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Vector operations and applications
(reminder)
(reminder)
We became familiar with several concepts and operations in connection with vectors in our earlier studies.
The magnitude of a vector means its length. The magnitude of the zero vector is \latex{ 0 }, its direction can be arbitrary.
If the points of the plane or the space are designated by vectors starting from a fixed point, the reference point, then these vectors are called position vectors.
The magnitude of a vector means its length. The magnitude of the zero vector is \latex{ 0 }, its direction can be arbitrary.
If the points of the plane or the space are designated by vectors starting from a fixed point, the reference point, then these vectors are called position vectors.
We know the ways to determine of the sum and the difference of vectors. We also know how to multiply vectors by real numbers (Figure 1).
We clarified that the following rules are satisfied for these operations:
We clarified that the following rules are satisfied for these operations:
- The addition of vectors is a commutative and associative operation (Figure 2).
\latex{\overrightarrow{a} +\overrightarrow{b} =\overrightarrow{b} +\overrightarrow{a};}
\latex{(\overrightarrow{a}+\overrightarrow{b} )+\overrightarrow{c} =\overrightarrow{a} +(\overrightarrow{b}+\overrightarrow{c} ).}
\latex{\overrightarrow{a}}
\latex{\overrightarrow{a}+\overrightarrow{b}}
\latex{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}
\latex{\overrightarrow{b}}
\latex{\overrightarrow{b}+\overrightarrow{c}}
\latex{\overrightarrow{c}}
Figure 2
\latex{\overrightarrow{a}}
\latex{\overrightarrow{b}}
\latex{\overrightarrow{a}+\overrightarrow{b}}
\latex{\overrightarrow{a}-\overrightarrow{b}}
\latex{\overrightarrow{a}}
\latex{\lambda \times \overrightarrow{a}}
Figure 1
- The following properties hold when multiplying by a real number:
\latex{\mu \times (\lambda \times \overrightarrow{a} )=(\mu \times \lambda )\times \overrightarrow{a} =\lambda \times (\mu \times \overrightarrow{a} );}
\latex{(\lambda +\mu )\times \overrightarrow{a} =\lambda \times \overrightarrow{a} +\mu \times \overrightarrow{a} ;}
\latex{\lambda \times (\overrightarrow{a} +\overrightarrow{b} )=\lambda \times \overrightarrow{a} +\lambda \times \overrightarrow{b}.}
\latex{(\lambda +\mu )\times \overrightarrow{a} =\lambda \times \overrightarrow{a} +\mu \times \overrightarrow{a} ;}
\latex{\lambda \times (\overrightarrow{a} +\overrightarrow{b} )=\lambda \times \overrightarrow{a} +\lambda \times \overrightarrow{b}.}
The vector
\latex{\overrightarrow{c} =\alpha \times \overrightarrow{a} +\beta \times \overrightarrow{b}}
created from the vectors \latex{\overrightarrow{a}} and \latex{\overrightarrow{b}} according to the interpretation aboveis called the linear combination of the vectors \latex{\overrightarrow{a}} and \latex{\overrightarrow{b}}.
We have shown that if \latex{\overrightarrow{a}} and \latex{\overrightarrow{b}} are non-parallel vectors and \latex{\overrightarrow{c}} isan arbitrary vector coplanar with them, then \latex{\overrightarrow{c}} can be expressed in the form
\latex{\overrightarrow{c} =\alpha \times \overrightarrow{a} +\beta \times \overrightarrow{b} ,}
where \latex{\alpha} and \latex{\beta} are uniquely defined numbers.
The corresponding spatial version of this property is: if \latex{\overrightarrow{a} ,\overrightarrow{b}} and \latex{\overrightarrow{c}} arevectors not parallel with a plane and \latex{\overrightarrow{d}} is an arbitrary vector, then \latex{\overrightarrow{d}} can be expressed in the form
\latex{\overrightarrow{d} =\alpha \times \overrightarrow{a} +\beta \times \overrightarrow{b} +\gamma \times \overrightarrow{c} ,}
where \latex{\alpha} and \latex{\beta} are unambiguously defined real numbers.
The position vector of the point \latex{ R } dividing the line segment \latex{ AB } in theratio \latex{ AR : RB = } \latex{p\div q} can be expressed with the help of the position vectors pointing to the end-points of the line segment (Figure 3):
\latex{\overrightarrow{r} =\frac{q\times \overrightarrow{a} +p\times \overrightarrow{b} }{p+q}.}
\latex{A}
\latex{p}
\latex{R}
\latex{q}
\latex{B}
\latex{\overrightarrow{a}}
\latex{\overrightarrow{b}}
\latex{\overrightarrow{r}}
Figure 3
Example 1
In a regular hexagon \latex{ ABCDEF } let the position vectors pointing from its vertex \latex{ A } to the adjacent vertices be: \latex{\overrightarrow{a}} and \latex{\overrightarrow{b}} . Let us express the position vectors pointing to the other vertices in terms of these position vectors.
