MTH 252 Lecture Notes - Lecture 2: Parallelogram, Cross Product, Solution Set

26 views2 pages

skyferret769

6 Sep 2016

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

= cos(x), given that x is the angle between the two vectors. = 0 means that and are perpendicular. (this is the only such case that and would be perpendicular. Given , in , find that is perpendicular to both and. +y+z = 0 in this, y=-z, so substitute to solve for x. So for any real number r : r. (1,2,-2) = (r,2r,-2r) is the solution set. It automates choosing values (using 2 vectors, produces a third) It chooses the third vector smoothly : calculates area. The vector produced by x is perpendicular to both and . Something else to note is that x = -(x) This means that the same values will be produced, but will have opposite signs. Using the same algebra from day one, we can derive that: sin(x) = This allows us to calculate area of parallelograms.

Related Questions

[1] 1. (a) Solve 2 – 31 < 7. Solution: We have – 7<-3 < 7 which gives -4 0 on (-0, 1] U [2,). Next, we need to see where Vr2 – 3r +2 -1 > 0. If r2 - 3.c + 2 – 1 > 0, then V.22 – 3.0 +2 > . Observe that this is definitely true if r < 0. If r > 0, then squar- ing both sides we get 22 - 3.+2 > 1 2 > 3. V A VICO Thus, the domain of f is (-00, ). [2] (c) Find the exact value of tan(sec-14). Solution: Let y = sec-14. Then, sec y = 4. From the diagram we get that tan(sec-? 4) = tan y = 15 [2] (d) Find all r such that sin(2.x) = COS I. Solution: We have 2 sin r cos r = COS I, SO 0 = 2 sin Cos I -Cos I = cos .r (2 sin c - 1). Hence, sin(2x) = cost when cos x = 0, or when sin r = Thus, r= +ik, kez, I= +2nk, ke Z, or r= +2nk, kez [2] (e) State the precise mathematical definition of lim f(x) = 0. Solution: For every e > 0 there exists a 8 >0 such that if 0 < x-al <, then f(c) >

You know its nice to get the answer, but I'm trying to learn these so please explain it in detail please.

1) Find an equation of the plane: The plane that passes through the point (â2, 3, 1) and contains the line of intersection of the planes x + y â z = 5 and 3x â y + 5z = 4

2) Find an equation of the plane: The plane that passes through the line of intersection of the planes x â z = 2 and y + 4z = 3 and is perpendicular to the plane x + y â 2z = 5

3) Find the point at which the line intersects the given plane: x = 4 â t, y = 3 + t, z = 2t; x â y + 3z = 5

4) Find the cosine of the angle between the planes : x + y + z = 0 and x + 4y + 3z = 7

5) Find an equation for the plane consisting of all points that are equidistant from the points (7, 0, â2) and (9, 6, 0).

6) Find the distance from the point to the given plane : (1, â9, 5), 3x + 2y + 6z = 5

7) Find the distance between the given parallel planes. 3x â 2y + z = 9, 6x â 4y + 2z = 3

MTH 252 Lecture Notes - Lecture 2: Parallelogram, Cross Product, Solution Set

Document Summary

Get access

Related Documents

MATH 251 Midterm: MATH 251 TAMU 251-Spring 11 Exam 1s11 Solutions

MATH 230 Midterm: Math 230 Exam1-Solutions

Mathematics 1229A/B Study Guide - Final Guide: Elementary Matrix, Diagonal Matrix, Triangular Matrix

Related Questions