Sean Fitzpatrick |
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University of Lethbridge |
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Compute the cross product of \(\vec{v}=\la 3, -1, 4\ra\) and \(\vec{w} = \la 2,0,-1\ra\text{.}\)
Find a vector equation for the line through \(P=(1,2,3)\) and \(Q=(3,-1,4)\)
Points on a line in space are given in terms of a parameter (usually \(t\) — we can think of motion in a straight line, with \(t\) as time).
The vector equation of a line through \(P_0=(x_0,y_0,z_0)\) in the direction of \(\vv = \la a,b,c\ra\) is
Sometimes see \(\vec{r}\) or \(\vec{r}(t)\) instead of \(\vec{x}\text{.}\)
Equating coefficients in the vector equation gives the parametric equations:
Find the vector equations of the lines:
Through the point \(P_0 = (2,-5,1)\) in the direction of \(\vv = \la 6,-7,3\ra\)
Through the points \(P=(4,5,-3)\) and \(Q=(-1,8,2)\text{.}\)
Through the point \(P_0=(4,5,-9)\) and parallel to the line
Two lines in \(\R^3\) are parallel if their direction vectors are parallel.
Not all parallel lines intersect: some are skew.
Checking for intersection leads to a system of equations. (Use a different parameter for each line.)
Determine if the following pairs of lines are parallel, skew, or if they intersect.
Given vectors \(\uu\) (not equal to \(\vec{0}\)) and \(\vv\text{,}\) often useful to write \(\vv\) as the sum of a vector parallel to \(\uu\text{,}\) and a vector orthogonal to \(\uu\text{.}\)
Vector \(\vec{a}\) called the projection of \(\vv\) onto \(\uu\text{.}\) Notation and formula:
Given \(\vv = \la 3,-1\ra\) and \(\ww = \la 2,5\ra\text{,}\) find vectors \(\vec{a}\) and \(\vec{b}\) such that:
\(\vec{a}\) is parallel to \(\vec{v}\)
\(\vec{b}\) is orthogonal to \(\vec{v}\)
\(\vec{a}+\vec{b} = \vec{w}\)
Find the distance from the point \(P=(3,1,2)\) to the line
Find the distance between the parallel lines
Two ways to describe a plane:
A point, and two vectors parallel to the plane.
A point, and one vector perpendicular to the plane.
Second option is simpler. Suppose \(\vec n = \la a,b,c\ra\) is perpendicular to the plane. Suppose also \(P_0 = (x_0,y_0,z_0)\) and \(P=(x,y,z)\) are in the plane. Then \(\orr{P_0P}\) is parallel to the plane, so \(\vec n \dotp \orr{P_0P}=0\text{.}\)
Find the equation of the plane:
Through \(P_0=(2,-1,3)\) and perpendicular to \(\vec n = \la 5, -3, 4\ra\)
Perpendicular to the line \(\vec{r}(t) = \la 3+2t, -5+3t, -2-9t\ra\text{,}\) and containing the origin.
Containing the points \(P=(1,0,4), Q=(2,-1,3), R=(5,2,4)\)
Containing the lines