Math 53 Summer 2018 Homework Assignment 6

James Rowan June 25, 2018

Assigned: Monday, June 25 Due: Thursday, June 28

0 Reading

Stewart Section 12.4

1 Textbook Exercises

Graded for completion. This portion of the homework is meant to help you practice important computational skills.

• Computing cross products: Stewart Section 12.4: exercises 2, 4, 8, 10

Solution to exercise 10. We know, since the cross product distributes over vector addition, that

k × (i − 2j) = k × i + k × (−2)j.

Since a × (kb) = ka × b for any vectors a, b ∈ R3 and scalar k, we know k × i + k × (−2)j = k × i − 2k × j.

Since k × i = j and k × j = −i,

k × (i − 2j) = j + 2i .

• Geometric interpretations of the cross product: Stewart Section 12.4: exercises 14, 16, 20, 28, 30

Solution to exercise 14. The length of u × v is |u||v| sin 45◦, so the length √ √ 2 of the cross product is 4 · 5 · 2 = 2 5. Using the right-hand rule, we can see that the vector u × v is directed up out of the page.

1 Solution to exercise 28. We see that the vector from P to Q is h3, 3, 3i − h1, 0, 2i = h2, 3, 1i and that the vector from P to S is h5, 2, 7i − h1, 0, 2i = h4, 2, 5i. Moreover, since the vector from P to R is h6, 5, 6i, which is the sum of the vectors from P to Q and from P to S, we see that the vectors from P to Q and from P to S form two adjacent sides of the parallelogram. The area of this parallelogram is then the length of the cross product of h2, 3, 1i and h4, 2, 5i. We compute:

i j k

h2, 3, 1i × h4, 2, 5i = 2 3 1 = 13i − 6j − 8k.

4 2 5 √ √ √ The length of this vector is 132 + 62 + 82 = 169 + 36 + 64 = 269

• The scalar triple product: Stewart Section 12.4: exercises 34, 38

Solution to exercise 38. The four points will lie in the same plane if and only if the scalar triple product of the displacement vectors from A to B, C, and D is zero. These displacement vectors from A are h2, −4, 4i, h4, −1, −2i, and h2, 3, −6i. The scalar triple product of these three vectors is

2 −4 4

4 −1 −2 = 2(6 + 6) − (−4)(−24 + 4) + 4(12 + 2) = 24 − 80 + 56 = 0,

2 3 −6 so the four points are coplanar.

• Properties of cross products: Stewart Section 12.4: exercise 50

Solution to exercise 50. We denote the vectors in coordinates as a = ha1, a2, a3i, b = hb1, b2, b3i, c = hc1, c2, c3i. We have:

a × (b × c) = a × ((b2c3 − b3c2)i − (b1c3 − b3c1)j + (b1c2 − b2c1)k)

i j k

= a1 a2 a3

b2c3 − b3c2 −(b1c3 − b3c1) b1c2 − b2c1

= (a2b1c2 − a2b2c1 + a3b1c3 − a3b3c1)i − (a1b1c2 − a1b2c1 − a3b2c3 + a3b3c2)j

+ (−a1b1c3 + a1b3c1 − a2b2c3 − a2b3c2)k

= (a2c2 + a3c3)b1i − (a2b2 + a3b3)c1i

+ (a1c1 + a3c3)b2j − (a1b1 + a3b3)c2j

+ (a1c1 + a2c2)b3k − (a1b1 + a2b2)c3k = (a • c)b − (a • b)c,

adding and subtracting the remaining terms a1b1c1i + a2b2c2j + a3b3c3k from the right-hand side of the second-to-last equality above.

2 2 True/false questions

Graded for correctness (+1 for correct, +0 for blank or incorrect). While we will not grade your reasoning, I recommend writing it down for yourself.

1. The expression “the cross product of h1, 2i and h0, 3i” makes sense exactly as written.

False . The cross product is deﬁned for vectors in R3, not for vectors in the plane.

2. If a, b, and c are vectors in R3 such that a × b = a × c, then we must have b = kc for some scalar k.

False . If a × b = a × c, then b and c need to be coplanar (so that the directions of their cross products with a agree), but they don’t have to be parallel. For example, consider the vectors a = i, b = j, and c = i + j.A quick computation shows that a × b = k and a × c = k, but b and c are not parallel. 3. The cross product of 0 (the zero vector) with any other vector is 0.

