Question

Calc III Tutorial: Lines & Planes in 3 Dimensions

Answer 1

for line L in space through point P0, parallel to vector v, L can be defined as the set of all points P that make the vector P0P parallel to v (will draw this below)

Answer 2

\({\bf{Parametrization:}}\)
the above line L can also be expressed as the sum of the position vector for P and the position vector for P0

r(t) = r0 + tv where t is all real values
for parametric equations be sure to define your domain for t

for point P0 = (x0,y0,z0) parallel to v1 i + v2j + v3k

x = x0 + t(v1), y = y0 + t(v2), z = z0 + t(v3)
if you are given a point and a vector simply plug them in. if you are given two points, simply find the vector between the two points, then plug it in. in this case it doesn't really matter which point you use as the "base point" in the formula as both points are on the line at different t values.

to parametrize a line segment, simply restrict the domain from 0 ≤ t ≤ 1

Answer 3

\({\bf{Decomposing~the~Parametric~Function:}}\)

\[r0 + tv = r0 + t|v|\frac{ v }{ |v| }\] for all real t

recall that you can re-write vector v as the product of its magnitude |v| times its direction v/|v|. in this case, t * |v| is time * speed which is also equal to distance. so adding t|v| in the direction of v/|v| is the same as extending a point r0 to make a line

Answer 4

\({\bf{Distance~from~Point~to~Line:}}\)


if we wish to express this as a cross product simply multiply the numerator and denominator by |v|. as you recall, PSsin(theta)|v| is the cross product of PS and v

\[PSsin(\theta) = \frac{ PSsin(\theta) |v| }{ |v| } = \frac{ |PS × v| }{ |v| }\]

Answer 5

\({\bf{Equation~for~a~Plane:}}\) for a plane M normal to vector n, any vector P0P on the plane will also be normal to n. if we define vector n as n = Ai + Bj + Ck, point P0(x0,y0,z0) and P(x,y,z), the plane equation is

A(x-x0) + B(y-y0) + C(z-z0) = 0, or, when terms are combined,
Ax + By + Cz = D where D = Ax0 + By0 + Cz0

so if you are given a point P(x0,y0,z0) on the plane perpendicular to n = Ax0 + By0 + Cz0 just plug the appropriate components into A(x-x0) + B(y-y0) + C(z-z0) and simplify

if you are given three points, simply create two vectors like you would for a triangle, calculate the cross product, creating a new vector. then pick one of the points you were given, your new vector, and plug the components into A(x-x0) + B(y-y0) + C(z-z0) as you did before.

Two nonparallel planes will intersect at a vector. the cross product btwn the two plane equations will yield a vector parallel to the vector of intersection.

Answer 6

cont: from the previous scenario, if you want to find *the* line where two planes intersect, plug in some random component like z = 0 into both plane equations, solve for the other two variables, to generate a point common to both planes. generate the parametric equation using 1. that point and 2. the cross product you obtained earlier

to find where a point intersects a plane, simply take the parametric equation of the line and plug the components into the plane equation, solving for t, then plugging t back into the line equation

Answer 7

Source material is section 12.5 of Thomas' Calculus, Early Transcendentals, 14th edition by Hass, Heil, Weir, et. al.