Scalar product of vectors, properties. Length of vectors. Angle between vectors.

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

Length of a vector.

Let a vector $\overline a=(x, y, z)$ be represented by its coordinates in a rectangular basis. Then its length can be calculated using the formula $$|\overline a|=\sqrt{x^2+y^2+z^2}.$$

Scalar product of vectors.

If the coordinates of points $A(x_1, y_1, z_1) $ and $B(x_2, y_2, z_2)$ are given, then the coordinates of the vector $\overline{AB}$ can be found using the formulas $$\overline{AB}=(x_2-x_1, y_2-y_1, z_2-z_1).$$ The scalar product of non-zero vectors $a_1$ and $a_2$ is defined as $$(a_1, a_2)=|a_1||a_2|\cos(\widehat{a_1, a_2}).$$

For scalar multiplication, the notation $a_1a_2$ is also used alongside the notation $(a_1,a_2)$.

Geometric properties of scalar multiplication:

1) $a_1\perp a_2\Leftrightarrow a_1a_2=0$ (condition for vectors to be perpendicular).

2) If $\varphi=(\widehat{a_1, a_2}),$ то $$0\leq\varphi<\frac{\pi}{2}\Leftrightarrow a_1a_2>0; \qquad\qquad \frac{\pi}{2}<\varphi\leq\pi\Leftrightarrow a_1 a_2<0.$$

Algebraic properties of scalar multiplication:

1) $a_1a_2=a_2a_1;$

2) $(\lambda a_1)a_2=\lambda (a_1 a_2);$

3) $a(b_1+b_2)=ab_1+ab_2.$

If vectors $a_1(X_1, Y_1, Z_1)$ and $a_2(X_2, Y_2, Z_2)$ are represented by their coordinates in a rectangular basis, then the scalar product is equal to $$a_1a_2=X_1X_2+Y_1Y_2+Z_1Z_2. $$

From this formula, in particular, follows the formula for determining the cosine of the angle between vectors:

$$\cos(\widehat{a_1, a_2})=\frac{a_1 a_2}{|a_1||a_2|}=\frac{X_1X_2+Y_1Y_2+Z_1Z_2}{\sqrt{X_1^2+Y_1^2+Z_1^2}\sqrt{X_2^2+Y_2^2+Z_2^2}}.$$

Examples.

2.65. $|a_1|=3; |a_2|=4; (\widehat{a_1,a_2})=\frac{2\pi}{3}.$ Compute:

а) $a_1^2=a_1a_1;$

б) $(3a_1-2a_2)(a_1+2a_2);$

в) $(a_1+a_2)^2.$

Solution.

а) $$a_1^2=(a_1, a_1)=|a_1||a_1|\cos(\widehat{a_1, a_1})=|a_1|^2=3^2=9.$$

б) $(3a_1-2a_2)(a_1+2a_2);$

Since the scalar product depends on the lengths of vectors and the angle between them, the given vectors can be arbitrarily chosen considering these characteristics. Let $a_1=(3; 0).$ Then, the vector $a_2,$ having a length $|a_2|=4,$ and forming an angle $\frac{2\pi}{3}$ with the positive semi-axis of the $x$-axis, has coordinates $x=|a_2|\cos\frac{2\pi}{3}=-\frac{4}{2}=-2;$

$y=|a_2|\sin\frac{2\pi}{3}=4\frac{\sqrt 3}{2}=2\sqrt 3$

a1a2

$3a_1-2a_2=3(3;0)-2(-2;2\sqrt 3)=(9;0)-(-4; 4\sqrt 3)=(13;-4\sqrt 3);$

$a_1+2a_2=(3; 0)+2(-2;2\sqrt 3) = (3; 0)+ (-4; 4\sqrt 3)= (-1; 4\sqrt 3).$

$(3a_1-2a_2)(a_1+2a_2)=(13; -4\sqrt 3)(-1; 4\sqrt 3) =-13-48=-61.$

в) $(a_1+a_2)^2.$

$a_1+a_2$=$(3; 0)+(-2; 2\sqrt 3)=(1; 2\sqrt 3).$

$(a_1+a_2)^2=(1; 2\sqrt3) (1; 2\sqrt 3)=1+12=13.$

Ответ: a) 9; б) -61; в) 13.

2.67. Calculate the length of the diagonals of the parallelogram constructed on the vectors $a=p-3q, $ $b=5p+2q,$ given that $|p|=2\sqrt{2}, |q|=3, (\widehat{p, q})=\frac{\pi}{4}.$

Solution.

Method 1.

From triangle $ABC$, we have $AC=AB+BC=a+b=p-3q+5p+2q=6p-q.$

Knowing the length of vectors $p$ and $q$ and the angle between these vectors, we can find the length of vector $AC$ using the cosine theorem:

$|AC|^2=|6p|^2+|q|^2-12pq\cos\widehat{(6p, q)}=288+9-72=225.$

Hence, $|AC|=15.$

From triangle $ABD$, we have: $BD=AD-AB=b-a=5p+2q-p+3q=4p+5q.$

Using the cosine theorem, we find the length of vector $BD$:

$|BD|^2=|4p|^2+|5q|^2-8p5q\cos \widehat{(6p, q)}=$ $128+225+240=593.$

Hence, $|BD|=\sqrt{593}.$

Method 2.

Let $q=(3; 0).$ Then, the vector $p,$ having a length $|p|=2\sqrt 2,$ and forming an angle $\frac{\pi}{4}$ with the positive half-axis of the $OX$ axis, has coordinates

$x=|p|\cos\frac{\pi}{4}=2\sqrt 2\frac{1}{\sqrt 2}=2; $

$y=|p|\sin\frac{\pi}{4}=2\sqrt 2\frac{1}{\sqrt 2}=2.$

The vector $BC=AD=b.$

From triangle $ABC$ we have

$AC=AB+BC=a+b=p-3q+5p+2q=6p-q=$ $=6(2;2)-(3;0)=(12; 12)-(3;0)=(9; 12).$

Therefore, $|AC|=\sqrt{81+144}=\sqrt{225}=15.$

From triangle $ABD$ we have

$BD=AD-AB=b-a=5p+2q-p+3q=4p+5q=$ $=4(2; 2)+5(3;0)=(8; 8)+(15; 0)=(23; 8).$

Thus, $|BD|=\sqrt{23^2+8^2}=\sqrt {593}.$

Answer: $15, \sqrt {593}.$

2.68. Determine the angle between vectors $a$ and $b$ if it is known that $(a-b)^2+(a+2b)^2=20$ and $|a|=1, |b|=2.$

Answer: $2\pi/3$

Homework:

2.66.

$|a_1|=3; |a_2|=5. $ Determine at what value of $\alpha$ the vectors $a_1+\alpha a_2$ and $a_1-\alpha a_2$ will be perpendicular.

Answer: $\alpha=\pm\frac{3}{5}$

2.69.

In triangle $ABC$, $\overline{AB}=3e_1-4e_2;$ $\overline{BC}=e_1+5e_2.$ Calculate the length of its altitude $\overline{CH}$, if it is known that $e_1$ and $e_2$ are mutually perpendicular units.

Answer: $\frac{19}{5}.$

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