Planes and Vectors in 3-Space

Planes in 3-Space

• A plane in 3-space is fixed by:-

  • A position vector r_0 and a direction vector d, or

  • Three points (a vector can be obtained by subtracting these two points).

• Vector parametric equation: r = r_0 + sd, where s \in R is an arbitrary parameter.

  • A line is 1-dimensional and needs only 1 parameter.

Vector Parametric Equation for a Plane

• r = (x, y, z)

• r = r_1 + sd + te

  • A plane is 2-dimensional.

  • Requires two direction vectors d and e and two parameters s, t \in R

Determining a Plane in R3

To uniquely determine a plane in R^3, any of the following will do:

1. One point on the plane and two independent vectors parallel to the plane

2. Three points on the plane (which do not all lie in a straight line)

3. One point on the plane and one vector perpendicular to the plane

4. One consistent linear (scalar) equation in three unknowns, ax + by + cz = d

Finding the Equation of a Plane

1. One Point and Two Parallel Vectors

• Given a point on the plane and two vectors parallel to the plane, the vector parametric form is: r = r_0 + sd + te, \quad s, t \in R

  • r_0 takes us to a point on the plane.

  • Adding sd + te (a linear combination of d and e) takes us to any point on the plane.

• The plane contains the point given by the position vector r_0.

• The vectors d and e are parallel to the plane (direction vectors).

• s and t are two arbitrary parameters (the plane is 2-dimensional).

Example

• Plane through the point (1, 2, 3) with direction vectors (4, 5, 6) and (7, 8, 9).

• Vector parametric form: r = (1, 2, 3) + s(4, 5, 6) + t(7, 8, 9)
  (x, y, z) = (1, 2, 3) + s (4, 5, 6) + t (7, 8, 9)

• Scalar parametric equations:
  x = 1 + 4s + 7t
  y = 2 + 5s + 8t
  z = 3 + 6s + 9t

• A plane is a 2-dimensional object that can exist in R^3 or higher, but is usually considered in R^3.

2. Planes via Three Points

• Given three points A, B, and C with position vectors a, b, c:
  r_0 = a
  d = b - a
  e = c - a

• Vector parametric equation for the plane: r = a + s(b - a) + t(c - a), \quad s, t \in R

Exercise 31. Determining a Plane

• Find the plane through the points (1, 2, 1), (5, 0, -1), and (3, -1, -1).

• Choose r_0 = (1, 2, 1).
  d = (5, 0, -1) - (1, 2, 1) = (4, -2, -2)
  e = (3, -1, -1) - (1, 2, 1) = (2, -3, -2)

• Vector parametric equation for the plane:
  r = (1, 2, 1) + s(4, -2, -2) + t(2, -3, -2), \quad s, t \in R

• This is not the only possible answer.

3. One Point and One Normal Vector

• Given a vector n perpendicular to the plane and a point r_0 on the plane.

• If r = (x, y, z) is any other point on the plane, then r - r_0 is parallel to the plane and perpendicular to n.

• Point-normal form of a plane:
  (r - r_0) \cdot n = 0

• r_0 is the position vector of a fixed point on the plane, and n is a normal vector to the plane.

• This can be used as a test to see if a point x lies in the plane: if x is on the plane, then (x - r_0) \cdot n = 0.

Exercise 32. Determining a Plane

• Find a vector point-normal equation for the plane through the point (-2, 5, 4) and perpendicular to n = (3, 1, -2).
  ((x, y, z) - (-2, 5, 4)) \cdot (3, 1, -2) = 0

• Expanding this:
  (x + 2, y - 5, z - 4) \cdot (3, 1, -2) = 0
  3(x + 2) + 1(y - 5) + (-2)(z - 4) = 0
  3x + 6 + y - 5 - 2z + 8 = 0
  3x + y - 2z = -9

• The coordinates of n become the coefficients of x, y, and z in the scalar equation for the plane.

4. Scalar Form of a Plane

• Definition 33: An equation of the form:
  ax + by + cz = d
  is the equation of a plane in 3-space in scalar form with normal n = (a, b, c).

