# s11wk02su - Math 23 B Dodson Week 1 Homework(Continued 12.2...

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This preview shows page 1 out of 4 pages. Unformatted text preview: Math 23 B. Dodson Week 1 Homework: (Continued) 12.2 vectors: unit, standard unit, notations 12.3 dot product: orthogonal, proj, comp 12.4 cross product: formula, properties Problem 12.2.19a: Find |a| and a − 2b when a =< 6, 2, 3 >, b =< −1, 5, −2 > . Solution: The length |a| = √ 36 + 4 + 9 = 7, and a − 2b =< 6, 2, 3 > −2 < −1, 5, −2 > =< 8, −8, 7 > . Problem 12.2.25: Find a unit vector u that has the same direction as a = 8i − j + 4k. [variation/continuation: ﬁnd a vector of length 4 in the opposite direction.] Solution: The length |8i − j + 4k | = so the unit vector is u = √ 64 + 1 + 16 = 1 a= |a| 1 9 √ 81 = 9, 1 8i − j + 4k = 8 i − 9 j + 4 k. 9 9 2 Likewise, the vector with length 4 is − 4 8i − j + 4k . 9 Problem 12.3.23bc: Determine whether the given vectors are othogonal, parallel or neither. (b) a =< 4, 6 >, b =< −3, 2 > . (c) a = −i + 2j + 5k, b = 3i + 4j − k. Solution: (b) Recall that vectors are parallel if one is a scalar multiple of the other, a = cb, for a scalar c. This says < 4, 6 >= c < −3, 2 >=< −3c, 2c >, so 4 = −3c, and 6 = 2c. The second equation gives c = 3, but that’s not a solution of the 4 ﬁrst equation (with c = − 3 ), so there’s no solution c to the vector equation. Conclusion: the vectors in (b) are NOT parallel. On the other hand, the dot product a · b =< 4, 6 > · < −3, 2 > = 4 · (−3) + 6 · 2 = −12 + 12 = 0, so the cosine of the angle between these vectors is 0, and the vectors ARE orthogonal (perpendicular!). part (c): a = cb gives −1 = 3c, 2 = 4c and 5 = −c. We see c = −5 is inconsistent with the ﬁrst two equations, so the vectors are not parallel. We also check a · b = −i + 2j + 5k · 3i + 4j − k = −3 + 8 − 5 = 0, so the vectors are orthogonal again. 3 Week 1 Homework (continued): 12.3 dot product: orthogonal, proj, comp 12.4 cross product: formula, properties Problem 12.3.37: Find the scalar and vector projections of b on a when a =< 4, 2, 0 >, and b =< 1, 1, 1 > . Solution: The length |a| = √ √ 16 + 4 = 2 5, and the dot product a · b = 4 + 2 + 0 = 6, so the scalar projection is compa (b) = a·b 6 = √ . Note: this is a scalar! |a| 25 For the vector projection, proja (b) = compa (b) · ua = a·b 6 3 ·a = < 4, 2, 0 >= < 4, 2, 0 > . |a|2 20 10 Again, note, this one is a vector. Problem 12.4.3: Find the cross product of a = 2i + j − k and b = j + k . Solution: a × b = 2 0 1 1 where 21 2 −1 1 −1 >, , ,− −1 =< 01 01 11 1 a c b = ad − bc. d =< 1 + 1, −(2 − 0), 2 − 0 >=< 2, −2, 2 > . We check that this vector is perpendicular to both a and b by ﬁnding the dot products. 4 Homework (Continued): ﬁnish 12.4 cross product: properties Start of week 2 hw: 12.5 lines, planes Problem 12.4.25: For points P (1, 0, 0), Q(0, 2, 0) and R(0, 0, 3) (a) Find a vector orthogonal to the plane through P, Q, R. (b) Find the area of the triangle P QR. Solution: The vector P Q =< 0, 2, 0 > − < 1, 0, 0 >=< −1, 2, 0 > is in the plane through P, Q, R; and so is P R =< −1, 0, 3 > . We use P Q × P R for the normal vector. P Q × P R = −1 −1 =< 2 0 0 3 20 −1 0 −1 2 ,− , > 03 −1 3 −1 0 =< 6, 3, 2 > . (b) The area of the triangle is half the area of the parallelogram, so Area(P QR) = 1 2 · | < 6, 3, 2 > | = 7 . 2 ...
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