 12th Grade

Video Transcript

Suppose that vector 𝐀 equals one, one, three and vector 𝐁 equals four, eight, negative eight fix two sides of a triangle. What is the area of this triangle, to the nearest hundredth?

Okay, so in this example, these two vectors 𝐀 and 𝐁, we’re told, define two sides of a triangle. If we connect the tips of these two vectors like this, we’ve got our triangle. And our question asks us to solve for its area. We want to combine the vectors 𝐀 and 𝐁 in some way that we can solve for this. And to start doing that, let’s recall that if two such vectors define two adjacent sides of a parallelogram like we see here, then the magnitude of the cross product of these two vectors equals that area.

Returning to our vectors 𝐀 and 𝐁 given in this problem statement, we can say that if we were to take the magnitude of their cross product, we would solve for this entire interior area. And notice that that is twice as big as the area of the triangle we actually want to solve for. Written mathematically, the magnitude of 𝐀 cross 𝐁 is twice the area of our triangle. Therefore, if we calculate the magnitude of 𝐀 cross 𝐁 divided by two, we’ll solve for the value of interest.

As our first step, let’s calculate the cross product 𝐀 cross 𝐁. In general, the cross product of two three-dimensional vectors, we’ve called them 𝐀 and 𝐁 here, is equal to the determinant of this three-by-three matrix. The first row of the matrix is populated by the unit vectors 𝐢, 𝐣, and 𝐤. The second and third rows has the corresponding components of our two three-dimensional vectors. So in our scenario with the given vectors 𝐀 and 𝐁, their cross product is the determinant of this matrix with the components of 𝐀 and 𝐁 yet to be filled in.

Let’s start with the components of our first vector, vector 𝐀. We see from the problem statement that it has components one, one, and three. And then moving on to our second vector, we see the components of 𝐁 are four, eight, and negative eight. Our task now is to compute the determinant of this matrix. We’ll do it component by component, starting with the 𝐢-component. Crossing out the row and column of our matrix that contain this element, we can say that this component is given by the determinant of this two-by-two matrix. That determinant equals one times negative eight or negative eight minus three times eight or 24.

We then move on to the 𝐣-component equal to negative the determinant of this matrix. One times negative eight once again is negative eight. And from that, we subtract three times four or 12. Finally, the 𝐤-component of this cross product equals the determinant of this matrix. One times eight is eight. And from that, we subtract one times four or four. We find then that the cross product of 𝐀 and 𝐁 is equal to negative 32𝐢 plus 20𝐣 plus four 𝐤. Now that we’ve figured this out, we want to solve for the magnitude of this resulting vector.

Clearing a bit of space on screen, we can remind ourselves that if we have a three-dimensional vector, the magnitude of that vector equals the square root of the sum of the squares of each of its three components. This tells us that the magnitude of 𝐀 cross 𝐁 equals the square root of the sum of negative 32 squared, 20 squared, and four squared. This is equal to the square root of 1440. From here, all we need to do to calculate the area of our triangle is divide this result by two. When we enter this expression on our calculator to the nearest hundredth, we get a result of 18.97.

And we know that this number refers to an area. So even though we don’t know its units, we’ll report our answer as 18.97 units of area. That’s the size of this triangle where vectors 𝐀 and 𝐁 fix two of its sides.