• At least one of a, b, or c must be non-zero.

Notes

• This form is sometimes called a scalar point-normal form and can be obtained by expanding the vector point-normal form:
  (r - r0) \cdot n = 0 r \cdot n - r0 \cdot n = 0
  r \cdot n = r_0 \cdot n

Exercise 34. Scalar Form from Point-Normal Form

• Expand the vector point-normal form for the plane from the previous exercise, containing the point given by the position vector r_0 = (-2, 5, 4) and with a normal vector n = (3, 1, -2), to obtain the scalar point-normal form (done on slide 58).

• Extra: Give a normal vector for the plane x - 3y + 6z = 0.
  n = (1, -3, 6)

• Does this plane go through the origin? Yes. Substitute (0, 0, 0).
  0 - 3(0) + 6(0) = 0

Changing from One Type of Equation to Another

Exercise 35. Scalar Form to Vector Parametric Form

• Write the plane x + 2y + 3z = 4 in (a) vector parametric form and (b) vector point-normal form.

• (a) From scalar form to vector parametric form, we need to introduce 2 parameters.

• Let y = s and z = t, where s, t \in R
  Then, x + 2s + 3t = 4
  x = 4 - 2s - 3t
  (x, y, z) = (4 - 2s - 3t, s, t) = (4 - 2s - 3t, 0 + s + 0, 0 + 0 + t) = (4, 0, 0) + s(-2, 1, 0) + t(-3, 0, 1)
  r = (4, 0, 0) + sd + te

• (b) To write in vector point-normal form, identify n and then find a point on the plane.
  n = (1, 2, 3)

• If y = 0 and z = 0, then x + 0 + 0 = 4, so x = 4.

• Then r0 = (4, 0, 0) is a point on the plane. (r - (4, 0, 0)) \cdot (1, 2, 3) = 0 (r - r0) \cdot n = 0

Quick Quiz: Projections

1. \text{proj}_d x is the vector component of x in the direction of d.

2. Is \text{proj}_d x parallel to d or to x? Parallel to d.

3. What is the vector component of x orthogonal to d? x - \text{proj}_d x

4. What calculation gives the shortest distance from the point x to the line r = sd?
   |x - \text{proj}_d x|

5. If x and d are orthogonal, what is \text{proj}_d x? \vec{0}

Exercise 36. Vector Parametric Form to Point-Normal and Scalar Form

• Find a vector point-normal form of the plane r = (1, 2, 3) + s(2, 0, 1) + t(4, 1, -2), s, t \in R and write it in scalar form.

• To find the vector point-normal form, we need a normal vector n.

• Since d and e are parallel to the plane, we must have n \cdot d = 0 and n \cdot e = 0.

• We could let n = (a, b, c) and solve the equations we get from the dot products, but is there an easier way?

1.7 The Cross Product

• In the previous exercise, it was time-consuming to find the normal vector orthogonal to two vectors (to find n if we know d and e).

• In 3-space, there is a special way of finding a vector n = (a, b, c) that is orthogonal to two given vectors x = (x1, x2, x3) and y = (y1, y2, y3).

• Definition 37: If x = (x1, x2, x3) and y = (y1, y2, y3) are two vectors in 3-space, then the vector
  x \times y = (x2y3 - x3y2, -(x1y3 - x3y1), x1y2 - x2y1)
  is called the cross product of x and y.

• Using column vectors:
  \begin{bmatrix} i \ j \ k \end{bmatrix} \begin{bmatrix} x1 \ x2 \ x3 \end{bmatrix} \begin{bmatrix} y1 \ y2 \ y3 \end{bmatrix} = [ x2y3-x3y2, -(x1y3-x3y1), x1y2-x2y1]

Example

• If x = (1, 2, 3) and y = (4, 5, 6), then the components of the cross product are obtained as follows:
  \newline
  i-th component: 2 \times 6 - 5 \times 3 = 12 - 15 = -3
  \newline
  j-th component: -(1 \times 6 - 4 \times 3) = -(6 - 12) = 6
  \newline
  k-th component: 1 \times 5 - 4 \times 2 = 5 - 8 = -3
  \newline
  Then x \times y = (-3, 6, -3) = -3(1, -2, 1).

