Notes of Formulas, Concepts for Test #1 Mat 240
10.1/10.2
(1) Component Form of a vector running from initial point (x1,y1) to terminal point (x2,y2) is written <a,b> = <x2 - x1,y2 - y1>. (a,b) is the terminal point of the vector now, and the initial point is the origin (0,0).
When we graph the vector so that it's original
point is the origin, we say that it is in
standard position.
problems: #1,#4, #12, p. 723,
#48,49,51, p. 733
(2) The magnitude of a vector in component form is :
, in 2-space, problems: #31,32, p. 724.
in 3-space. #79,80, p. 733
(3) We add vectors by adding the corresponding components, and forming a new vector:
<u1,u2,u3> + <v1,v2,v3> = <u1+v1,u2+v2,u3+v3>. (We leave off the third part in ).
Geometrically, we call u + v the resultant, and it obeys the so-called "parallelogram rule": u + v is the diagonal of the parallelogram with u and v as two of it's sides.
problems: #26,#28, p. 724, #61,62, p. 733
(4) When we multiply a vector by a constant, we call the constant
a scalar, and we simply multiply all of the components of the vector
by the scalar.
k<v1,v2,v3> = <kv1, kv2, kv3>. Geometrically, the scalar multiple is a vector
in the same direction as v. It extends v by a factor of k if k > 1, and it shortens v by a factor of k if k < 1.
problems: #17, p. 724
(5) A unit vector is a vector of length 1. We can find a vector of length 1
in the same direction as a given vector v by taking this scalar multiple:
. problems: #37,38, p. 724
(6) There are three special unit vectors (two in ) called i = <1,0,0>, j = <0,1,0>, and
k = <0,0,1>. We can express any vector in the plane or 3-space as a linear combination of the standard unit vectors.
<a,b,c> = ai + bj + ck. , or <a,b> = ai + bj .
(7) Trigonometric Representation of a vector. We can represent a vector using trigonometry if we know the angle that v makes with the x-axis, and the length of v.
v = cos i + sin j.
This form is useful in some 'force problems' we did.
problems: #52,53, p. 724, #80,81, p. 725.
(8) In addition, you should know how to plot points in 3-space. problems: #1-6, p. 732.
10.3
(9) The dot product of two vectors uv is obtained by multiplying the corresponding coordinates, then adding to get a real number:
<u1,u2,u3 <v1,v2,v3> = u1v1 + u2v2 + u3v3 (3-space). problems: #3a,2a,6a, p. 740
<u1,u2 <v1,v2> = u1v1 + u2v2 (2-space).
(10)One use we had for the dot product was to find the angle between two vectors:
cos We can then take arccos of the right-hand side to find the actual angle.
problems: #12,13,15,16, p. 741.
An important side-effect of this rule is that orthogonal vectors have dot product 0.
(11) Another use of the dot product was to solve the configuration known as the projection problem. Given a vector u and another v, what we can do is 'decompose' u (which means break it into the sum of two vectors) into two components, one in the direction of v, which we call w1, and another orthogonal to v, which we call w2.
u = w1 +w2.
We can find w1= projvu = by theorem 10.6.
And then we find w2 = u - w1.
problems: #45,47,48,67, pps. 742-743
10.4
(12)The cross-productu v is formed by evaluating the 3 x 3 determinant:
= (u2v3 - u3v2)i - (u1v3 - u3v1)j + (u1v2 - u2v1)k
(13) One important concept that we use in connection with the cross-product is that u v is orthogonal to both u and v.
problems: #11-16, p. 750
10.5
(14) To find the equation of a line in space, we need a point on the line (x1,y1,z1), and a direction vector u = <a, b, c>. There are two forms of the equation of the line:
(a) Parametric Form: x = x1+ at, y = y1 + bt, z = z1 + ct.
(b) Symmetric Form: ,
There are two basic types of problems that you should know how to solve for the test:
(15) The easy type I problem, where you are given the point on the line, and the direction vector, and all you have to do is plug them into the formula(s). problems: #3-8, p. 759
(16) The type II problem, where you are given just two points on the line, and asked to find the equation(s) of the line. In this case, you
(i) find the vector containing these two in component form, and use it as the direction vector: <a,b,c> = <x2 - x1,y2 - y1, z2 - z1>.
(ii) Pick one of the given points, and plug it and <a,b,c> into the formula(s).
problems: #9-12, p. 759
(17) Finding the equation of a plane in space. To do this, you need a point (x1,y1,z1) in the plane, and a normal vector to the plane n = <a,b,c>. The formula is:
a(x - x1) + b(y - y1) + c(z - z1) = 0. After distributing, this can be simplified to:
ax + by + cz + d = 0.
Two types of problems that you might anticipate with this subject are:
(18) Type I - the easy one, where you are just given the point and the normal vector, and then you have to plug them into the formula.
problems: #29,30, p. 760.
(19) Type II - given three points in the plane, find the equation of the plane. In this case, you have got to:
(i) Find two vectors u and v in the plane, by pairing off points, and subtracting terminal - initial to find the vectors.
(ii)Then find the normal vector as <a,b,c> = u v, which is orthogonal to these vectors.
(iii) Then pick one of the points to be (x1,y1,z1), and plug it and <a,b,c> into the formula.
problems: #33,34,35, p. 760.
(20) You should also be able to sketch a plane. We do this by finding the intercepts - the points where the plane intersects the x-, y-, and z-axis. To find a particular intercept, set the other two coordinates equal to 0, and solve for the remaining one:
(x,0,0) : x-intercept,
(0,y,0): y-intercept,
(0,0,z): z-intercept.
Then graph these points, and draw the triangle cutting between them by connecting the points - this is a sketch of the plane, the part that cuts between the three axes.
problems: #51-55, p. 761