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Find the Cartesian equation of a plane passing through a point and parallel to a given plane

Let M be a plane which passes through the point (1,2,-3) and is parallel to the plane given by the equation 3x-y+2z = 4. Find the Cartesian equation of M. Further, find the distance between the two planes.


Let M' denote the plane given by the equation 3x-y+2z = 4. Since M and M' are parallel, we know they share a common normal vector. Therefore, the Cartesian equation of M is of the form

    \[ 3x - y + 2z = d. \]

Since (1,2,-3) is on M, we have

    \[ d = 3 - 2 - 6 = -5. \]

Hence, the Cartesian equation of M is

    \[ 3x - y + 2z = -5. \]

The distance between M and M' is then

    \[ \left| \frac{d_1 - d_2}{\lVert N \rVert} \right| = \frac{9}{\sqrt{14}}. \]

Find geometric properties of a plane with given Cartesian equation

Consider a plane with the Cartesian equation

    \[ x + 2y - 2z + 7 = 0. \]

Find each of the following:

  1. A normal vector to the plane with length 1.
  2. The x,y,z intercepts of the plane.
  3. The distance from the plane to the origin.
  4. The point on the plane, Q, nearest to the origin.

  1. From the Cartesian equation we know that a normal vector to the plane is N' = (1,2,-2). Making this of unit length, we have a unit normal N given by

        \[ N = \frac{N'}{\lVert N \rVert} = \frac{1}{3} (1,2,-2) = \left( \frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right). \]

  2. Since the Cartesian equation gives us x + 2y - 2z = -7 we have the x-intercept is -7, the y-intercept is -\frac{7}{2} and the z-intercept is \frac{7}{2}.
  3. The distance from the plane to the origin is given by

        \[ d = \frac{7}{\lVert (1,2,-2) \rVert} = \frac{7}{3}. \]

  4. The point Q on the plane which is nearest to the origin is in the direction of the normal vector, and at a distance of \frac{7}{3}, therefore,

        \[ Q = \frac{7}{3} \left( \frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right) = \left( \frac{7}{9}, \frac{14}{9}, -\frac{14}{9} \right). \]

Give a Cartesian for planes through given points spanned by given vectors

Consider the vectors

    \[ A = 2 \mathbf{i} + 3 \mathbf{j} - 4 \mathbf{k}, \qquad B = \mathbf{j} + \mathbf{k}. \]

  1. Find a nonzero vector N perpendicular to both A and B.
  2. Find a Cartesian equation for the plane through (0,0,0) which is spanned by A and B.
  3. Find a Cartesian equation for the plane through (1,2,3) which is spanned by A and B.

  1. Since A and B are independent, we can take

        \[ N = A \times B = (3 - (-4), 0 - 2, 2 - 0) = (7,-2,2). \]

  2. From part (a) we have N = (7,-2,2) is perpendicular to both A and B, so a Cartesian equation for the plane is given by

        \[ 7x - 2y + 2z = d \]

    Further, since the point (0,0,0) is on the plane, we must have d = 0. Hence, the Cartesian equation for the plane is

        \[ 7x - 2y + 2z = 0. \]

  3. Again, we have a Cartesian equation for the plane given by

        \[ 7x - 2y + 2z = d. \]

    Since (1,2,3) is on the plane we must have

        \[ d = 7(1) - 2(2) + 2(3) \quad \implies \quad d = 9. \]

    Hence, the Cartesian equation for the plane is given by

        \[ 7x - 2y + 2z = 9. \]

Give a vector based proof of Heron’s formula for computing the area of a triangle

Let S denote the area of a triangle with sides of lengths a,b,c. Heron’s formula states that

    \[ S = \sqrt{s (s-a)(s-b)(s-c)}, \qquad \text{where} \qquad s = \frac{a+b+c}{2}. \]

We prove this formula using vectors via the following steps. Assume the triangle has vertices O,A, and B with

    \[ \lVert A \rVert = a, \qquad \lVert B \rVert = b, \qquad \lVert B - A \rVert = c. \]

  1. Using the identities

        \[ \lVert A \times B \rVert^2 = \lVert A \rVert^2 \lVert B \rVert^2 - (A \cdot B)^2, \qquad -2A \cdot B = \lVert A-B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 \]

    prove the formula

        \[ 4S^2 = a^2 b^2 - \frac{1}{4} (c^2 - a^2 - b^2)^2 = \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2). \]

  2. Simplify the formula in part (a) to obtain the formula

        \[ S^2 = \frac{1}{16}(a+b+c)(a+b-c)(c-a+b)(c+a-b), \]

    and use this to deduce Heron’s formula.


