Equations of lines:

Slope-intercept form: x_{2} = mx_{1} + b in R^{2} (in R^{3}, this is the equation of a plane)

General form: ax_{1} + bx_{2} = c in R^{2} (in R^{3}, this is the equation of a plane)

Point-Normal Vector Form: n · (x − p) = 0 in R^{2} (in R^{3}, this is the equation of a plane)

Point-Normal Form: a(x_{1} −p_{1})+b(x_{2} −p_{2}) = 0 in R^{2} (in R^{3}, this is the equation of a plane)

Point-Parallel Vector Form: x = p + tv in R^{m}

Parametric equations: x_{1} = p_{1} + tv_{1}

x_{2} = p_{2} + tv_{2} . . .

x_{m} = p_{m} + tv_{m} in R^{m} Equations of planes:

General form: ax_{1} + bx_{2} + cx_{3} = d in R^{3}

Point-Normal Vector Form: n · (x − p) = 0 in R^{3}

Point-Normal Form:

a(x_{1} − p_{1}) + b(x_{2} − p_{2}) + c(x_{3} − p_{3}) = 0 in R^{3}

Plane containing the vectors v and w anchored at 0 = plane containing the points 0, v and w = plane containing the point 0 & vectors v & w:

x = tv + sw

Plane containing the point p and the vectors v & w: x = p + tv + sw

Parametric equations: x_{1} = p_{1} + tv_{1} + sw_{1}

x_{2} = p_{2} + tv_{2} + sw_{2} . . .

x_{m} = p_{m} + tv_{m} + sw_{m} in

R^{m} 2.1 Hyperplanes:

The solution set of a system of linear equations in R^{m} is called a hyperplane. The number of free variables is the dimension of the hyperplane.

If n = (a_{1}, ..., a_{m}), p = (p_{1}, ..., p_{m}), x = (x_{1}, ..., x_{m}),

then n · (x − p) = 0

n · x − n · p = 0
n · x = n · p

(a_{1}, ..., a_{m}) · (x_{1}, ..., x_{m}) = (a_{1}, ..., a_{m}) · (p_{1}, ..., p_{m}) a_{1}x_{1} + a_{2}x_{2} + ... + a_{m}x_{m} = a_{1}p_{1} + a_{2}p_{2} + ... + a_{m}p_{m}

[ a_{1} a_{2} ... a_{m}  a_{1}p_{1} + a_{2}p_{2} + ... + a_{m}p_{m}]

Thus, the solution set of n · (x − p) = 0 in R^{m} where n ≠ 0 is m − 1 dimensional.

Thus in R^{2}, n · (x − p) = 0 is the equation of a line.

Thus in R^{3}, n · (x − p) = 0 is the equation of a plane.

The smallest hyperplane containing the vectors v_{1}, v_{2}, ..., v_{k} anchored at 0

= the hyperplane containing the points 0, v_{1}, v_{2}, ..., v_{k}

= the hyperplane containing the point 0 and the vectors v_{1}, v_{2}, ..., v_{k}:

x = t_{1}v_{1} + t_{2}v_{2} + ... + t_{k}v_{k}

The smallest hyperplane containing the point p and the vectors v_{1}, v_{2}, ..., v_{k}:

x = p + t_{1}v_{1} + t_{2}v_{2} + ... + t_{k}v_{k}

Parametric equations:

x_{1} = p_{1} + t_{1}v_{11} + t_{2}v_{21} + ... + t_{k}v_{k1}

x_{2} = p_{2} + t_{1}v_{12} + t_{2}v_{22} + ... + t_{k}v_{k2}

. . .

x_{m} = p_{m} + t_{1}v_{1m} + t_{2}v_{2m} + ... + t_{k}v_{km}in R^{m}

Defn: If u, v, and w are vectors in 3-space, then:

the scalar triple product of u, v, and w =

u · (v × w) = det [ \begin{matrix}u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \\ w_{1} & w_{2} & w_{3} \end{matrix} ]

Thus, u · (v × w) = w · (u × v) = v · (w × u).

Given x = t(2, 4, 1) + s(−1, 0, 3).

Note this is a plane in R^{3} in point-parallel vector form.

1.) Find the line containing the point (7, 8, 9) and perpendicular to this plane given above.

First find a vector perpendicular to the plane, x = t(2, 4, 1) + s(−1, 0, 3).

This is equivalent to finding a vector perpendicular to the vectors (2, 4, 1) and (-1, 0, 3). Thus we can use the cross product.

det [ \begin{matrix}i & j & k \\ 2 & 4 & 1 \\ −1 & 0 & 3 \end{matrix} ]

= i det [ \begin{matrix}4 & 1 \\ 0 & 3 \end{matrix} ] − j det [ \begin{matrix}2 & 1 \\ −1 & 3 \end{matrix} ] + k det [ \begin{matrix}2 & 4 \\ −1 & 0 \end{matrix} ]

= 12i − 7j + 4k = (12, −7, 4)

Since the line is perpendicular to the plane, x = t(2, 4, 1) + s(-1, 0, 3), the line is perpendicular to the vectors (2, 4, 1) and (-1, 0, 3) and thus the vector (12, -7, 4) describes the direction of the line. Since (7, 8, 9) is a point in the line, an equation of the line in point parallel vector form is

x = (7, 8, 9) + t(12, -7, 4)

Given x = t(2, 4, 1) + s(−1, 0, 3).

Note this is a plane in R^{3} in point-parallel vector form.

2.) Write the above plane in point-normal vector form.

First find a vector perpendicular to the plane, x = t(2, 4, 1) + s(−1, 0, 3). We did this for problem 1. The vector (12, -7, 4) is perpendicular to this plane.

We also need a point in the plane. Setting t = 0, s = 0, we note that (0, 0, 0) = 0(2, 4, 1)+0(−1, 0, 3) is a point in the plane.

Thus, an equation of this plane in point normal vector form is

(12, −7, 4) · (x − (0, 0, 0)) = 0

or

(12, −7, 4) · x = 0

Note there are many equivalent equations that also describe this plane. For example, since (12, -7, 4) is perpendicular to the plane, so is (-120, 70, -40).

Setting t = 1, s = 0, we note that (2, 4, 1) = 1(2, 4, 1)+0(−1, 0, 3) is a point in the plane.

Thus (−120, 70, −40) · (x − (2, 4, 1)) = 0

is also an equation of this plane in point-normal form.