Tag: vectors

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

by RoRi

May 13, 2016

Consider the vectors

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

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

Since $A$ and $B$ are independent, we can take
$N = A \times B = (3 - (-4), 0 - 2, 2 - 0) = (7,-2,2).$
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

$7x - 2y + 2z = 0.$
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

$7x - 2y + 2z = 9.$

Using Cramer’s rule, solve the given system of equations

by RoRi

May 13, 2016

Use Cramer’s rule to solve the system of equations:

$\begin{align*} x + y &= 5, \\ x + z &= 2, \\ y + z &= 5. \end{align*}$

From Cramer’s rule we have,

$\begin{align*} x &= \frac{ \begin{vmatrix*}[r] 5 & 1 & 0 \\ 2 & 0 & 1 \\ 5 & 1 & 1 \end{vmatrix*} }{ \begin{vmatrix*}[r] 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix*} } = \frac{-2}{-2} = 1 \\[10pt] y &= \frac{ \begin{vmatrix*}[r] 1 & 5 & 0 \\ 1 & 2 & 1 \\ 0 & 5 & 1 \end{vmatrix*} }{ \begin{vmatrix*}[r] 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix*} } = \frac{-8}{-2} = 4 \\[10pt] z &= \frac{ \begin{vmatrix*}[r] 1 & 1 & 5 \\ 1 & 0 & 2 \\ 0 & 1 & 5 \end{vmatrix*} }{ \begin{vmatrix*}[r] 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix*} } = \frac{-2}{-2} = 1. \end{align*}$

Solve a system of equations using Cramer’s rule

by RoRi

May 13, 2016

Use Cramer’s rule to solve the system of equations:

$\begin{align*} x + y + 2z &= 4, \\ 3x - y - z &= 2, \\ 2x + 5y + 3z &= 3. \end{align*}$

From Cramer’s rule we have,

$\begin{align*} x &= \frac{ \begin{vmatrix*}[r] 4 & 1 & 2 \\ 2 & -1 & -1 \\ 3 & 5 & 3 \end{vmatrix*} }{ \begin{vmatrix*}[r] 1 & 1 & 2 \\ 3 & -1 & -1 \\ 2 & 5 & 3 \end{vmatrix*} } = \frac{25}{25} = 1, \\[10pt] y &= \frac{ \begin{vmatrix*}[r] 1 & 4 & 2 \\ 3 & 2 & -1 \\ 2 & 3 & 3 \end{vmatrix*} }{ \begin{vmatrix*}[r] 1 & 1 & 2 \\ 3 & -1 & -1 \\ 2 & 5 & 3 \end{vmatrix*} } = \frac{-25}{25} = -1, \\[10pt] z &= \frac{ \begin{vmatrix*}[r] 1 & 1 & 4 \\ 3 & -1 & 2 \\ 2 & 5 & 3 \end{vmatrix*} }{ \begin{vmatrix*}[r] 1 & 1 & 2 \\ 3 & -1 & -1 \\ 2 & 5 & 3 \end{vmatrix*} } = \frac{50}{25} = 2. \end{align*}$

Solve a given system of equations using Cramer’s rule

by RoRi

May 13, 2016

Use Cramer’s rule to solve the system of equations:

$\begin{align*} x + 2y + 3z &= 5,\\ 2x - y + 4z &= 11,\\ -y + z &= 3. \end{align*}$

From Cramer’s rule we have

$\begin{align*} x &= \frac{ \begin{vmatrix*}[r] 5 & 2 & 3 \\ 11 & -1 & 4 \\ 3 & -1 & 1 \end{vmatrix*} }{ \begin{vmatrix*}[r] 1 & 2 & 3 \\ 2 & -1 & 4 \\ 0 & -1 & 1 \end{vmatrix*} } = \frac{-7}{-7} = 1\\[10pt] y &= \frac{ \begin{vmatrix*}[r] 1 & 5 & 3 \\ 2 & 11 & 4 \\ 0 & 3 & 1 \end{vmatrix*} }{ \begin{vmatrix*}[r] 1 & 2 & 3 \\ 2 & -1 & 4 \\ 0 & -1 & 1 \end{vmatrix*} } = \frac{7}{-7} = -1 \\[10pt] z & = \frac{ \begin{vmatrix*}[r] 1 & 2 & 5 \\ 2 & -1 & 11 \\ 0 & -1 & 3 \end{vmatrix*} }{ \begin{vmatrix*}[r] 1 & 2 & 3 \\ 2 & -1 & 4 \\ 0 & -1 & 1 \end{vmatrix*} } = \frac{-14}{-7} = 2. \end{align*}$