  1. Proof. We know the area of the triangle (in terms of the vectors A and B) is

        \[ S = \frac{1}{2} \lVert A \times B \rVert. \]

    Therefore,

        \begin{align*}  4S^2 &= \lVert A \times B \rVert^2 \\  &= \lVert A \rVert^2 \lVert B \rVert^2 - (A \cdot B)^2 \\  &= a^2 b^2 - \frac{1}{4} ( \lVert A-B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 ) \\  &= a^2 b^2 - \frac{1}{4} (c^2 - a^2 - b^2)^2 \\  &= \frac{1}{4} (4a^2 b^2 - (c^2 - a^2 - b^2)^2) \\  &= \frac{1}{4} (2ab - (c^2 - a^2 - b^2))(2ab + (c^2 - a^2 - b^2)) \\  &= \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2). \qquad \blacksquare \end{align*}

  2. Proof. We simplify the formula in part (a),

        \begin{align*}  && 4S^2 &= \frac{1}{4} (2ab - c^2 + a^2 + b^2)(2ab + c^2 - a^2 - b^2) \\  \implies && S^2 &= \frac{1}{16} ((a+b)^2 - c^2)(c^2 - (a-b)^2) \\  \implies && S^2 &= \frac{1}{16} (a+b+c)(a+b-c)(c-a+b)(c+a-b) \\  \implies && S^2 &= \frac{a+b+c}{2} \cdot \frac{a+b-c}{2} \cdot \frac{c-a+b}{2} \cdot \frac{c+a-b}{2} \\[9pt]  \implies && S^2 &= s(s-c)(s-a)(s-b) \\  \implies && S &= \sqrt{s (s-a)(s-b)(s-c)}. \qquad \blacksquare \end{align*}

Prove a formula for the perpendicular distance between a point and a line

  1. Assume that B \neq C. Prove that the perpendicular distance from A to the line passing through the points B and C is given by the formula

        \[ \frac{\lVert (A - B) \times (C - B) \rVert}{\lVert B - C\rVert}. \]

  2. Compute the distance in the case

        \[ A = (1,-2,-5), \qquad B = (-1,1,1), \qquad C = (4,5,1). \]


  1. Proof. We know the area of the parallelogram determined by (A-B) and (C-B) is given by

        \[ \lVert (A-B) \times (C-B) \rVert. \]

    But this is exactly twice the area of the triangle with base C-B and height the perpendicular line from A to the line through B and C. Hence, h is the perpendicular distance from A to the line through B and C, and

        \[ \lVert (A-B) \times (C-B) \rVert = 2 \cdot \left( \frac{1}{2} h \lVert B-A \rVert \right) \quad \implies \quad h = \frac{\lVert (A-B) \times (C-B) \rVert}{\lVert B-C \rVert}. \qquad \blacksquare \]

  2. When A = (1,-2,-5), \ B = (-1,1,1), \ C = (4,5,1) we have

        \begin{align*}  \frac{\lVert (A-B) \times (C - B) \rVert}{\lVert B-C \rVert} &= \frac{\lVert (2,-3,-6) \times (5,4,0) \rVert}{\lVert (-5,-4,0) \rVert} \\[9pt]  &= \frac{\lVert (24,-30,23) \rVert}{\lVert (-5,-4,0) \rVert} \\[9pt]  &= \frac{\sqrt{2005}}{\sqrt{41}}.  \end{align*}

Prove a vector formula for the volume of a tetrahedron

  1. Consider a tetrahedron with vertices A,B,C,D. Prove that the volume of the tetrahedron is given by the formula

        \[ V = \frac{1}{6} |(B-A) \cdot (C -A) \times (D-A)|. \]

  2. Compute the volume in the case that

        \[ A = (1,1,1), \quad B = (0,0,2), \quad C = (0,3,0), \quad D = (4,0,0). \]