True . This can be seen by a direct computation with the determinant formula, or also by using the formula |a×b| = |a||b| sin θ; since the length of the zero vector is 0, the cross product will also have length zero, and thus be the zero vector. 4. If |a| = |b| and a × b = 0, then a = b.

False . If b = −a, then the angle between a and b will be π, so that its sine will be 0, and |b| will be equal to |a|, then a × b will be the zero vector but a will not be equal to b. 5. Using the vector tools developed so far, if we are given coordinates for three points P , Q, and R in space, we can ﬁnd all of the following things: (a) The lengths of the three sides of the triangle P QR (b) The cosines of the three angles of the triangle P QR (c) The area of the triangle P QR (d) A vector perpendicular to the plane containing P , Q, and R.

True . Let p, q, r denote vectors from the origin to P , Q, R. The lengths of the vectors p − q, q − r, and r − p will be the lengths of the three sides of P QR. By taking dot products of pairs of these vectors and using the fact that a • b = |a||b| cos θ, we can then ﬁnd the cosines of the three angles of the triangle P QR. By taking the cross product of p − q and

3 r − q, we can find a vector perpendicular to the plane containing P , Q, 1 and R, and moreover 2 times its length will give us the area of P QR. 6. When we use the scalar triple product a • (b × c) to find the volume of the parallelepiped with sides given by a, b, and c, we need to take the absolute value at the end because if we make changes to the order of a, b, and c in the scalar triple product, we might flip its sign.

True . We know that c × b = −b × c, and that u • (kv) = ku • v for any scalar k, so

a • (c × b) = a • −(b × c) = −a • (b × c),

showing that changing the order of the vectors in the scalar triple product will change the sign. Taking the absolute value at the end thus ensures we always get a positive value for volume.

3 Problems

Justify your answer, clearly explaining your reasoning. Correct answers with incorrect or unclear explanations will not receive full credit.

1. (sholeace formula) Suppose we are given the coordinates A = (x1, y1, 0), 3 B = (x2, y2, 0), C = (x3, y3, 0) of the vertices of a triangle ABC in R , and suppose that this triangle lies in the plane z = 0 with the origin (0, 0, 0) lying in the interior of the triangle.

(a) Sketch a picture in the plane z = 0 of the triangle ABC along with the vectors a, b, c from the origin to the three vertices A, B, C. The sketch below illustrates the situation:

4 We see that the triangle ABC has been split up into three smaller triangles OAB, OBC, and OCA. (b) Using the picture from part a) and the cross product, ﬁnd the area of the triangle ABC in terms of x1, x2, x3, y1, y2, and y3.

We ﬁnd the area of ABC by ﬁnding the areas of OAB, OBC, and OCA and adding them up. We see that the area of OAB is equal to 1 1 2 |a × b|, and similarly that the area of OBC is 2 |b × c| while the 1 area of OCA is 2 |c × a|. Evaluating a × b in coordinates, we see

i j k

a × b = x1 y1 0 = (x1y2 − x2y2)k.

x2 y2 0

Similarly, we ﬁnd b×c = (x2y3 −x3y2)k, and c×a = (x3y1 −x1y3)k. Adding up the lengths of these vectors and dividing by 2, we get that 1 the area of ABC is 2 |x1y2 − x2y2| + |x2y3 − x3y2| + |x3y1 − x1y3| .

5 Note: as long as the correct order/orientation is chosen so that all the cross products point in the same direction, this can also be written 1 as 2 (x1y2 − x2y2) + (x2y3 − x3y2) + (x3y1 − x1y3) . The term “shoelace formula” comes from the following way of writing down the multiplications done in this method:

(c) (optional challenge problem—will not be graded) Does this formula generalize to polygons in the plane with more than three vertices? What about ones which do not contain the origin in the interior?

We can similarly break down larger polygons containing the origin into triangles and use the cross product to ﬁnd the area of each of these pieces.

6 If the polygon does not contain the origin, then we can still use a method like this, using the fact that the cross product gives a directed area (up or down), so that areas in this decomposition outside the polygon are cancelled out. Try sketching a picture of this situation— it is a bit tricky!