Vector and Mixed Product of Vectors.

Vector Product of Vectors.

An ordered triple of non-coplanar vectors $e_{1}, e_{2}, e_{3}$ is called right-handed if, to an observer situated within the angle formed by these vectors, the shortest rotations from $e_{1}$ to $e_{2}$ and from $e_{2}$ to $e_{3}$ appear to be counterclockwise. Otherwise, the triple $(e_{1}, e_{2}, e_{3})$ is called left-handed.

The vector product of vector $a_{1}$ by vector $a_{2}$ is defined as a vector, denoted by the symbol $[a_{1}, a_{2}]$ (or $a_{1} \times a_{2}$ ), determined by the following three conditions:

The length of the vector $[a_{1}, a_{2}]$ equals the area of the parallelogram constructed on vectors $a_{1}$ and $a_{2}$ , i.e., $| [a_{1}, a_{2}] | = | a_{1} | | a_{2} | \sin (\hat{a_{1}, a_{2}})$

The vector $[a_{1}, a_{2}]$ is perpendicular to the plane of vectors $a_{1}$ and $a_{2}$ ;

The ordered triple $a_{1}, a_{2}, [a_{1}, a_{2}]$ is right-handed.

From the definition of the vector product, it follows that $(\hat{a_{1}, a_{2}}) = \frac{π}{2} \Leftrightarrow [a_{1}, a_{2}] = 0.$

Algebraic properties of the vector product:

$[a_{1}, a_{2}] = - [a_{2}, a_{1}];$

$[λ a_{1}, a_{2}] = λ [a_{1}, a_{2}];$

$[a_{1} + a_{2}, b] = [a_{1}, b] + [a_{2}, b] .$

If $a_{1} (X_{1}, Y_{1}, Z_{1})$ and $a_{2} (X_{2}, Y_{2}, Z_{2})$ are vectors, given by their coordinates in a right-handed orthogonal basis, then the decomposition of the vector product $[a_{1}, a_{2}]$ in the same basis is expressed as $[a_{1}, a_{2}] = (Y_{1} Z_{2} - Z_{1} Y_{2}) i - (X_{1} Z_{2} - Z_{1} X_{2}) j + (X_{1} Y_{2} - Y_{1} X_{2}) k,$ or, using symbolic notation (with the concept of the 3rd order determinant) $[a_{1}, a_{2}] = | \begin{matrix} i & j & k \\ X_{1} & Y_{1} & Z_{1} \\ X_{2} & Y_{2} & Z_{2} \end{matrix} | .$

Examples.

1. $| a_{1} | = 1, | a_{2} | = 2, (\hat{a_{1}, a_{2}}) = 2 π / 3.$ Calculate:

a) $| [a_{1}, a_{2}] |$

b) $| [2 a_{1} + a_{2}, a_{1} + 2 a_{2}] |$

c) $| [a_{1} + 3 a_{2}, 3 a_{1} - a_{2}] | .$

Solution.

a) $| [a_{1}, a_{2}] | = | a_{1} | | a_{2} | \sin (\hat{a_{1}, a_{2}}) = 2 \sin (2 π / 3) = 2 \frac{\sqrt{3}}{2} = \sqrt{3} .$

b) $[2 a_{1} + a_{2}, a_{1} + 2 a_{2}] = [2 a_{1}, a_{1} + 2 a_{2}] + [a_{2}, a_{1} + 2 a_{2}] =$

$= 2 [a_{1}, a_{1} + 2 a_{2}] + [a_{2}, a_{1} + 2 a_{2}] = - 2 [a_{1} + 2 a_{2}, a_{1}] - [a_{1} + 2 a_{2}, a_{2}] =$

$= - 2 [a_{1}, a_{1}] - 2 [2 a_{2}, a_{1}] - [a_{1}, a_{2}] - [2 a_{2}, a_{2}] = - 4 [a_{2}, a_{1}] - [a_{1}, a_{2}] =$

$= 4 [a_{1}, a_{2}] - [a_{1}, a_{2}] = 3 [a_{1}, a_{2}] .$

$| [2 a_{1} + a_{2}, a_{1} + 2 a_{2}] | = 3 | [a_{1}, a_{2}] | = 3 \sqrt{3} .$

c) $[a_{1} + 3 a_{2}, 3 a_{1} - a_{2}] = [a_{1}, 3 a_{1} - a_{2}] + 3 [a_{2}, 3 a_{1} - a_{2}] =$

