Exam_1_20C_Solution

Exam_1_20C_Solution - Math 20C (Shenk). Summer, 2010. Exam...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 20C (Shenk). Summer, 2010. Exam 1 Solution. 1. 2. 3. 2 A · B = 2, 1, 3 · −1, 4, k = −2 + 4 + 3k = 2 + 3k • A · B = 0 ⇐⇒ 2 + 3k = 0 • k = − 3 √ C · D = 1, 2, 2 · 1, 2, −2 = 1(1) + 2(2) + 2(−2) = 1 • |C| = 12 + 22 + 22 = 3 • √ C·D 1 1 = = 1 • θ = cos–1 ( 9 ) |D| = 12 + 22 + 22 = 3 • cos θ = 9 |C||D| 3(3) − → − → One solution: OQ = 3, 3, 3 • P S = −1 − 1, 4 − 1, 2 − 1 = −2, 3, 1 − →− →− →− →− → OR = OQ + QR = OQ + P S = 3, 3, 3 + −2, 3, 1 = 1, 6, 4 • R = (1, 6, 4) x =3+t y =1−t z = 2 + 2t 4. One answer: Normal vector: n =< 1, −1, 2 > • Line: 5. − → One answer: P Q = −2 − 1, 3 − 4, 0 − 2 = −3, −1, −2 • Line: 6. − → One solution: P Q = 4 − 2, 3 − 3, 5 − 3 = 2, 0, 3 i jk x = 1 − 3t y =4−t z = 2 − 2t − → • P R = 2 − 2, 6 − 3, 4 − 2 = 0, 3, 2 − →− → 03 23 20 i− [Normal vector ] = n = P Q × P R = 2 0 3 = j+ k 32 02 03 032 = [0(2) − 3(3))]i − [(2(2) − 3(0)]j + [2(3) − 0(0)]k = −9i − 4j + 6k = −9, −4, 6 • − → − → Check: P Q · n = 2(−9) + 0(−4) + 3(6) = 0 • P R · n = 0(−9) + 3(−4) + 2(6) = 0 • Plane (using P ): −9(x − 2) − 4(y − 3) + 6(z − 2) = 0 7. 8. • 123 01 05 15 +3 −2 1, 2, 3 · [ 0, 1, 5 × 4, 0, 1 ] = 0 1 5 = 1 40 41 01 401 = 1[(1)(1)−(5)(0)]−2[(0)(1)−(5)(4)]+3[(0)(0)−(1)(4)] = 1(1)−2(−20)+3(−4) = 1+40−12 = 29 Figure A8 y 3 2 1 1 Figure A6 1 2 3 x Math 20C, Summer, 2010 9. Exam 1 Solution, p. 2 d • v (t ) = t − t2 , t2 + t = 1 − 2t, 2t + 1 • [Speed at time t] dt √ √ = |v(t)| = (1 − 2t)2 + (2t + 1)2 = 1 − 4t + 4t2 + 4t2 + 4t + 1 = 2 + 8t2 (b) The speed is a minimum at t = 0. • v(0) = 1, 1 • Figure A9 (a) r(t) = t − t2 , t2 + t y 3 v(0) = 1, 1 Figure A9 −3 x ...
View Full Document

Ask a homework question - tutors are online