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π΄π΅πΆπ· is a parallelogram with the vector π΄π΅ having components negative one, one, three and the vector π΄π· having components three, four, one.
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Find the area of π΄π΅πΆπ·.
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Give your answer to one decimal place.
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Weβre told in the question that π΄π΅πΆπ· is a parallelogram.
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So letβs draw it.
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Hereβs our parallelogram.
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Now, we just need to label its vertices.
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Once weβve called one of the vertices π΄, then we have only two choices for where π΅ goes.
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It has to be adjacent to π΄.
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So π΅ either has to go here or here.
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And once weβve chosen where π΄ and π΅ go, weβve got no choice for πΆ and π·.
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πΆ has to be adjacent to π΅.
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And thereβs only one spot left for π·.
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So starting from π΄, we can walk around the parallelogram, visiting π΅ first, then πΆ, and lastly π· before we end up at π΄ again.
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In other words, we can visit the vertices in the same order they come up in the name of the parallelogram π΄π΅πΆπ·.
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Okay, well what else weβre told in the problem text?
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Weβre given the components of the vector π΄π΅.
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Letβs mark them on our picture and also the components of the vector π΄π·.
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And what weβre looking for?
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We want to find the area of the parallelogram π΄π΅πΆπ·.
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To find this area, we use the fact that the magnitude of the cross product of two vectors π’ and π£ is the area of the parallelogram whose adjacent sides are π’ and π£.
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Weβre looking for the area of the parallelogram whose adjacent sides have components negative one, one, three and three, four, one.
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And the rule above tells us that this is the magnitude of the cross product of the two vectors.
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So letβs clear some room and find this magnitude.
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Before finding the magnitude of the cross product, of course weβre gonna have to find the cross product itself.
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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 π₯-, π¦-, and π§-directions π, π, and π.
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The second row contains the components of the first vector in our cross products negative one, one, and three.
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And the third row contains the components of the second vector in our cross product three, four, and one.
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We can expand the determinant along the first row.
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And evaluating each two by two, this is how meant we get negative 11π plus 10π minus seven π, which we can write in components as negative 11, 10, negative seven.
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This is the cross product of the vectors that weβre looking for the area, which is the magnitude of this cross product.
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So weβre looking for the magnitude of the vector with components negative 11, 10, and negative seven.
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And the magnitude of the vector is just the square root of the sum of squares of the components.
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So we have the square root of negative 11 squared plus 10 squared plus negative seven squared.
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We can write this exactly as the square root of 270 or as three times the square root of 30.
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But weβre not asked for the exact answer.
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Weβre asked for the answer correct to one decimal place.
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And correct to one decimal place, the square root of 270 is 16.4.
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So our parallelogram π΄π΅πΆπ· has an area of 16.4 units correct to one decimal place.
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We found this by computing the magnitude of the cross product of the vectors along two of the adjacent sides of the parallelogram.
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And we were lucky that we were given the components of two adjacent sides in the problem.
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We werenβt given for example the components of one of the diagonals either π΄πΆ or π΅π·.
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And to make sure that the vectors we were given were along two adjacent sides.
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We had to be careful how we label the vertices of our parallelogram.
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The parallelogram Iβve just drawn is not a correct diagram for a parallelogram π΄π΅πΆπ·.
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Going around the vertices in order, we might get π΄π΅π·πΆ or π΄πΆπ·π΅, but not π΄π΅πΆπ·.
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Had we used this incorrect diagram, we would have got an incorrect answer and thatβs why it was worth spending some time at the beginning to label our vertices correctly.