$= - [3 a_{1} - a_{2}, a_{1}] - 3 [3 a_{1} - a_{2}, a_{2}] =$

$= - [3 a_{1}, a_{1}] + [a_{2}, a_{1}] - 9 [a_{1}, a_{2}] + 3 [a_{2}, a_{2}] = - 10 [a_{1}, a_{2}] .$

$| [a_{1} + 3 a_{2}, 3 a_{1} - a_{2}] | = | 10 [a_{1}, a_{2}] | = 10 \sqrt{3} .$

Answer: a) $\sqrt{3};$ b) $3 \sqrt{3};$ c) $10 \sqrt{3} .$

2. Simplify the expressions:

a) $[i, j + k] - [j, i + k] + [k, i + j + k];$

b) $[a + b + c, c] + [a + b + c, b] + [b - c, a];$

c) $[2 a + b, c - a] + [b + c, a + b];$

d) $2 i [j, k] + 3 j [i, k] + 4 k [i, j] .$

Solution.

a) $[i, j + k] - [j, i + k] + [k, i + j + k] = - [j + k, i] + [i + k, j] - [i + j + k, k] =$

$= - [j, i] - [k, i] + [i, j] + [k, j] - [i, k] - [j, k] - [k, k] =$

$= [i, j] + [i, k] + [i, j] - [j, k] - [i, k] - [j, k] = 2 [i, j] - 2 [j, k] =$

$2 | \begin{matrix} i & j & k \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} | - 2 | \begin{matrix} i & j & k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} | = 2 k - 2 i .$

b) $[a + b + c, c] + [a + b + c, b] + [b - c, a] =$

$= [a, c] + [b, c] + [c, c] + [a, b] + [b, b] + [c, b] + [b, a] - [c, a] =$

$= [a, c] + [b, c] + [a, b] - [b, c] - [a, b] + [a, c] = 2 [a, c] .$

c) $[2 a + b, c - a] + [b + c, a + b] = 2 [a, c - a] + [b, c - a] + [b, a + b] + [c, a + b] =$

$= - 2 [c - a, a] - [c - a, b] - [a + b, b] - [a + b, c] =$

$= - 2 [c, a] + 2 [a, a] - [c, b] + [a, b] - [a, b] - [b, b] - [a, c] - [b, c] = [a, c] .$

Answer: а) $2 (k - i);$ b) $2 [a, c];$ c) $[a, c] .$

3. Calculate the area of the triangle with vertices $A (1, 1, 1), B (2, 3, 4)$ and $C (4, 3, 2) .$

Solution.

$S △ A B C = \frac{1}{2} | [\overset{―}{B A}, \overset{―}{B C}] | .$

$\overset{―}{B A} = (1 - 2; 1 - 3; 1 - 4) = (- 1; - 2; - 3) .$

$\overset{―}{B C} = (4 - 2; 3 - 3; 2 - 4) = (2; 0; - 2) .$

$[\overset{―}{B A}, \overset{―}{B C}] = | \begin{matrix} i & j & k \\ - 1 & - 2 & - 3 \\ 2 & 0 & - 2 \end{matrix} | = 4 i - 8 j + 4 k .$

$S △ A B C = \frac{1}{2} | [\overset{―}{B A}, \overset{―}{B C}] | = \frac{1}{2} \sqrt{16 + 64 + 16} = \frac{\sqrt{96}}{2} = 2 \sqrt{6} .$

Answer: $2 \sqrt{6} .$

4. For the given vectors $a (2, 0, 3), b (- 3, 5, 4), c (3, 4, - 1)$ , calculate the projection of the vector $[a, b]$ onto the vector $(a, b) c .$

Solution.

Find the vectors $d = [a, b]$ and $k = (a, b) c :$

$d = [a, b] = | \begin{matrix} i & j & k \\ 2 & 0 & 3 \\ - 3 & 5 & 4 \end{matrix} | = - 15 i - 17 j + 10 k = (- 15, - 17, 10);$

$k = (a, b) c = (- 6 + 12) (3, 4, - 1) = 6 (3, 4, - 1) = (18, 24, - 6) .$

$P r_{k} d = \frac{(d, k)}{| k |} = \frac{- 270 - 408 - 60}{\sqrt{324 + 576 + 36}} = \frac{- 738}{\sqrt{936}} = \frac{- 738}{6 \sqrt{26}} = \frac{- 123}{\sqrt{26}} .$

Answer: $\frac{- 123}{\sqrt{26}} .$

5. Find the vector $[a, a + b] + [a, [a, b]],$ если $a (2, 1, - 3)$ $b (1, - 1, 1) .$

Solution.

