Tag: vectors

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

by RoRi

May 5, 2016

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:

$(A \times B) \times (C \times D) = (A \times B \cdot D)C - (A \times B \cdot C) D$ .
$A \times (B \times C) + B \times (C \times A) + C \times (A \times B) = O$ .
$A \times (B \times C) = (A \times B) \times C$ if and only if $B \times (C \times A) = O$ .
$(A \times B) \cdot (C \times D) = (B \cdot D)(A \cdot C) - (B \cdot C)(A \cdot D)$ .

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

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

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

by RoRi

May 5, 2016

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

by RoRi

May 4, 2016

Find all of the vectors $a \mathbf{i} + b \mathbf{j} + c \mathbf{k}$ which satisfy
$(a \mathbf{i} + b \mathbf{j} + c \mathbf{k}) \cdot \mathbf{k} \times (6 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k}) = 3.$
Find the shortest length vector $a \mathbf{i} + b \mathbf{j} + c \mathbf{k}$ which satisfies the relation in part (a).

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.
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

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

Prove an identity of cross products and the unit coordinate vectors

by RoRi

May 4, 2016

Prove that we have the identity:

$\mathbf{i} \times (A \times \mathbf{i}) + \mathbf{j} \times (A \times \mathbf{j}) + \mathbf{k} \times (A \times \mathbf{k}) = 2A.$

Proof. Let $A = (a_1, a_2, a_3)$ and compute,

$\begin{align*} \mathbf{i} &\times (A \times \mathbf{i}) + \mathbf{j} \times (A \times \mathbf{j}) + \mathbf{k} (A \times \mathbf{k}) \\ &= \mathbf{i} \times (0, a_3, -a_2) + \mathbf{j} \times (-a_3, 0, a_1) + \mathbf{k} \times (a_2, -a_1, 0) \\ &= (0,a_2, a_3) + (a_1, 0, a_3) + (a_1, a_2, 0) \\ &= 2(a_1, a_2, a_3) \\ &= 2A. \end{align*}$

Prove a formula for the cross product of vectors in terms of the unit coordinate vectors

by RoRi

May 4, 2016

Prove that we have the formula

$A \times B = A \cdot (B \times \mathbf{i}) \mathbf{i} + A \cdot (B \times \mathbf{j}) \mathbf{j} + A \cdot (B \times \mathbf{k}) \mathbf{k}.$

Proof. Let $A = (a_1, a_2, a_3)$ and $B = (b_1, b_2, b_3)$ . Then we compute,

$\begin{align*} A &\cdot (B \times \mathbf{i}) \mathbf{i} + A \cdot (B \times \mathbf{j}) \mathbf{j} + A \cdot (B \times \mathbf{k}) \mathbf{k} \\ &= A \cdot (0, b_3, -b_2) \mathbf{i} + A \cdot (-b_3, 0, b_1) \mathbf{j} + A \cdot (b_2, -b_1, 0) \mathbf{k} \\ &= (a_2 b_3 - a_3 b_2) \mathbf{i} + (a_3 b_1 - a_1 b_3) \mathbf{j} + (a_1 b_2 - a_2 b_1) \mathbf{k} \\ &= A \times B. \qquad \blacksquare \end{align*}$

Compute the volume of a parallelpiped determined by given vectors

by RoRi

May 4, 2016

Compute the volume of the parallelpiped determined by the vectors $\mathbf{i} + \mathbf{j}, \ \mathbf{j} + \mathbf{k}, \ \mathbf{k} + \mathbf{i}$ .

The volume is given by the scalar triple product

$V = (1,1,0) \cdot (0,1,1) \times (1,0,1) = (1,1,0) \cdot (1,1,1) = 2.$

Find all values of t so that (1,t,1), (t,1,0), (0,1,t) are dependent

by RoRi

May 4, 2016

Find all $t \in \mathbb{R}$ such that the vectors

$(1,t,1), \qquad (t,1,0), \qquad (0,1,t)$

are linearly dependent.

The vectors $(1,t,1), \ (t,1,0), \ (0,1,t)$ are linearly dependent if and only if the scalar triple product

$(1,t,1) \cdot (t,1,0) \times (0,1,t) = 0.$

So, we compute

$\begin{align*} (1,t,1) \cdot (t,1,0) \times (0,1,t) &= (1,t,1) \cdot (t - 0, 0 - t^2, t - 0) \\ &= (1,t,1) \cdot (t,-t^2,t) \\ &= (1)(t) + (t)(-t^2) + (1)(t) \\ &= 2t-t^3. \end{align*}$

So, the vectors are dependent for all $t$ such that $2t-t^3 = 0$ . Thus the vectors are dependent for

$t = 0, \ \pm \sqrt{2}.$

Compute the scalar triple product of given vectors

by RoRi

May 3, 2016

For each of the following sets of vectors $A,B,C$ compute the scalar triple product.