  1. Proof. We know the volume of a tetrahedron is given by \frac{1}{3} A_0 h (where A_0 denotes the altitude of the tetrahedron). We know (page 490 of Apostol) that the volume of the parallelepiped with base formed by vector A,B and height formed by vector C is given by A \times B \cdot C. In this case we have that the base of the tetrahedron is formed by the vectors (B-A) and (C-A), and the height is formed by the vector (D-A). Further, we know that the area of the base described by the vectors (B-A) and (C-A) is one half that of the parallelepiped whose base is given by vectors A and B (since the base of the parallelepiped described by vectors A and B is a rectangle, and the base of the tetrahedron is the triangle formed by cutting this rectangle along the diagonal). Therefore we have

        \begin{align*}  V &= \frac{1}{3} \left( \frac{1}{2} \right) (B -A)\cdot (C-A) \times (D - A) \\  &= \frac{1}{6} \big( (B-A) \cdot (C-A) \times (D-A) \big). \qquad \blacksquare \end{align*}

  2. Using the formula in part (a) with the given values of A,B,C,D we have

        \begin{align*}  V &= \frac{1}{6} \big( (-1,-1,1) \cdot (-1,2,-1) \times (3,-1,-1) \big) \\   &= \frac{1}{6} \big( (-1,-1,1) \cdot (-2-1, -3-1, 1-6) \big) \\  &= \frac{1}{6} \big( (-1,-1,1) \cdot (-3,-4,-5) \big) \\  &= \frac{1}{6} (3 + 4 - 5) \\  &= \frac{1}{3}. \end{align*}

Prove some vector identities using the “cab minus bac” formula

In the previous exercise (Section 13.14, Exercise #9) we proved the “cab minus bac” formula:

    \[ A \times (B \times C) = (C \cdot A) B - (B \cdot A) C. \]

Using this formula prove the following identities:

  1. (A \times B) \times (C \times D) = (A \times B \cdot D)C - (A \times B \cdot C) D.
  2. A \times (B \times C) + B \times (C \times A) + C \times (A \times B) = O.
  3. A \times (B \times C) = (A \times B) \times C if and only if B \times (C \times A) = O.
  4. (A \times B) \cdot (C \times D) = (B \cdot D)(A \cdot C) - (B \cdot C)(A \cdot D).

  1. Proof. Using the cab minus bac formula with A \times B in place of A, C in place of B, and D in place of C we have

        \begin{align*}  (A \times B) \times (C \times D) &= (D \cdot (A \times B))C - (C \cdot (A \times B))D \\  &= (A \times B \cdot D) C - (A \times B \cdot C) D. \qquad \blacksquare \end{align*}

  2. Proof. Applying the cab minus back formula to each of the three terms in the sum we have

        \begin{align*}  A \times (B \times C) &= (C \cdot A) B - (B \cdot A)C \\  B \times (C \times A) &= (A \cdot B) C - (C \cdot B)A \\  C \times (A \times B) &= (B \cdot C) A - (A \cdot C)B. \end{align*}

    So, putting these together we have

        \begin{align*}  A &\times (B \times C) + B \times (C \times A) + C \times (A \times B) \\  &= (C \cdot A)B - (B\cdot A)C + (A \cdot B)C - (C \cdot B)A + (B \cdot C)A - (A \cdot C)B \\  &= O. \qquad \blacksquare \end{align*}

  3. Proof. From cab minus bac we have

        \[ A \times (B \times C) = (C \cdot A)B - (B \cdot A)C. \]

    Furthermore, since (A \times B) \times C = -C \times (A \times B), we can apply bac minus cab to get

        \begin{align*}  (A \times B) \times C &= -C \times (A \times B) &(\text{Thm 13.12(a)}) \\  &= - ((B \cdot C)A - (A \cdot C) B) \\  &= (A \cdot C)B - (B \cdot C)A \\  &= (C \cdot A)B - (B \cdot C)A + (A \cdot B)C - (A \cdot B)C \\  &= (C \cdot A)B - (B \cdot A)C + B \times (C \times A) \\  &= A \times (B \times C) + B \times (C \times A). \end{align*}

    Therefore,

        \[ (A \times B) \times C = A \times (B \times C) \iff B \times (C \times A) = O. \qquad \blacksquare\]

  4. Proof. From a previous exercise (Section 13.14, Exercise #7(d)) we know the identity A \cdot B \times C = C \cdot A \times B. In this case we have A \times B in place of A, C in place of B and D in place of C. This gives us

        \begin{align*}  (A \times B) \cdot (C \times D) &= D \cdot ((A \times B) \times C)\\  &= ((A \times B) \times C) \cdot D \\  &= (-C \times (A \times B)) \cdot D \\  &= (-(B \cdot C)A + (A \cdot C)B) \cdot D \\  &= (B \cdot D)(A \cdot C) - (B \cdot C)(A \cdot D). \qquad \blacksquare \end{align*}