$a + b = (2 + 1; 1 - 1; - 3 + 1) = (3; 0; - 2);$

$[a, a + b] = | \begin{matrix} i & j & k \\ 2 & 1 & - 3 \\ 3 & 0 & - 2 \end{matrix} | = - 2 i - 5 j - 3 k;$

$[a, b] = | \begin{matrix} i & j & k \\ 2 & 1 & - 3 \\ 1 & - 1 & 1 \end{matrix} | = - 2 i - 5 j - 3 k;$

$[a, [a, b]] = | \begin{matrix} i & j & k \\ 2 & 1 & - 3 \\ - 2 & - 5 & - 3 \end{matrix} | = - 18 i + 12 j - 8 k;$

$[a, a + b] + [a, [a, b]] = (- 2; - 5; - 3) + (- 18; 12; - 8) = (- 20; 7; - 11) .$

Answer: $(- 20; 7; - 11) .$

Mixed Product of Vectors.

The mixed product of an ordered triple of vectors $a_{1}, a_{2}, a_{3}$ is defined as the number $[a_{1}, a_{2}] a_{3} .$

Geometric properties of the mixed product:

1) If $V$ is the volume of the parallelepiped built on vectors $a_{1}, a_{2},$ and $a_{3},$ then

$[a_{1}, a_{2}] a_{3} = V$ if the triple of vectors $(a_{1}, a_{2}, a_{3})$ is right-handed;

$[a_{1}, a_{2}] a_{3} = - V$ if the triple of vectors $(a_{1}, a_{2}, a_{3})$ is left-handed.

2) For three vectors $a_{1}, a_{2}, a_{3}$ to be coplanar, it is necessary and sufficient that $[a_{1}, a_{2}] a_{3} = 0.$

The primary algebraic property of the mixed product is that a cyclic permutation of the vectors does not change its magnitude, i.e., $[a_{1}, a_{2}] a_{3} = a_{1} [a_{2}, a_{3}] = [a_{3}, a_{1}] a_{2} .$

This property allows for the notation $[a_{1}, a_{2}] a_{3} = a_{1} a_{2} a_{3} .$

The mixed product expressed through the coordinates of vectors in a right-handed orthogonal basis is written as $a_{1} a_{2} a_{3} = | \begin{matrix} X_{1} & Y_{1} & Z_{1} \\ X_{2} & Y_{2} & Z_{2} \\ X_{3} & Y_{3} & Z_{3} \end{matrix} | .$

Examples.

1. The vectors $a_{1}, a_{2}, a_{3}$ form a right-handed set, are mutually perpendicular, and $| a_{1} | = 4, | a_{2} | = 2, | a_{3} | = 3.$ Calculate $a_{1} a_{2} a_{3} .$

Solution.

$a_{1} a_{2} a_{3} = [a_{1}, a_{2}] a_{3} = | [a_{1}, a_{2}] | | a_{3} | c o s (\hat{[a_{1}, a_{2}], a_{3}}) .$

$\cos (\hat{[a_{1}, a_{2}], a_{3}}) = \cos 0 = 1;$

$| [a_{1}, a_{2}] | = | a_{1} | | a_{2} | \sin (\hat{a_{1}, a_{2}}) = | a_{1} | | a_{2} |;$

Therefore, $a_{1} a_{2} a_{3} = | a_{1} | | a_{2} | | a_{3} | = 24.$

Answer: 24.

2. Determine whether the vectors $a_{1}, a_{2},$ and $a_{3}$ form a basis in the set of all vectors, if

a) $a_{1} (2, 3, - 1), a_{2} (1, - 1, 3), a_{3} (1, 9, - 11);$

b) $a_{1} (3, - 2, 1), a_{2} (2, 1, 2), a_{3} (3, - 1, - 2) .$

Solution.

