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The distance between two points. The angle of inclination of two vectors
Example 1
Let us calculate the distance between the points \latex{A(-2; 3)} and \latex{B(1; 7)}.
Solution
As the distance between the points \latex{ A } and \latex{ B } is equal to the length of \latex{\overrightarrow{AB}}, the solution can be reduced to the definition of magnitude. Let \latex{\overrightarrow{a}} denote the position vector of the point \latex{ A } and \latex{\overrightarrow{b}} denote the position vector of the point \latex{ B } (Figure 6).
Since
\latex{\overrightarrow{AB}=\overrightarrow b-\overrightarrow a,\;\;\overrightarrow a=-2}\latex{\text{i}}\latex{+3}\latex{\text{j}}, \latex{\overrightarrow b=}\latex{\text{i}}\latex{+7}\latex{\text{j}},
\latex{\overrightarrow{AB}=(1-(-2))}\latex{\text{i}}\latex{+(7-3)}\latex{\text j}\latex{=3}\latex{\text i}\latex{+4}\latex{\text j}, thus \latex{\lvert\overrightarrow{AB}\rvert=\sqrt{3^2+4^2}=5}.
This method for solving the example can be generalised, and so the following theorem holds:
\latex{\overrightarrow{AB}}
\latex{B(1; 7)}
\latex{A(-2; 3)}
\latex{\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow a}
\latex{(3; 4)}
\latex{\overrightarrow{b}}
\latex{ -2 }
\latex{ 1 }
\latex{ 3 }
\latex{ 3 }
\latex{ 4 }
\latex{ 7 }
x
y
Figure 6
THEOREM: The distance between the points \latex{A(a_1; a_2)} and \latex{B(b_1; b_2)} is
\latex{AB=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2}}.
\latex{A(a_1; a_2),\;\;B(b_1, b_2)\\AB=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2}}
Proof
The position vector of the point \latex{ A } is \latex{\overrightarrow a(a_1; a_2),} the position vector of the point
\latex{ B } is \latex{\overrightarrow b(b_1; b_2)}, \latex{\overrightarrow{AB}=\overrightarrow b-\overrightarrow a=(b_1-a_1)}\latex{\text i}\latex{+(b_2-a_2)}\latex{ j}.
Thus based on the formula relating to the magnitude:\latex{AB=\lvert\overrightarrow{AB}\rvert=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2}}.
Example 2
Two vertices of a regular triangle are \latex{A(–3; 5)}, \latex{B(–4; 7)}. Let us calculate the length of the side and the perpendicular height of the triangle, along with its area.
Solution
Let \latex{ a } denote the length of the side of the triangle. Based on the previous theorem:
\latex{a=AB=\sqrt{(-4-(-3))^2+(7-5)^2}=\sqrt{(-1)^2+2^2}=\sqrt 5\approx2.24}.
We know from earlier that the perpendicular height of the regular triangle with side \latex{ a } is
\latex{m=\frac{a\times\sqrt3}2}, its area is \latex{T=\frac{a^2\times\sqrt3}4} (Figure 7).
By substituting the value obtained for a into these formulae
\latex{m=\frac{a\times\sqrt3}2=\frac{15}2\approx1.94}; and \latex{T=\frac{a^2\times\sqrt3}4=\frac{5\times\sqrt3}4\approx2.17}.
\latex{m=\frac{a\times\sqrt3}{2}}
\latex{A=\frac{a^2\times\sqrt3}{2}}
\latex{ a }
\latex{ a }
\latex{ m }
\latex{ a }
Figure 7
Example 3
What is the angle included between the vectors \latex{\overrightarrow a (4; 6)} and \latex{\overrightarrow b (7; –1)}?
Solution
We have alreadycome across such a question while studying trigonometry.
We are going to use the two expressions of the dot product of the two vectors to determine the angle included: the definition and the form expressed with the coordinates.
We are going to use the two expressions of the dot product of the two vectors to determine the angle included: the definition and the form expressed with the coordinates.
\latex{(4; 6)}
\latex{(7; -1)}
\latex{\overrightarrow a}
\latex{\overrightarrow b}
\latex{ 6 }
\latex{ 2 }
\latex{ 1 }
\latex{ 0 }
\latex{ -1 }
\latex{ 4 }
Figure 8
If \latex{\varphi} denotes the angle included between the two vectors (Figure 8), then on one hand
\latex{\overrightarrow a\times\overrightarrow b=4\times7+6\times(-1)=22}.
By comparing the two results and rearranging for cosj we obtain the following:
\latex{\cos \varphi=\frac{22}{10\times\sqrt{26}}=\frac{11}{5\times\sqrt{26}}\approx0.4315}, which implies \latex{\varphi\approx 64.44^{\circ}}.
\latex{\overrightarrow a\times\overrightarrow b=\lvert\overrightarrow a\rvert\times\lvert\overrightarrow b\rvert\times\cos{\cos{\varphi}}=\sqrt{4^2+6^2}\times\sqrt{7^2+(-1)^2}\times\cos{\varphi}=\\ =\sqrt{52}\times\sqrt{50}=10\times\sqrt{26}\times\cos{\varphi},}
on the other hand\latex{\overrightarrow a\times\overrightarrow b=4\times7+6\times(-1)=22}.
\latex{\cos \varphi=\frac{22}{10\times\sqrt{26}}=\frac{11}{5\times\sqrt{26}}\approx0.4315}, which implies \latex{\varphi\approx 64.44^{\circ}}.