$A = (3,0,0), \qquad B = (0,4,0), \qquad C = (0,0,8)$ .
$A = (2,3,-1), \qquad B = (3,-7,5), \qquad C = (1,-5,2)$ .
$A = (2,1,3), \qquad B = (-3,0,6), \qquad C = (4,5,-1)$ .

We compute
$[ABC] = A \cdot B \times C = (3,0,0) \cdot (32,0,0) = 96+0+0 = 96.$
We compute
$[ABC] = A \cdot B \times C = (2,3,-1) \cdot (11,-1,-8) = 2(11) + 3(-1) + (-1)(-8)= 27.$
We compute
$[ABC] = A \cdot B \times C = (2,1,3) \cdot (-30,21,-15) = -60 + 21 - 45 = -84.$

Prove statements about a vector satisfying a given vector equation

by RoRi

May 3, 2016

Consider unit length, orthogonal vectors $A,B \in \mathbb{R}^3$ , and a vector $P$ such that

$P \times B = A-P.$

Prove the following.

$P$ and $B$ are orthogonal and the length of $P$ is $\frac{1}{2} \sqrt{2}$ .
The vectors $P,B,P \times B$ form a basis for $\mathbb{R}^3$ .
$(P \times B) \times B = -P$ .
$P = \frac{1}{2}A - \frac{1}{2} (A \times B)$ .

Proof. We compute,
$\begin{align*} P \times B = A - P && \implies && B \cdot (P \times B) &= B \cdot A - B \cdot P \\ && \implies && 0 = B \cdot P \end{align*}$

since $B \cdot (P \times B) = 0$ and $B \cdot A = 0$ since $A$ and $B$ are orthogonal by assumption. Thus, $B$ and $P$ are orthogonal. Next,

$\begin{align*} \lVert P \times B \rVert^2 &= \lVert P\rVert^2 \lVert B \rVert^2 - (P \cdot B)^2 \\ &= \lVert P \rVert^2 \end{align*}$

since $\lVert B \rVert^2 = 1$ by hypothesis and $P \cdot B = 0$ . Hence, from the vector equation we have

$\begin{align*} P \times B = A - P && \implies && \lVert P \times \rVert^2 &= \lVert A-P \rVert^2 \\ && \implies && \lVert P \rVert^2 &= \lVert A \rVert^2 - \lVert P \rVert^2 \\ && \implies && 2 \lVert P \rVert^2 &= \lVert A \rVert^2 \\ && \implies && 2 \lVert P \rVert^2 &= 1 \\ && \implies && \lVert P \rVert^2 &= \frac{1}{2} \\ && \implies && \lVert P \rVert &= \frac{\sqrt{2}}{2}. \qquad \blacksquare \end{align*}$
Proof. Since $P$ and $B$ are orthogonal (part (a)), we know the vectors $P,B, P \times B$ are independent. Thus, they form a basis for $\mathbb{R}^3$ since any three independent vectors in $\mathbb{R}^3$ are a basis $. \qquad \blacksquare$
Proof. We compute, the vector $(P \times B) \times B$ is given by
$(p_2 b_3 - p_3 b_2, p_3 b_1 - p_1 b_3, p_1 b_2 - p_2 b_1) \times B.$

Then the three coordinates of this cross product are given by

$\begin{align*} &(p_3 b_1 - p_1 b_3)b_3 - (p_1 b_2 - p_2 b_1)b_2 \\ &(p_1 b_2 - p_2 b_1)b_1 - (p_2 b_3 - p_3 b_2)b_3 \\ &(p_2 b_3 - p_3 b_2)b_2 - (p_3 b_1 - p_1 b_3)b_1. \end{align*}$

Expanding these out we obtain the coordinates

$\begin{align*} &p_3 b_1 b_3 - p_1 b_3^2 - p_1 b_2^2 + p_2 b_1 b_2 \\ &p_1 b_1 b_2 - p_2 b_1^2 - p_2 b_3^2 + p_3 b_2 b_3 \\ &p_2 b_2 b_3 - p_3 b_2^2 - p_3 b_1^2 + p_1 b_1 b_3. \end{align*}$

Since $\lVert B \rVert = 1$ we know $b_1^2 + b_2^2 + b_3^2 = 1$ and since $P \cdot B = 0$ we know $p_1 b_1 + p_2 b_2 + p_3 b_3 = 0$ . So, simplifying the expressions above, for each of the coordinates we have