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

by RoRi

May 9, 2016

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

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

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

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

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,

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

by RoRi

May 9, 2016

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
$\frac{\lVert (A - B) \times (C - B) \rVert}{\lVert B - C\rVert}.$
Compute the distance in the case
$A = (1,-2,-5), \qquad B = (-1,1,1), \qquad C = (4,5,1).$

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

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

by RoRi

May 9, 2016

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

$V = \frac{1}{6} |(B-A) \cdot (C -A) \times (D-A)|.$
Compute the volume in the case that
$A = (1,1,1), \quad B = (0,0,2), \quad C = (0,3,0), \quad D = (4,0,0).$

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

by RoRi

May 9, 2016

Prove or disprove:

$A \times \big( A \times (A \times B) \big) \cdot C = -\lVert A \rVert^2 A \cdot B \times C.$

Proof. We can compute using the identities we have derived in the previous exercises of this section,

$\begin{align*} A \times \big( A \times (A \times B) \big) &= A \times \big( (B \cdot A) A - (A \cdot A) B \big) \cdot C \\ &= \big( (A \times A)(B \cdot A) - \lVert A \rVert^2 (A \times B) big) \cdot C \\ &= -\lVert A \rVert^2 ((A \times B) \cdot C) \\ &= -\lVert A \rVert^2 ((A \cdot B) \times C). \qquad \blacksquare \end{align*}$

Prove an identity relating scalar triple products of vectors A,B,C

by RoRi

May 9, 2016

Prove the following identity holds for vectors $A,B,C \in \mathbb{R}^3$ .

$(A \times B) \cdot (B \times C) \times (C \times A) = (A \cdot B \times C)^2.$

Proof. From a previous exercise (Section 13.14, Exercise #9(d)), we know

$(A \times B) \cdot (C \times D) = (B \cdot D)(A \cdot C) - (B \cdot C)(A \cdot D).$

So, with $B \times C$ in place of $C$ and $C \times A$ in place of $D$ in this formula we have

$\begin{align*} (A \times B) \cdot \big((B \times C) \times (C \times A) \big) &= (B \cdot C \times A)(A \cdot B \times C) - (B \cdot B \times C)(A \cdot C \times A) \\ &= (B \times C \cdot A)(A \cdot B \times C) \\ &= (A \cdot B \times C)^2.\qquad \blacksquare \end{align*}$

Compute the cross product of given vectors in terms of the unit coordinate vectors

by RoRi

May 5, 2016

Let $A,B,C,D \in \mathbb{R}^3$ such that

$A \times C \cdot B = 5, \quad A \times D \cdot B = 3, \quad C + D = \mathbf{i} + 2 \mathbf{j} + \mathbf{k}, \quad C - D = \mathbf{i} - \mathbf{k}.$

In terms of the unit coordinate vectors $\mathbf{i}, \mathbf{j}, \mathbf{k}$ compute the cross product

$(A\times B) \times (C \times D).$

Using part (a) of the previous exercise and equation (13.10) on page 490 of Apostol ( $A \times B \cdot C = A \cdot B \times C$ ) we compute

$\begin{align*} (A \times B) \times (C \times D) &= (A \times B \cdot D)C - (A \times B \cdot C)D \\ &= (A \cdot B \times C) C - (A \cdot B \times C) D \\ &= (A \cdot C \times B) D - (A \cdot B \times B) C \\ &= 5D - 3C. \end{align*}$

Then we use the other given relations

$\begin{align*} && C + D &= \mathbf{i} + 2 \mathbf{j} + \mathbf{k} \\ \implies && 2D &= 2 \mathbf{j} + 2 \mathbf{k} \\ \implies && 5D &= 5\mathbf{j} + 5 \mathbf{k} \\ && C-D &= \mathbf{i} - \mathbf{k} \\ \implies && 2C &= 2 \mathbf{i} + 2 \mathbf{j} \\ \implies && 3C &= 3 \mathbf{i} + 3 \mathbf{j}. \end{align*}$

Which all implies

$5D - 3C = -3 \mathbf{i} + 2 \mathbf{j} + 5 \mathbf{k}.$