Example 4
Let us calculate the area of the triangle defined by the points \latex{A(-2; 2)}, \latex{B(3; 3)}, \latex{C(8; -2)}.
Solution
Let us denote the interior angle at the vertex \latex{ A } of the triangle by a (Figure 9). The area of the triangle can be expressed with the help of the known formula:
\latex{T=\frac{AB\times AC\times\sin\alpha}{2}}.
To calculate the area we have to determine the lengths of the sides \latex{ AB } and \latex{ AC } and also \latex{\sin\alpha}.
\latex{AB=\sqrt{(3-(-2))^2+(3-2)^2}=\sqrt{26}},
\latex{AC=\sqrt{(8-(-2))^2+(-2-2)^2}=\sqrt{116}=2\times\sqrt{29}}.
\latex{\alpha} is the angle included between the vectors \latex{\overrightarrow{AB}} and \latex{\overrightarrow{AC}}, thus \latex{\cos{\alpha}} can be determined with the help of the dot product of the two vectors.
Since \latex{\overrightarrow{AB}\;(5; 1)} and \latex{\overrightarrow{AC}\;(10; -4)},
Since \latex{\overrightarrow{AB}\;(5; 1)} and \latex{\overrightarrow{AC}\;(10; -4)},
\latex{\cos{\alpha}=\frac{\overrightarrow{AB}\times\overrightarrow{AC}}{AB\times AC}=\frac{5\times10+1\times(-4)}{2\times\sqrt{26}\times\sqrt{29}}=\frac{46}{2\times\sqrt{26}\times\sqrt{29}}=\frac{23}{\sqrt{26}\times\sqrt{29}}}.
\latex{\overrightarrow{AC}}
\latex{B(3; 3)}
\latex{A(-2; 2)}
\latex{C(8; -2)}
\latex{\overrightarrow{AB}}
\latex{\alpha}
\latex{ -2 }
\latex{ 0 }
\latex{ -2 }
\latex{ 1 }
\latex{ 3 }
\latex{ 8 }
x
y
Figure 10
Since \latex{\alpha} is an interior angle of the triangle, \latex{\sin\alpha \gt 0}, thus
\latex{\sin{\alpha}=\sqrt{1-\cos^{\alpha}}=\sqrt{1-\frac{529}{26\times29}}=\frac{15}{\sqrt{26}\times\sqrt{29}}}.
\latex{T=\frac{AB\times AC\times\sin{\alpha}}{2}=\frac{\sqrt{26}\times2\times\sqrt{29}}{2}=15.}
\latex{\sin{\alpha}=\sqrt{1-\cos^{\alpha}}=\sqrt{1-\frac{529}{26\times29}}=\frac{15}{\sqrt{26}\times\sqrt{29}}}.
Thus the area of the triangle \latex{ ABC } is
\latex{T=\frac{AB\times AC\times\sin{\alpha}}{2}=\frac{\sqrt{26}\times2\times\sqrt{29}}{2}=15.}
Exercises
{{exercise_number}}. Give the magnitude of \latex{\overrightarrow a} , if
- \latex{\overrightarrow a(1; 1)};
- \latex{\overrightarrow a(2; 3)};
- \latex{\overrightarrow a(3; -2)};
- \latex{\overrightarrow a(-7; -4)};
- \latex{\overrightarrow a\left(\frac3 5; -\frac4 5\right)};
- \latex{\overrightarrow a\left(\sqrt5; -\sqrt6\right)}.
{{exercise_number}}. How far away is the point \latex{ A } from the origin, if
- \latex{A(0; -3)};
- \latex{A(2; 2)};
- \latex{A(5; 12)};
- \latex{A(4; -3)};
- \latex{A(-10; -8)};
- \latex{A(-\sqrt7; -3)}.
{{exercise_number}}. Calculate the distance \latex{ AB }, if
- \latex{A(0; 1),\;\;B(3; 2)};
- \latex{A(4; 1),\;\;B(-1; 6)};
- \latex{A(-2; -5),\;\;B(7; -10)};
- \latex{A(8; -7),\;\;B(-4; 5)}.
{{exercise_number}}. The vertices of a triangle are
- \latex{A(0; 0),\;\;B(5; 1),\;\;C(2; 6)};
- \latex{A(0; 2),\;\;B(5; 0),\;\;C(3; 3)};
- \latex{A(4; 3),\;\;B(-5; -1),\;\;C(1; -3)}.
Determine the perimeter of the triangles in each case.
{{exercise_number}}. Calculate the angle included between the vectors \latex{\overrightarrow a} and \latex{\overrightarrow b}, if
- \latex{\overrightarrow a(1; 1),\;\;\overrightarrow b(3; 0)};
- \latex{\overrightarrow a(3; 5),\;\;\overrightarrow b(1; 4)};
- \latex{\overrightarrow a(-4; 7),\;\;\overrightarrow b(6; -9)}.
{{exercise_number}}. Calculate the area of the triangles given in exercise \latex{ 4 }.
{{exercise_number}}. The coordinates of three vertices of the parallelogram \latex{ ABCD } labelled in positive orientation are\latex{ A }(\latex{ 2; 3 }), \latex{ B }(\latex{ 5; 0 }), \latex{ C }(\latex{ 10; 1 }).
- Calculate the perimeter and the area of the parallelogram.
- Give the coordinates of the vertex \latex{D}.
{{exercise_number}}. Give the value of \latex{ y } if the angle included between the vectors \latex{\overrightarrow p(3; 4)} and \latex{ \overrightarrow q (6; y)} is \latex{60^{\circ}}.
