Vector and Mixed Product of Vectors.

Vector Product of Vectors.

Literature: Collection of Problems in Mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.

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., $|[a_1, a_2]| = |a_1||a_2|\sin(\widehat{a_1, a_2})$

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 $(\widehat{a_1,a_2}) = \frac{\pi}{2} \Leftrightarrow [a_1, a_2] = 0.$

Algebraic properties of the vector product:

$[a_1, a_2] = -[a_2, a_1];$

$[\lambda a_1, a_2] = \lambda[a_1, a_2];$

$[a_1 + a_2, b] = [a_1, b] + [a_2, b].$

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 $$[a_1, a_2]=(Y_1Z_2-Z_1Y_2)i-(X_1Z_2-Z_1X_2)j+(X_1Y_2-Y_1X_2)k,$$or, using symbolic notation (with the concept of the 3rd order determinant) $$[a_1, a_2]=\begin{vmatrix}i& j& k\\X_1& Y_1&Z_1\\X_2&Y_2&Z_2\end{vmatrix}.$$

Examples.

2.98. $|a_1| = 1, |a_2| = 2, (\widehat{a_1, a_2}) = 2\pi/3.$ Calculate:

a) $|[a_1, a_2]|$

b) $|[2a_1 + a_2, a_1 + 2a_2]|$

c) $|[a_1 + 3a_2, 3a_1 - a_2]|.$

Solution.

a) $|[a_1, a_2]|=|a_1||a_2|\sin(\widehat{a_1, a_2})=2\sin({2\pi/3})=2\frac{\sqrt {3}}{2}=\sqrt 3.$

b) $[2a_1+a_2, a_1+2a_2]=[2a_1, a_1+2a_2]+[a_2, a_1+2a_2]=$

$=2[a_1, a_1+2a_2]+[a_2, a_1+2a_2]=-2[a_1+2a_2, a_1]-[a_1+2a_2, a_2]=$

$=-2[a_1, a_1]-2[2a_2, a_1]-[a_1, a_2]-[2a_2, a_2]=-4[a_2, a_1]-[a_1,a_2]=$

$=4[a_1, a_2]-[a_1, a_2]=3[a_1, a_2].$

$|[2a_1+a_2, a_1+2a_2]|=3|[a_1, a_2]|=3\sqrt 3.$

c) $[a_1+3a_2, 3a_1-a_2]=[a_1, 3a_1-a_2]+3[a_2, 3a_1-a_2]=$

$=-[3a_1-a_2, a_1]-3[3a_1-a_2, a_2]=$

$=-[3a_1, a_1]+[a_2, a_1]-9[a_1, a_2]+3[a_2, a_2]=-10[a_1, a_2].$

$|[a_1+3a_2, 3a_1-a_2]|=|10[a_1, a_2]|=10\sqrt 3.$

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

2.100. 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) $[2a+b, c-a]+[b+c, a+b];$

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

Solution.

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

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

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

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

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

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

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

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

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

Solution.

$S \triangle ABC=\frac{1}{2}|[\overline{BA}, \overline{BC}]|.$

$\overline{BA}=(1-2; 1-3; 1-4)=(-1; -2; -3).$

$\overline{BC}=(4-2; 3-3; 2-4)=(2; 0; -2).$

$[\overline{BA}, \overline{BC}]=\begin{vmatrix}i&j&k\\-1&-2&-3\\2&0&-2\end{vmatrix}=4i-8j+4k.$

$S \triangle ABC=\frac{1}{2}|[\overline{BA}, \overline{BC}]|=\frac{1}{2}\sqrt{16+64+16}=\frac{\sqrt {96}}{2}=2\sqrt{6}.$

Answer: $2\sqrt{6}.$

2.110. 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 $k = (a, b)c:$

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

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

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

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

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

Solution.

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

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

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

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

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

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 $[a_1, a_2]a_3 = a_1a_2a_3.$

The mixed product expressed through the coordinates of vectors in a right-handed orthogonal basis is written as$$a_1a_2a_3=\begin{vmatrix}X_1&Y_1&Z_1\\X_2&Y_2&Z_2\\X_3&Y_3&Z_3\end{vmatrix}.$$

Examples.

2.124. 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 $a_1a_2a_3.$

Solution.

$a_1a_2a_3=[a_1, a_2]a_3=|[a_1, a_2]||a_3|cos(\widehat{[a_1, a_2], a_3}).$

$\cos(\widehat{[a_1, a_2], a_3})=\cos 0 =1;$

$|[a_1, a_2]|=|a_1||a_2|\sin(\widehat{a_1, a_2})=|a_1||a_2|;$

Therefore,$a_1a_2a_3=|a_1||a_2||a_3|=24.$

Answer: 24.

