### Video Transcript

Madison has fitted a triangular shade sail in her garden, as shown in the diagram. Find the area of the shade sail. Give your answer correct to two decimal places.

There are many ways to solve this question. Weβre going to see an approach which uses vectors. The key thing here is to see this triangular shade sail as half of a parallelogram. And so the area of the triangular shade sail is half the area of the parallelogram. And hopefully we know how to find the area of a parallelogram using vectors.

If we call the vectors along the two adjacent sides of the parallelogram π’ and π£, as shown, then the area of the parallelogram is the magnitude of the vector π’ cross π£. Putting this together with the previous statement, the area of the shade sail is then half the magnitude of π’ cross π£.

Okay, but how are we going to find the cross product of π’ and π£ without their components? Weβre not told the components of π’ and π£ in the question. We have to make some choices about the directions of the π₯-, π¦-, and π§-axes, which will determine the directions of the unit π, π, and π vectors.

The bottom left-hand corner of our diagram shows three mutually perpendicular directions. So we might as well choose those directions to be π, π, and π in some order. And as is somewhat conventional, I choose π and π to be the horizontal vectors and π to point upwards. With this choice, itβs relatively easy to see what π£ is. Itβs pointing in the π-direction. And its length is eight meters. So it must be eight π.

But how about the vector π’? How do we find its components? Well, luckily, we can see a path from the initial point or tail of π’ to its terminal point or tip. We can go down to the ground, along the ground to the other wall, along the base of that wall, and then up the wall. And so the vector π’ is just the sum of these vectors. The first of these vectors points in the opposite direction to the vector π. And looking at the other end of the wall, we can see that its magnitude must be two. So this vector must be negative two π.

How about the next vector which points in the opposite direction to the vector π and has a magnitude of six? It must be negative six π. The third vector which has a magnitude of three meters and points in the π-direction must be three π. And finally, the fourth vector pointing in the π-direction has magnitude four, the height of that wall, and so is four π.

Now we just need to simplify this sum, to find that it is negative six π plus three π plus two π. Now perhaps itβs a good idea to write both vectors in component form because weβll need them in this form to find the cross product, which is after all what we used to find the area of the sail.

Letβs clear some room and compute this cross product. Okay, now that we have some room, letβs compute this cross product and hence find the area. The cross product of π’ and π£ is the determinant of the three-by-three matrix whose first row contains the unit vectors π, π, and π, the second row contains the components of π’, and the third row contains the components of π£. We expand this determinant along the first row and compute the values of the individual two-by-two determinants. Again, we can write this in component form if we wanted to.

Now that we found π’ cross π£, letβs clear some more room and find its magnitude. In fact, we want to find half its magnitude. And how do we find the magnitude of a vector? Well, thatβs the square root of the sum of squares of the components.

And now this is just a matter of either simplification or putting it into our calculator. We get a half times 16 root 10, which is of course eight root 10. And as this quantity is an area where all the lengths were given in meters, this area has units of meters squared. So this is the exact area of the triangular shade sail.

But we only need this correct to two decimal places. So again, we turn to our calculator. Eight root 10 is 25.29822 dot dot dot. And to two decimal places, this gives us 25.30. The area of Madisonβs triangular shade sail is 25.30 meters squared, correct to two decimal places.

To find this, we used the fact that the area of a triangle in 3D space is half the magnitude of the cross product of the vectors along two of its sides.