Prove the “cab minus bac” formula

The “cab minus bac” formula is the vector identity

    \[ A \times (B \times C) = (C \cdot A)B - (B \cdot A)C. \]

Let B = (b_1, b_2, b_3) and C = (c_1, c_2, c_3). Prove that

    \[ \mathbf{i} \times (B \times C) = c_1 B - b_1 C. \]

This is the “cab minus bac” formula in the case A = \mathbf{i}. Prove similar formulas for the special cases A = \mathbf{j} and A = \mathbf{k}. Put these three results together to prove the formula in general.


Proof. For the case A = \mathbf{i} we have

    \begin{align*}  \mathbf{i} \times (B \times C) &= \mathbf{i} \times (b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1) \\  &= (0, -b_1 c_2 + b_2 c_1, b_3 c_1 - b_1 c_3) \\  &= c_1 B - b_1 C. \end{align*}

Similarly, for A = \mathbf{j} and A = \mathbf{k} we have

    \begin{align*}  \mathbf{j} \times (B \times C) &= \mathbf{j} \times (b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1) \\  &= (b_1 c_2 - b_2 c_1, 0, -b_2 c_3 + b_3 c_2) \\  &= c_2 B - b_2 C \\  \mathbf{k} \times (B \times C) &= \mathbf{k} \times (b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1) \\  &= (-b_3 c_1 + b_1 c_3, b_2 c_3 - b_3 c_2, 0) \\  &= c_3 B - b_3 C. \end{align*}

So, if A = a_i \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} is any vector in \mathbb{R}^3 then we have

    \begin{align*}  A \times (B \times C) &= a_1 \mathbf{i} \times (B \times C) + a_2 \mathbf{j} \times (B \times C) + a_3 \mathbf{k} \times (B \times C) \\  &= a_1 (c_1 B - b_1 C) + a_2 (c_2 B - b_2 C) + a_3 (c_3 B - b_3 C) \\  &= (a_1 c_1 + a_2 c_2 + a_3 c_3) B - (a_1 b_1 + a_2 b_2 + a_3 b_3) C \\  &= (A \cdot C)B - (A \cdot B) C. \qquad \blacksquare \end{align*}

Find vectors which satisfy given relations

  1. Find all of the vectors a \mathbf{i} + b \mathbf{j} + c \mathbf{k} which satisfy

        \[ (a \mathbf{i} + b \mathbf{j} + c \mathbf{k}) \cdot \mathbf{k} \times (6 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k}) = 3. \]

  2. Find the shortest length vector a \mathbf{i} + b \mathbf{j} + c \mathbf{k} which satisfies the relation in part (a).

  1. We can compute,

        \begin{align*}  && (a \mathbf{i} + b \mathbf{j} + c \mathbf{k}) \cdot ((0,0,1) \times (6,3,4)) &= 3 \\  \implies && (a,b,c) \cdot (-3,6,0) &=3 \\  \implies && -3a + 6b &= 3 \\  \implies && a &= 2b-1. \end{align*}

    So, b,c can take any value, and then we must have a = 2b-1 for the relation to be satisfied. Therefore, any vector

        \[ (2b-1) \mathbf{i} + b \mathbf{j} + c \mathbf{k} \]

    satisfies the relations.

  2. So, we know the vectors satisfying the relation are of the form (2b-1)\mathbf{i} + b \mathbf{j} + c \mathbf{k}. This means we want to minimize

        \[ (2b-1)^2 + b^2 + c^2 = 5b^2 -4b + 1 + c^2. \]

    This is minimal when c = 0 (since c^2 > 0 for any other value of c). Then we want to find the value of b which minimizes 5b^2 - 4b + 1. Taking the derivative and setting it equal to 0 we have

        \[ 10b - 4 =  0 \qquad \implies \qquad b = \frac{2}{5} \quad \implies \quad a = -\frac{1}{5}. \]

    Hence, the vector of minimal length which satisfies the given relation is

        \[ -\frac{1}{5} \mathbf{i} + \frac{2}{5} \mathbf{j}. \]