A basis is any ordered triple of non-coplanar vectors. Let's check whether our vectors are coplanar, i.e., whether the condition $a_{1} a_{2} a_{3} = 0$ holds.

a) $a_{1} a_{2} a_{3} = | \begin{matrix} 2 & 3 & - 1 \\ 1 & - 1 & 3 \\ 1 & 9 & - 11 \end{matrix} | = 2 | \begin{matrix} - 1 & 3 \\ 9 & - 11 \end{matrix} | - 3 | \begin{matrix} 1 & 3 \\ 1 & - 11 \end{matrix} | - | \begin{matrix} 1 & - 1 \\ 1 & 9 \end{matrix} | =$

$= - 32 + 42 - 10 = 0.$

The vectors are coplanar, i.e., they do not form a basis.

b) $a_{1} a_{2} a_{3} = | \begin{matrix} 3 & - 2 & 1 \\ 2 & 1 & 2 \\ 3 & - 1 & - 2 \end{matrix} | = 3 | \begin{matrix} 1 & 2 \\ - 1 & - 2 \end{matrix} | + 2 | \begin{matrix} 2 & 2 \\ 3 & - 2 \end{matrix} | + | \begin{matrix} 2 & 1 \\ 3 & - 1 \end{matrix} | =$ $= 0 - 20 - 5 = - 25.$

The vectors are not coplanar, i.e., they do form a basis.

Answer. a) do not form; b) form.

3. Prove that for any $a, b,$ and $c$ , the vectors $a - b, b - c,$ and $c - a$ are coplanar. What is the geometric meaning of this fact?

Solution.

Let $a = (x_{a}, y_{a}, z_{a}), b = (x_{b}, y_{b}, z_{b}), c = (x_{c}, y_{c}, z_{c})$ .

Then, $[a - b, b - c] (c - a) = | \begin{matrix} x_{a} - x_{b} & y_{a} - y_{b} & z_{a} - z_{b} \\ x_{b} - x_{c} & y_{b} - y_{c} & z_{b} - z_{c} \\ x_{c} - x_{a} & y_{c} - y_{a} & z_{c} - z_{a} \end{matrix} | =$

using the property of determinants, we add the first row to the second row, the determinant remains unchanged:

$= | \begin{matrix} x_{a} - x_{b} & y_{a} - y_{b} & z_{a} - z_{b} \\ x_{a} - x_{c} & y_{a} - y_{c} & z_{a} - z_{c} \\ x_{c} - x_{a} & y_{c} - y_{a} & z_{c} - z_{a} \end{matrix} | = 0$ (Using the property of determinants again -- the second and third rows are proportional, so the determinant equals zero.)

Thus, the vectors $a - b, b - c,$ and $c - a$ are coplanar. The geometric meaning is that the vectors $a - b, b - c,$ and $c - a$ lie in parallel planes.

4. In a tetrahedron with vertices at points $A (1, 1, 1), B (2, 0, 2), C (2, 2, 2),$ and $D (3, 4, - 3)$ , calculate the height $h = | \overset{―}{D E} | .$

Solution.

Let's calculate the volume of the tetrahedron using the formula: $V = \frac{1}{6} | \overset{―}{A B} \overset{―}{A C} \overset{―}{A D} | :$

$\overset{―}{A B} = (1, - 1, 1);$

$\overset{―}{A C} = (1, 1, 1);$

$\overset{―}{A D} = (2, 3, - 4);$

$\overset{―}{A B} \overset{―}{A C} \overset{―}{A D} = | \begin{matrix} 1 & - 1 & 1 \\ 1 & 1 & 1 \\ 2 & 3 & - 4 \end{matrix} | = | \begin{matrix} 1 & 1 \\ 3 & - 4 \end{matrix} | + | \begin{matrix} 1 & 1 \\ 2 & - 4 \end{matrix} | + | \begin{matrix} 1 & 1 \\ 2 & 3 \end{matrix} | =$

$= - 7 - 6 + 1 = - 12.$

$V = \frac{1}{6} | \overset{―}{A B} \overset{―}{A C} \overset{―}{A D} | = 2.$

The volume can also be calculated using the well-known formula from high school geometry:

$V = \frac{1}{3} S_{O C H} h = \frac{1}{3} S_{△ A B C} | D E | = \frac{1}{6} | [\overset{―}{A B}, \overset{―}{A C}] | | D E | .$

$[\overset{―}{A B}, \overset{―}{A C}] = | \begin{matrix} i & j & k \\ 1 & - 1 & 1 \\ 1 & 1 & 1 \end{matrix} | = i | \begin{matrix} - 1 & 1 \\ 1 & 1 \end{matrix} | - j | \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} | + k | \begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix} | = - 2 i + 2 k .$

$| [\overset{―}{A B}, \overset{―}{A C}] | = \sqrt{2^{2} + 2^{2}} = \sqrt{8} .$

Thus, $2 = \frac{1}{6} \sqrt{8} | D E | .$ Отсюда $| D E | = \frac{12}{\sqrt{8}} = 3 \sqrt{2} .$

Answer: $3 \sqrt{2} .$

5 Prove that the four points $A (1, 2, - 1), B (0, 1, 5), C (- 1, 2, 1),$ and $D (2, 1, 3)$ lie in the same plane.

Solution.

Four points $A, B, C,$ and $D$ lie in the same plane if the vectors $\overset{―}{A B}, \overset{―}{A C},$ and $\overset{―}{A D}$ are coplanar.

Let's check if these vectors are coplanar:

$\overset{―}{A B} = (- 1, - 1, 6);$

$\overset{―}{A C} = (- 2, 0, 2);$

$\overset{―}{A D} = (1, - 1, 4) .$

$\overset{―}{A B} \overset{―}{A C} \overset{―}{A D} = | \begin{matrix} - 1 & - 1 & 6 \\ - 2 & 0 & 2 \\ 1 & - 1 & 4 \end{matrix} | = - | \begin{matrix} 0 & 2 \\ - 1 & 4 \end{matrix} | + | \begin{matrix} - 2 & 2 \\ 1 & 4 \end{matrix} | + 6 | \begin{matrix} - 2 & 0 \\ 1 & - 1 \end{matrix} | =$ $= - 2 - 10 + 12 = 0.$

Therefore, the vectors $\overset{―}{A B}, \overset{―}{A C},$ and $\overset{―}{A D}$ are coplanar, and the points $A, B, C,$ and $D$ lie in the same plane.

Homework.

1. Given $| a | = | b | = 5,$ and $(\hat{a, b}) = π / 4$ , calculate the area of the triangle constructed on the vectors $a - 2 b$ and $3 a + 2 b$ .

Answer: $50 \sqrt{2} .$

2. Prove that for any vectors $a$ , $p$ , $q$ , and $r$ , the vectors $[a, p]$ , $[a, q]$ , and $[a, r]$ are coplanar.

3. Given vectors $a_{1} (3, - 1, 2)$ and $a_{2} (1, 2, - 1)$ . Find the coordinates of the vectors

a) $[a_{1}, a_{2}];$

b) $[2 a_{1} + a_{2}, a_{2}];$

в) $[2 a_{1} - a_{2}, 2 a_{1} + a_{2}] .$

Answer: a) $(- 3, 5, 7);$ b) $(- 6, 10, 14);$ в) $(- 12, 20, 28) .$

4. In the triangle with vertices $A (1, - 1, 2)$ , $B (5, - 6, 2)$ , and $C (1, 3, - 1)$ , find the height $h = | B D |$ .

Answer: 5.

5. For the given vectors $a (2, 1, - 1)$ , $b (1, 2, 1)$ , $c (2, - 1, 3)$ , and $d (3, - 1, 2)$ , calculate the projection of the vector $a + c$ onto the vector $[b - d, c]$ .

Answer: $\sqrt{6}$ .

6. Find the coordinates of the vector $x$ if it is perpendicular to the vectors $a_{1} (2, - 3, 1)$ and $a_{2} (1, - 2, 3)$ , and also satisfies the condition $x (i + 2 j - 7 k) = 10$ .

Answer: $(7, 5, 1) .$

7. Vectors $a$ , $b$ , $c$ form a left triple $| a | = 1$ , $| b | = 2$ , $| c | = 3$ , $(\hat{a, b}) = π / 6$ , $c ⊥ a$ , $c ⊥ b$ . Find $a b c$ .

Answer: $- 3 / 2$ .

8. Prove the identity $(a + b + c) (a - 2 b + 2 c) (4 a + b + 5 c) = 0$ .

9. Calculate the volume of the tetrahedron $O A B C$ if $\overset{―}{O A} = 3 i + 4 j$ , $\overset{―}{O B} = - 3 j + k$ , $\overset{―}{O C} = 2 j + 5 k$ .