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

Length of a vector.

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

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 $\overset{―}{A B}$ can be found using the formulas $\overset{―}{A B} = (x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1}) .$ The scalar product of non-zero vectors $a_{1}$ and $a_{2}$ is defined as $(a_{1}, a_{2}) = | a_{1} | | a_{2} | \cos (\hat{a_{1}, a_{2}}) .$

For scalar multiplication, the notation $a_{1} a_{2}$ is also used alongside the notation $(a_{1}, a_{2})$ .

Geometric properties of scalar multiplication:

1) $a_{1} ⊥ a_{2} \Leftrightarrow a_{1} a_{2} = 0$ (condition for vectors to be perpendicular).

2) If $φ = (\hat{a_{1}, a_{2}}),$ then $0 \leq φ < \frac{π}{2} \Leftrightarrow a_{1} a_{2} > 0; \frac{π}{2} < φ \leq π \Leftrightarrow a_{1} a_{2} < 0.$

Algebraic properties of scalar multiplication:

1) $a_{1} a_{2} = a_{2} a_{1};$

2) $(λ a_{1}) a_{2} = λ (a_{1} a_{2});$

3) $a (b_{1} + b_{2}) = a b_{1} + a b_{2} .$

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 $a_{1} a_{2} = X_{1} X_{2} + Y_{1} Y_{2} + Z_{1} Z_{2} .$

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

$\cos (\hat{a_{1}, a_{2}}) = \frac{a_{1} a_{2}}{| a_{1} | | a_{2} |} = \frac{X_{1} X_{2} + Y_{1} Y_{2} + Z_{1} Z_{2}}{\sqrt{X_{1}^{2} + Y_{1}^{2} + Z_{1}^{2}} \sqrt{X_{2}^{2} + Y_{2}^{2} + Z_{2}^{2}}} .$

Examples.

1. $| a_{1} | = 3; | a_{2} | = 4; (\hat{a_{1}, a_{2}}) = \frac{2 π}{3} .$ Compute:

a) $a_{1}^{2} = a_{1} a_{1};$

b) $(3 a_{1} - 2 a_{2}) (a_{1} + 2 a_{2});$

c) $(a_{1} + a_{2})^{2} .$

Solution.

a) $a_{1}^{2} = (a_{1}, a_{1}) = | a_{1} | | a_{1} | \cos (\hat{a_{1}, a_{1}}) = | a_{1} |^{2} = 3^{2} = 9.$

b) $(3 a_{1} - 2 a_{2}) (a_{1} + 2 a_{2});$

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 $\frac{2 π}{3}$ with the positive semi-axis of the $x$ -axis, has coordinates $x = | a_{2} | \cos \frac{2 π}{3} = - \frac{4}{2} = - 2;$

$y = | a_{2} | \sin \frac{2 π}{3} = 4 \frac{\sqrt{3}}{2} = 2 \sqrt{3}$

a1a2

$3 a_{1} - 2 a_{2} = 3 (3; 0) - 2 (- 2; 2 \sqrt{3}) = (9; 0) - (- 4; 4 \sqrt{3}) = (13; - 4 \sqrt{3});$

$a_{1} + 2 a_{2} = (3; 0) + 2 (- 2; 2 \sqrt{3}) = (3; 0) + (- 4; 4 \sqrt{3}) = (- 1; 4 \sqrt{3}) .$

$(3 a_{1} - 2 a_{2}) (a_{1} + 2 a_{2}) = (13; - 4 \sqrt{3}) (- 1; 4 \sqrt{3}) = - 13 - 48 = - 61.$

c) $(a_{1} + a_{2})^{2} .$

$a_{1} + a_{2}$ = $(3; 0) + (- 2; 2 \sqrt{3}) = (1; 2 \sqrt{3}) .$

$(a_{1} + a_{2})^{2} = (1; 2 \sqrt{3}) (1; 2 \sqrt{3}) = 1 + 12 = 13.$

Answer: a) 9; b) -61; c) 13.

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

Solution.

Method 1.

From triangle $A B C$ , we have $A C = A B + B C = a + b = p - 3 q + 5 p + 2 q = 6 p - q .$

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

$| A C |^{2} = | 6 p |^{2} + | q |^{2} - 12 p q \cos \hat{(6 p, q)} = 288 + 9 - 72 = 225.$

Hence, $| A C | = 15.$

From triangle $A B D$ , we have: $B D = A D - A B = b - a = 5 p + 2 q - p + 3 q = 4 p + 5 q .$

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

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

Hence, $| B D | = \sqrt{593} .$

Method 2.

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

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

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

The vector $B C = A D = b .$

From triangle $A B C$ we have

$A C = A B + B C = a + b = p - 3 q + 5 p + 2 q = 6 p - q =$ $= 6 (2; 2) - (3; 0) = (12; 12) - (3; 0) = (9; 12) .$

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

From triangle $A B D$ we have

$B D = A D - A B = b - a = 5 p + 2 q - p + 3 q = 4 p + 5 q =$ $= 4 (2; 2) + 5 (3; 0) = (8; 8) + (15; 0) = (23; 8) .$

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

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

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

Answer: $2 π / 3$

Homework:

$| a_{1} | = 3; | a_{2} | = 5.$ Determine at what value of $α$ the vectors $a_{1} + α a_{2}$ and $a_{1} - α a_{2}$ will be perpendicular.

Answer: $α = \pm \frac{3}{5}$

In triangle $A B C$ , $\overset{―}{A B} = 3 e_{1} - 4 e_{2};$ $\overset{―}{B C} = e_{1} + 5 e_{2} .$ Calculate the length of its altitude $\overset{―}{C H}$ , if it is known that $e_{1}$ and $e_{2}$ are mutually perpendicular units.

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

Tags: scalar multiplication, vector, length of vectors, angle between vectors, scalar product

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