If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
query_09_6
Question: Write the equation of the plane 3(x-2) - 2(y-1) - 3(z-5) = 0 in standard form.
Your solution:
Confidence rating:
Given Solution:
The standard form A x + B y + C z + D = 0 is easily found by applying the distributive law:
We get
3 x - 6 - 2 y + 2 - 3 z + 15 = 0
which we simplify to get
3 x - 2 y - 3 z + 11 = 0.
Self-critique (if necessary):
Self-critique rating:
Question: Find the equation of the plane containing the point P(-1,3,2) and having normal vector N = 3j - 1k.
Your solution:
Confidence rating:
Given Solution:
(x, y, z) lies on the plane if and only if the vector (x + 1) `i + (y - 3) `j + (z - 2) `k, from P to (x, y, z), is perpendicular to N.
This condition is
((x + 1) `i + (y - 3) `j + (z - 2) `k ) dot (3 `j - `k) = 0
giving us
3 ( y - 3 ) - (z - 2) = 0,
which simplifies to
3 y - z - 7 = 0.
Self-critique (if necessary):
Self-critique rating:
Question: Find two unit vectors perpendicular to the plane x + 3y - 4z = 2.
Your solution:
Confidence rating:
Given Solution:
A vector perpendicular to the plane is `i + 3 `j - 4 `k.
A unit vector in this direction is `i sqrt(26) / 26 + 3 j sqrt(26) / 26 - 4 k sqrt(26) / 26 = `i sqrt(26) / 26 + 3 j sqrt(26) / 26 - 2 k sqrt(26) / 13.
Another vector perpendicular to the plane is the negative of the preceding.
Self-critique (if necessary):
Self-critique rating:
Question: Find the distance between the point (-1,2,1) and the plane which contains the point (3,3,-2) and is normal to the vector N = -2i + j + 3k.
Your solution:
Confidence rating:
Given Solution:
A vector from the first point to the second is
`u = (3 - (-1) ) `i + (3 - 2) `j + (-2 - 1) `k = 4 `i + `j - 3 `k.
The component of this vector perpendicular to the plane is found by projecting `u onto the normal vector. The magnitude of the projection is ( `u dot `N / || `N || ) = -16 / sqrt(14), which can easily be simplified and approximated. This is the distance between the first point and the plane.
Note on vector projection:
We don't need it here, but the vector projection of `u onto `N is
( `u dot `N / || `N || ) * `N / || `N ||
= (-16 / sqrt(2^2 + 1^2 + 3^2) ) * (-2i + j + 3k) / sqrt(2^2 + 1^2 + 3^2)
= -16 / 14 * (-2i + j + 3k).
The magnitude of this vector is the requested distance.
Note that ( `u dot `N / || `N || ) is the magnitude of the projection of `u onto `N. This is multiplied by the unit vector `N / || `N || to get a vector of the appropriate magnitude in the direction of `N.
Self-critique (if necessary):
Self-critique rating:
Question: Find the distance between the lines (x+1)/(-2) = (y+2) / (-2) = (z+1)/(-1) and (x-4)/5 = (y+1)/2 = (z-1)/3
Your solution:
Confidence rating:
Given Solution:
The lines are in the directions of the respective vectors `u = -2 `i - 2 `j - `k and `v = 5 `i + 2 `j + 3 `k.
The distance between the lines is measured perpendicular to both lines, in the direction of `u X `v = -4 `i + `j + 6 `k.
Any vector from a point of one line to a point of the other will project onto this vector in such a way that the magnitude of the projection is equal to the distance between the lines.
The point (-1, -2, -1) is on the first line, and the point (4, -1, 1) is on the second. A vector from the first to the second is therefore
`w = 5 `i - `j - 2 `k
The magnitude of the projection of this vector onto `u X `v is
`w dot (`u X `v) / || `u X `v || = (5 * -4 + -1 * 1 + (-2) * 6 ) / sqrt(4^2 + 1^2 + 6^2) = -33 / sqrt(53), which can be put into standard form and approximted (approximate value is between 4 and 5, so the lines are between 4 and 5 units apart).
STUDENT QUESTION
I used
|PQ dot N|/|| N ||
The given solution says to use the formula `w dot (`u X `v) / || `u X `v ||, but isn't that the volume formula????????
INSTRUCTOR RESPONSE
`u X `v is the area of the base of a
parallellopiped defined by `u, `v and `w.
`w dot (`u X `v) is the volume of that parallellopiped, since w cos(theta) is
the altitude perpendicular to that base.
However in the interpretation used here
`u X `v is the vector normal to the two lines
(`u X `v) / || `u X `v || is a unit vector normal to the two lines
`w is a vector from a point on one line to a point on the other
and
`w dot (`u X `v) / || `u X `v ||
is the component of `w normal to the two lines, which is the distance between
the lines.
The given solution does not quote a formula, but rather indicates the geometry
of the solution. The notation is different than the formula given by your text,
but the interpretation is identical.
Self-critique (if necessary):
Self-critique rating:
Question: Find the equation of the sphere with center C(-2,7,1) and tangent the the plane x + 4y - 2z = 10.
Your solution:
Confidence rating:
Given Solution:
The sphere has equation (x - (-2)) ^ 2 + (y - 7) ^2 + (z - 1)^2 = r^2, where r is its presently unknown radius.
The sphere is tangent to the plane, which by the geometry of circles and spheres implies that a vector from the center of the sphere to the point of tangency is perpendicular to the plane. It follows that the magnitude of that vector is equal to the distance from the point to the plane.
So to find r we need only find the distance from (-2, 7, 1) to the plane x + 4 y - 2 z = 10.
We do this by finding some point, any point, on the plane, and projecting the vector from (-2, 7, 1) to that point onto the vector `i + 4 `j - 2 `k which is normal to the plane.
Self-critique (if necessary):
Self-critique rating: