Tag: dot products

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

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

Prove some properties of the scalar triple product

by RoRi

May 5, 2016

Use the properties of the cross product and the dot product to prove the following properties of the scalar triple product.

$(A+B) \cdot (A+B) \times C = 0$ .
$A \cdot B \times C = -B \cdot A \times C$ .
$A \cdot B \times C = -A \cdot C \times B$ .
$A \cdot B \times C = -C \cdot B \times A$ .

Proof. We have
$(A+B) \cdot (A+B) \times C = (A+B) \cdot ((A+B) \times C) = 0$

since $D \cdot (D \times E) = 0$ for any vectors $D,E$ (in this case $D = A+B$ and $E = C). \qquad blacksquare$
Proof. Using part (a), we have
$\begin{align*} (A+B) \cdot ((A+B) \times C) = 0 && \implies && (A+B) \cdot ((A \times C) + (B \times C)) &= 0 \\ && \implies && B \cdot (A \times C) + A \cdot (B \times C) &= 0 \\ && \implies && A \cdot (B \times C) &= -B \cdot (A \times C). \qquad \blacksquare \end{align*}$
Proof. We have
$A \cdot B \times C = A \cdot (-C \times B) = A \cdot (-1)(C \times B) = -A \cdot (C \times B). \qquad \blacksquare$
Proof. We have
$\begin{align*} A \cdot B \times C &= -A \cdot C \times B &(\text{part c}) \\ &= C \cdot A \times B &(\text{part b}) \\ &= -C \cdot B \times A &(\text{part c}). \qquad \blacksquare \end{align*}$

Compute the angle between (1,0,i,i,i) and (i,i,i,0,i) in C⁵

by RoRi

April 29, 2016

We define the angle $\theta$ between two vectors $A$ and $B$ in $\mathbb{C}^n$ by the formula

$\theta = \arccos \frac{\frac{1}{2} (A \cdot B+ \overline{A \cdot B})}{\lVert A \rVert \lVert B \rVert}.$

Compute the angle between $A = (1,0,i,i,i)$ and $B = (i,i,i,0,i)$ in $\mathbb{C}^5$ .

We compute as follows:

$\begin{align*} \theta &= \arccos \frac{\frac{1}{2} ( A \cdot B + \overline{A \cdot B})}{\lVert A \rVert \lVert B \rVert} \\[9pt] &= \arccos \left( \frac{\frac{1}{2}(2-i + (2 + i))}{4} \right) \\[9pt] &= \arccos \left( \frac{1}{2} \right) \\[9pt] &= \frac{\pi}{3}. \end{align*}$

Compute some dot products of vectors in C²

by RoRi

April 28, 2016

Let

$A = (1,i), \qquad B = (i,-i), \qquad C = (2i,1)$

be vectors in $\mathbb{C}^2$ . Compute the following dot products.

$A \cdot B$ ;
$B \cdot A$ ;
$(iA) \cdot B$ ;
$A \cdot (iB)$ ;
$(iA) \cdot (iB)$ ;
$B \cdot C$ ;
$A \cdot C$ ;
$(B+C) \cdot A$ ;
$(A-C) \cdot B$ ;
$(A - iB) \cdot (A + iB)$ .