Using a Diagram/Table

Using a Diagram/Table

Using a Diagram/Table

This method uses a visual aid to compute the cross product by arranging components in a pattern to facilitate determinant calculation.

Given vectors x = (x1, x2, x3) and y = (y1, y2, y3), the cross product x \times y can be computed as follows:

\begin{aligned}
x \times y &= \begin{vmatrix}
i & j & k \
x1 & x2 & x3 \ y1 & y2 & y3
\end{vmatrix} \
&= i(x2y3 - x3y2) - j(x1y3 - x3y1) + k(x1y2 - x2y1) \
&= (x2y3 - x3y2, -(x1y3 - x3y1), x1y2 - x2y1)
\end{aligned}

This visual method helps ensure accurate calculation of all components, minimizing errors in the cross product computation.

By following the diagram, adhere to formula requirements for an accurate cross product.

Using a Diagram/Table

\begin{bmatrix} i \ 1 \end{bmatrix} \begin{bmatrix} j \ 2 \end{bmatrix} \begin{bmatrix} k \ 3 \end{bmatrix} \begin{bmatrix} i \ 4 \end{bmatrix} \begin{bmatrix} j \ 5 \end{bmatrix}
i(2 \times 6 - 3 \times 5) - j(1 \times 6 - 3 \times 4) + k(1 \times 5 - 2 \times 4)
= i(12 - 15) - j(6 - 12) + k(5 - 8)
= -3i + 6j - 3k
= (-3, 6, -3)

Exercise 38. Cross Products

• Find the cross product of a = (1, 2, -2) and b = (3, 0, 1).
  a \times b = i(2 \times 1 - (-2 \times 0)) - j(1 \times 1 - (-2 \times 3)) + k(1 \times 0 - (2 \times 3))
  = i(2 - 0) - j(1 + 6) + k(0 - 6)
  = 2i - 7j - 6k
  = (2, -7, -6)

• What happens when we find the dot product of a \times b with a and b?

• (a \times b) \cdot a = (2, -7, -6) \cdot (1, 2, -2) = 2 - 14 + 12 = 0

• (a \times b) \cdot b = (2, -7, -6) \cdot (3, 0, 1) = 6 + 0 - 6 = 0

• The cross product of two vectors is always orthogonal to both these vectors.

Where Does the Formula for the Cross Product Come From?

• The cross product of vectors x and y:1.

  • Is orthogonal to both x and y, and

  • Has magnitude equal to the area of the parallelogram determined by x and y.

• The direction of the cross product is determined by the 'right-hand rule'.

  • If you point the fingers of your right hand in the direction of x and then curl your fingers in the direction of y, then your thumb is pointing in the direction of x \times y.

• The cross product is the unique vector in R^3 with these properties.

• \text{Area} = |x \times y|

• Note: y \times x points in the opposite direction to x \times y.

  • In fact, y \times x = -(x \times y).

Comments

• The cross product operation is not commutative.
  -

Using a Diagram/Table

This method uses a visual aid to compute the cross product by arranging components in a pattern to facilitate determinant calculation.

Given vectors x = (x1, x2, x3) and y = (y1, y2, y3), the cross product x \times y can be computed as follows:

\begin{aligned}
x \times y &= \begin{vmatrix}
i & j & k \
x1 & x2 & x3 \
y1 & y2 & y3
\end{vmatrix} \
&= i(x2y3 - x3y2) - j(x1y3 - x3y1) + k(x1y2 - x2y1) \
&= (x2y3 - x3y2, -(x1y3 - x3y1), x1y2 - x2y1)
\end{aligned}

This visual method helps ensure accurate calculation of all components, minimizing errors in the cross product computation.

By following the diagram, adhere to formula requirements for an accurate cross product.