$\begin{align*} p_3 b_1 b_3 - p_1 b_3^2 - p_1 b_2^2 + p_2 b_1 b_2 &= p_3 b_1 b_3 - p_1 (1-b_1^2) + p_2 b_1 b_2 \\ &= p_3 b_1 b_3 - p_1 + p_1 b_1^2 + p_2 b_1 b_2 \\ &= b_1(p_3 b_3 + p_1 b_1 + p_2 b_2) - p_1 \\ &= -p_1 \\[9pt] p_1 b_1 b_2 - p_2 b_1^2 - p_2 b_3^2 + p_3 b_2 b_3 &= p_1 b_1 b_2 - p_2 (1-b_2^2) + p_3 b_2 b_3 \\ &= p_1 b_1 b_2 - p_2 + p_2 b_2^2 + p_3 b_2 b_3 \\ &= b_2(p_1 b_1 + p_2 b_2 + p_3 b_3) - p_2 \\ &= -p_2 \\[9pt] p_2 b_2 b_3 - p_3 b_2^2 - p_3 b_1^2 + p_1 b_1 b_3 &= p_2 b_2 b_3 - p_3 (1-b_3^2) + p_1 b_1 b_3 \\ &= p_2 b_2 b_3 - p_3 + p_3 b_3^2 + p_1 b_1 b_3 \\ &= b_3 (p_2 b_2 + p_3 b_3 + p_1 b_3) - p_3 \\ &= -p_3. \end{align*}$

Hence, we indeed have

$(P \times B) \times B = (-p_1, -p_2, -p_3) = -P. \qquad \blacksquare$
Proof. We compute
$\begin{align*} (P \times B) \times B = -P && \implies && (A-P) \times B &= -P \\ && \implies && (A \times B) - (P \times B) &= -P \\ && \implies && (A \times B) - A + P &= -P \\ && \implies && -2P &= (A \times B) - A \\ && \implies && P &= \frac{1}{2}A - \frac{1}{2}(A \times B). \qquad \blacksquare \end{align*}$

Prove some facts about vectors in R³

by RoRi

May 3, 2016

Prove that $A,B,C \in \mathbb{R}^3$ are all on the same line if and only if
$(B -A) \times (C -A) = O.$
For two vectors $A,B$ with $A \neq B$ , prove that the line through $A$ and $B$ is the set of all vectors $P$ such that $(P-A) \times (P-B) = O$ .

Proof. Assume $A,B,C \in \mathbb{R}^3$ all lie on a line $L$ , and let $A = (a_1, a_2, a_3), \ B = (b_1, b_2, b_3), \ C = (c_1, c_2, c_3)$ . Then
$L = \{ A + tB \} \quad \implies \quad C = (a_1 + tb_1, a_2 + tb_2, a_3 + tb_3).$

So, we have

$B - A = (b_1 - a_1, b_2 - a_2, b_3 - a_3), \qquad C - A = (tb_1, tb_2, tb_3).$

Thus, $(B-A) \times (C-A)$ is the vector

$\begin{align*} &((b_2 - a_2)tb_3 - (a_3-b_3)tb_2, (a_3 - b_3)tb_1 - (a_1-b_1)tb_3, (a_1 - b_1)tb_2 - (a_2 - b_2)tb_1) \\ &= (0,0,0). \end{align*}$

Conversely, assume $(B-A) \times (C-A) = O$ . From Theorem 13.12(g) we know this is the case if and only if the vectors $B-A$ and $C-A$ are linearly dependent. Hence, there are nonzero $s,t$ such that

$s(B-A) + t(C-A) = O$

Since $t$ is nonzero we may divide by $t$ , and we obtain

$sB - sA + tC - tA = O \quad \implies \quad C = A + \frac{s}{t}(A-B).$

But this means $C$ is on the line passing through $A$ and $B. \qquad \blacksquare$
Proof. If $A \neq B$ , then we know there is a unique line passing through $A$ and $B$ , say
$L = \{ A + t(B-A) \}.$

Thus, any point $P \in L$ is given by

$P = (a_1 + t(b_1 - a_1), a_2 + t(b_2 - a_2), a_3 + t(b_3 - a_3) \}.$

Therefore, we have

$\begin{align*} (P-A) \times (P-B) &= t(b_1 - a_1, b_2 - a_2, b_3 - a_3) \times (1-t)(a_1 - b_1, a_2 - b_2, a_3 - b_3) \\ &= t(B-A) \times (B-A) - t(B-A) \times (B-A) \\ &= O. \qquad \blacksquare \end{align*}$