We compute,
$A \cdot B = (1)(-i) + (i)(i) = -1 -i.$
We compute,
$B \cdot A = (i)(1) + (-i)(-i) = -1 + i.$
We compute,
$(iA) \cdot B = (i, -1) \cdot (i, -i) = (i)(-i) + (-1)(i) = 1 - i.$
We compute,
$A \cdot (iB) = (1,i) \cdot (-1,1) = (1)(-1) + (i)(1) = -1 + i.$
We compute,
$(iA) \cdot (iB) = (i,-1)\cdot (-1,1) = (i)(-1) + (-1)(1) = -1-i.$
We compute,
$B \cdot C = (i)(-2i) + (-i)(1) = 2 - i.$
We compute,
$A \cdot C = (1)(-2i) + (i)(1) = -i.$
We compute,
$(B+C)\cdot A = (3i,1-i) \cdot(1,i) = (3i)(1) + (1-i)(-i) = -1 + 2i.$
We compute,
$(A-C) \cdot B = (1-2i, i-1) \cdot (i,-i) = (1-2i)(-i) + (i-1)(i) = -i - 2 -1 - i = -3-2i.$
We compute,
$(A-iB)\cdot(A+iB) = A\cdot A + A \cdot i B - iB \cdot A - B\cdot B = 2 + (-1+i) - (-1-i) -2 = 2i.$

Show that the cancellation law does not hold for the dot product

by RoRi

April 26, 2016

Prove or disprove the following statement about vectors in $\mathbb{R}^n$ :

If $A \cdot B = A \cdot C$ and $A \neq O$ , then $B = C$ .

This statement is false. Let $A = (1,1,1)$ , $B = (1,-1,0)$ and $C = (0,-1,1)$ . Then we have

$A \cdot B = (1)(1)+(1)(-1) + (1)(0) = 0, \qquad A \cdot C = (1)(0) + (1)(-1) + (1)(1) = 0$

but, $B \neq C$ .

Compute some vector algebra expressions

by RoRi

April 26, 2016

Let

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

Compute each of the following (and insert appropriate parentheses to make sensible expressions):

$A \cdot BC$ ;
$A\cdot B + C$ ;
$A + B \cdot C$ ;
$AB \cdot C$ ;
$A/B \cdot C$ .

First, we have $A \cdot BC$ means $(A \cdot B)C$ , then we compute
$\begin{align*} (A \cdot B)C &= ((2,4,-7) \cdot (2,6,3))(3,4,-5) \\ &= ((2)(2) + (4)(6)+ (-7)(3)) (3,4,-5)\\ &= (4 + 24 - 21)(3,4,-5)\\ &= 7(3,4,-5)\\ &= (21,28,-35). \end{align*}$
First, we have $A \cdot B + C$ means $A \cdot (B+C)$ , then we compute
$\begin{align*} A \cdot (B+C) &= (2,4,-7) \cdot ((2,6,3) +(3,4,-5)) \\ &= (2,4,-7) \cdot (5,10,-2)\\ &= (2)(5) + (4)(10) + (-7)(-2) \\ &= 10 + 40 + 14\\ &= 64. \end{align*}$
First, we have $A + B \cdot C$ means $(A+B)\cdot C$ , then we compute
$\begin{align*} (A+B) \cdot C &= ((2,4,-7) + (2,6,3)) \cdot (3,4,-5) \\ &= (4,10,-4) \cdot (3,4,-5)\\ &= (4)(3)+ (10)(4) + (-4)(-5) \\ &= 12 + 40 + 20 \\ &= 72. \end{align*}$
First, we have $AB \cdot C$ means $A(B \cdot C)$ , then we compute
$\begin{align*} A(B \cdot C) &= (2,4,-7)((2,6,3) \cdot (3,4,-5)) \\ &= (2,4,-7)((2)(3) + (6)(4) + (3)(-5)) \\ &= (2,4,-7)(6 + 24 - 15)\\ &= (2,4,-7)(15)\\ &= (30,60,-105). \end{align*}$
First, we have $A/B\cdot C$ means $A / (B \cdot C)$ , then we compute
$\begin{align*} A / (B \cdot C) &= (2,4,-7) / ((2,6,3) \cdot (3,4,-5)) \\ &= (2,4,-7) / ((2)(3) + (6)(4) + (3)(-5)) \\ &= (2,4,-7) / (6 + 24 - 15) \\ &= (2,4,-7) / (15) \\ &= \left( \frac{2}{15}, \frac{4}{15}, -\frac{7}{15} \right). \end{align*}$