2.127. 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 $a_1a_2a_3 = 0$ holds.

a) $a_1a_2a_3=\begin{vmatrix}2&3&-1\\1&-1&3\\1&9&-11\end{vmatrix}=2\begin{vmatrix}-1&3\\9&-11\end{vmatrix}-3\begin{vmatrix}1&3\\1&-11\end{vmatrix}-\begin{vmatrix}1&-1\\1&9\end{vmatrix}=$

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

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

b) $a_1a_2a_3=\begin{vmatrix}3&-2&1\\2&1&2\\3&-1&-2\end{vmatrix}=3\begin{vmatrix}1&2\\-1&-2\end{vmatrix}+2\begin{vmatrix}2&2\\3&-2\end{vmatrix}+\begin{vmatrix}2&1\\3&-1\end{vmatrix}=$ $=0-20-5=-25.$

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

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

2.129. 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, $[a-b, b-c](c-a)=\begin{vmatrix}x_a-x_b&y_a-y_b&z_a-z_b\\x_b-x_c&y_b-y_c&z_b-z_c\\x_c-x_a&y_c-y_a&z_c-z_a\end{vmatrix}=$

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

$=\begin{vmatrix}x_a-x_b&y_a-y_b&z_a-z_b\\x_a-x_c&y_a-y_c&z_a-z_c\\x_c-x_a&y_c-y_a&z_c-z_a\end{vmatrix}=0$ (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.

2.134. 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 $h=|\overline{DE}|.$

Solution.

Let's calculate the volume of the tetrahedron using the formula: $V=\frac{1}{6}|\overline{AB}\overline{AC}\overline{AD}|:$

$\overline{AB}=(1, -1, 1);$

$\overline{AC}=(1, 1, 1);$

$\overline{AD}=(2, 3, -4);$

$\overline{AB}\overline{AC}\overline{AD}=\begin{vmatrix}1&-1&1\\1&1&1\\2&3&-4\end{vmatrix}=\begin{vmatrix}1&1\\3&-4\end{vmatrix}+\begin{vmatrix}1&1\\2&-4\end{vmatrix}+\begin{vmatrix}1&1\\2&3\end{vmatrix}=$

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

$V=\frac{1}{6}|\overline{AB}\overline{AC}\overline{AD}|=2.$

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

$V=\frac{1}{3}S_{OCH}h=\frac{1}{3}S_{\triangle ABC}|DE|=\frac{1}{6}|[\overline{AB}, \overline{AC}]||DE|.$

$[\overline{AB}, \overline{AC}]=\begin{vmatrix}i&j&k\\1&-1&1\\1&1&1\end{vmatrix}=i\begin{vmatrix}-1&1\\1&1\end{vmatrix}-j\begin{vmatrix}1&1\\1&1\end{vmatrix}+k\begin{vmatrix}1&-1\\1&1\end{vmatrix}=-2i+2k.$

$|[\overline{AB}, \overline{AC}]|=\sqrt{2^2+2^2}=\sqrt{8}.$

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

Answer: $3\sqrt {2}.$

2.137. 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 $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are coplanar.

Let's check if these vectors are coplanar:

$\overline{AB}=(-1, -1, 6);$

$\overline{AC}=(-2, 0, 2);$

$\overline{AD}=(1, -1, 4).$

$\overline{AB}\overline{AC}\overline{AD}=\begin{vmatrix}-1&-1&6\\-2&0&2\\1&-1&4\end{vmatrix}=-\begin{vmatrix}0&2\\-1&4\end{vmatrix}+\begin{vmatrix}-2&2\\1&4\end{vmatrix}+6\begin{vmatrix}-2&0\\1&-1\end{vmatrix}=$ $=-2-10+12=0.$

Therefore, the vectors $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are coplanar, and the points $A, B, C,$ and $D$ lie in the same plane.

Homework.

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

Answer: $50\sqrt{2}.$

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

2.106. 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) $[2a_1+a_2, a_2];$

в) $[2a_1-a_2, 2a_1+a_2].$

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

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

Answer: 5.

2.111. 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: $\sqrt{6}$.

2.119. 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 $x(i+2j-7k)=10$.

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

2.125. Vectors $a$, $b$, $c$ form a left triple $|a|=1$, $|b|=2$, $|c|=3$, $(\widehat{a, b})=\pi/6$, $c\perp a$, $c\perp b$. Find $abc$.

Answer: $-3/2$.

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

2.132. Calculate the volume of the tetrahedron $OABC$ if $\overline{OA}=3i+4j$, $\overline{OB}=-3j+k$, $\overline{OC}=2j+5k$.

Answer: $17/2$.

2.133. Calculate the volume of the tetrahedron with vertices at points $A(2, -3, 5)$, $B(0, 2, 1)$, $C(-2, -2, 3)$, and $D(3, 2, 4)$.

Answer: $6.$

2.136. For what value of $\lambda$ will the vectors $a$, $b$, $c$ be coplanar?

a) $a(\lambda, 3, 1), b(5, -1, 2), c(-1, 5, 4);$

b) $a(1, 2\lambda, 1), b(1,\lambda,0), c(0, \lambda, 1).$

Answer: a) $-3$ b) for any $\lambda.$