Practice 1 MAT 240
(1) Find the component form, magnitude, and a unit vector in the direction of the vector with initial point ( -4, 0, 3) and terminal point ( 2, 3, 5).
(2) For the vectors u = < 6, 3, 2> and v = < 2, -2, 0>:
(a) find 2u – 3v
(b) Find the angle between the two vectors
(c) Give a vector which is normal to both u and v.
(3) A 200 pound dolly is on a ramp at an inclination of . How much force is required to keep the dolly stationary on the ramp?
(4) If u = < -2, 4 >, and v = < 3, x > and u is orthogonal to v, what is x?
(5) Find parametric and symmetric equations of the line containing the points (-4, 0, 3) and (2, 3, 5).
(6) (a) Find the equation of the plane containing the points
(0, 2, 0), (-4, 1, 3) and (2, 3, 5).
(b) Find the intercepts, and sketch the plane
(c) Find the distance between the plane in (6) and the point Q = (6, 2, -4).
(7) Tell the type of quadric surface given by the equation, and give a rough sketch, directed down the proper axis (if that is relevant) .
(a)
(b)
(c)
(8) What are the (xy-, xz-, and yz-) traces of the following quadric surfaces?
(a)
(b)
Solutions:
(1) Subtract the components of the initial point from those of the terminal point:
u = < 2 - - 4, 3 – 0, 5 – 3> = < 6, 3, 2 >
Magnitude: ||u|| =
Unit vector:
(2)
(a) 2< 6, 3, 2 > - 3< 2, -2, 0> = <12 – 6, 6 - - 6, 4 – 0> = < 6, 12, 4 >
(b)
(c) The cross product is normal to both vectors:
(3) Model the ramp as a unit vector with a trigonometric representation:
v = cos(30)i + sin(30)j, and then the force of the truck as F = -200j (this is because the weight of the truck is driven straight down by gravity) . The force required to keep the truck up is
projvF = =
which has magnitude 100 lbs.
(4) If they are orthogonal, then their dot product is zero – so dot them up and solve for x:
(5) We need a direction vector – take the vector with the given points as initial and terminal points: u = < 2 - - 4, 3 – 0, 5 – 3> = < 6, 3, 2 >
Then choose one of the points (I'm going to use (2, 3, 5) , and use the formulas
parametric:
symmetric:
(6) (a) Let P = (0, 2, 0), Q = (-4, 1, 3) and R = (2, 3, 5).
Use these points to find a couple of vectors in the plane:
Then find the cross-product of these to generate our normal vector:
Pick a point (I'm going to use (0, 2, 0)) and use the formula;
-8(x – 0) + 26(y – 2) - 2(z – 0) = 0,
-8x +26y- 2z= 52, divide through by -2:
4x – 13y + z = -2.
note: when a plane's equation is missing a variable (here, z) that means that it is parallel to the plane
(b) Intercepts: , ,
(c) Use one of the points in plane (i'm going to use (0,2,0) ) , the normal vector
n = <4, -13, 1> and then the formula:
(7) Get it into recognizable standard form:
(a) The power on y is 1, and thus the quadric surfaces that we have seen in these situations must be solved for the variable with power 1, if present:
, this is is an elliptic paraboloid pointed in the y-direction (it's the special variable.
(b) We can either subtract away the terms on the right or left – the constant term is going to be zero, and from our list of surfaces, that means we're going to need two positive terms – so move everything to the left.
x is the special variable and direction of this elliptic cone.
(c) Move the variables all to the left: The special variable (and direction) is z, for this hyperboloid of two sheets.
(8) (a) xy: hyperbola, (b) xz: hyperbola, (c) yz: none
(b) All traces are circles – this is an ellipsoid which happens to be a sphere as well.