Video Transcript
๐ด๐ต๐ถ๐ท is a parallelogram with the vector ๐ด๐ต having components negative one, one, three and the vector ๐ด๐ท having components three, four, one. Find the area of ๐ด๐ต๐ถ๐ท. Give your answer to one decimal place.
Weโre told in the question that ๐ด๐ต๐ถ๐ท is a parallelogram. So letโs draw it. Hereโs our parallelogram. Now, we just need to label its vertices. Once weโve called one of the vertices ๐ด, then we have only two choices for where ๐ต goes. It has to be adjacent to ๐ด. So ๐ต either has to go here or here. And once weโve chosen where ๐ด and ๐ต go, weโve got no choice for ๐ถ and ๐ท. ๐ถ has to be adjacent to ๐ต. And thereโs only one spot left for ๐ท. So starting from ๐ด, we can walk around the parallelogram, visiting ๐ต first, then ๐ถ, and lastly ๐ท before we end up at ๐ด again. In other words, we can visit the vertices in the same order they come up in the name of the parallelogram ๐ด๐ต๐ถ๐ท.
Okay, well what else weโre told in the problem text? Weโre given the components of the vector ๐ด๐ต. Letโs mark them on our picture and also the components of the vector ๐ด๐ท. And what weโre looking for? We want to find the area of the parallelogram ๐ด๐ต๐ถ๐ท.
To find this area, we use the fact that the magnitude of the cross product of two vectors ๐ข and ๐ฃ is the area of the parallelogram whose adjacent sides are ๐ข and ๐ฃ. Weโre looking for the area of the parallelogram whose adjacent sides have components negative one, one, three and three, four, one. And the rule above tells us that this is the magnitude of the cross product of the two vectors. So letโs clear some room and find this magnitude.
Before finding the magnitude of the cross product, of course weโre gonna have to find the cross product itself. And we can write this cross product as the determinant of a three-by-three matrix, whose first row contains the unit vectors in the ๐ฅ-, ๐ฆ-, and ๐ง-directions ๐, ๐, and ๐. The second row contains the components of the first vector in our cross products negative one, one, and three. And the third row contains the components of the second vector in our cross product three, four, and one.
We can expand the determinant along the first row. And evaluating each two by two, this is how meant we get negative 11๐ plus 10๐ minus seven ๐, which we can write in components as negative 11, 10, negative seven. This is the cross product of the vectors that weโre looking for the area, which is the magnitude of this cross product. So weโre looking for the magnitude of the vector with components negative 11, 10, and negative seven.
And the magnitude of the vector is just the square root of the sum of squares of the components. So we have the square root of negative 11 squared plus 10 squared plus negative seven squared. We can write this exactly as the square root of 270 or as three times the square root of 30. But weโre not asked for the exact answer. Weโre asked for the answer correct to one decimal place. And correct to one decimal place, the square root of 270 is 16.4. So our parallelogram ๐ด๐ต๐ถ๐ท has an area of 16.4 units correct to one decimal place.
We found this by computing the magnitude of the cross product of the vectors along two of the adjacent sides of the parallelogram. And we were lucky that we were given the components of two adjacent sides in the problem. We werenโt given for example the components of one of the diagonals either ๐ด๐ถ or ๐ต๐ท. And to make sure that the vectors we were given were along two adjacent sides. We had to be careful how we label the vertices of our parallelogram.
The parallelogram Iโve just drawn is not a correct diagram for a parallelogram ๐ด๐ต๐ถ๐ท. Going around the vertices in order, we might get ๐ด๐ต๐ท๐ถ or ๐ด๐ถ๐ท๐ต, but not ๐ด๐ต๐ถ๐ท. Had we used this incorrect diagram, we would have got an incorrect answer and thatโs why it was worth spending some time at the beginning to label our vertices correctly.