Solution
In Figure 4 \latex{\overrightarrow{AB} =\overrightarrow{a},\overrightarrow{AF} =\overrightarrow{b.}} Since the polygon is regular, the positionvector pointing to its centre is: \latex{\overrightarrow{AO} =\overrightarrow{a} +\overrightarrow{b}}
It can easily be seen that for the position vector of the point \latex{ C } thefollowing holds:
It can easily be seen that for the position vector of the point \latex{ C } thefollowing holds:
\latex{\overrightarrow{AC} =\overrightarrow{AO} +\overrightarrow{a} =\overrightarrow{a}+\overrightarrow{b} +\overrightarrow{a} =2\times \overrightarrow{a} +\overrightarrow{b}.}
At the same time the vector pointing to the point \latex{ D } is:
\latex{\overrightarrow{AD}=2\times \overrightarrow{AO}=2\times (\overrightarrow{a}+\overrightarrow{b} ).}
And the position vector of the point \latex{ E } is:
\latex{\overrightarrow{AE}=\overrightarrow{AO} +\overrightarrow{b} =\overrightarrow{a} +\overrightarrow{b}+\overrightarrow{b} =\overrightarrow{a} +2\times \overrightarrow{b}.}
\latex{\overrightarrow{b}}
\latex{\overrightarrow{a}}
\latex{\overrightarrow{a}+\overrightarrow{2b}}
\latex{\overrightarrow{a}+\overrightarrow{b}}
\latex{\overrightarrow{2a}+\overrightarrow{b}}
\latex{ A }
\latex{ F }
\latex{ E }
\latex{ D }
\latex{ C }
\latex{ B }
\latex{ O }
Figure 4
Example 2
In the plane of the quadrilateral \latex{ ABCD } let us construct the point, from which the sum of the position vectors pointing to the vertices of the quadrilateral is the zero vector.
Solution
We are looking for a reference point \latex{ O } the following holds for the position vectors starting from which: \latex{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\overrightarrow{0}}. Let us rearrange the sum:
\latex{\frac{\overrightarrow{OA}+\overrightarrow{OB} }{2} + \frac{\overrightarrow{OC}+\overrightarrow{OD} }{2} =\overrightarrow{0}}.
We know that the vectors
\latex{\frac{\overrightarrow{OA}+\overrightarrow{OB} }{2} =\overrightarrow{OE}} and \latex{\frac{\overrightarrow{OC}+\overrightarrow{OD} }{2} =\overrightarrow{OF}}
point to the midpoints of the corresponding sides, and the sum of these will be the zero vector only in the case they are opposite in direction, but equal in length.
Thus the suitable reference point in question can be found at the intersection point of the midlines of the quadrilateral. It is also the midpoint of the midlines. (Figure 5)
Thus the suitable reference point in question can be found at the intersection point of the midlines of the quadrilateral. It is also the midpoint of the midlines. (Figure 5)
\latex{ O }
\latex{ E }
\latex{ B }
\latex{ C }
\latex{ F }
\latex{ D }
\latex{ A }
Figure 5
⯁ ⯁ ⯁
In year \latex{ 10 } we proved that the position vector of the centroid of a triangle can be expressed in terms of the position vectors pointing to the verticesof the triangle (Figure 6):
\latex{\overrightarrow{s}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} }{3}}.
\latex{\overrightarrow{b}}
\latex{\overrightarrow{s}}
\latex{\overrightarrow{a}}
\latex{\overrightarrow{c}}
\latex{ S }
\latex{ B }
\latex{ O }
\latex{ C }
\latex{ A }
Figure 6
Example 3
Let us choose the centre of the circumscribed circle of the triangle as the reference centre. To which point is the sum of the position vectors of the vertices pointing?
Solution
Let us first add the position vectors pointing to the points \latex{ A } and \latex{ B }. The sum of these will be perpendicular to the side \latex{ AB } of the triangle. (Figure 7)
Let us add this vector to the position vector of the point \latex{ C }.
Based on the construction it can be seen that the point designated bythe vector \latex{(\overrightarrow{a}+\overrightarrow{b} )+\overrightarrow{c} =\overrightarrow{a} +\overrightarrow{b}+\overrightarrow{c}} lies on the straight line of the altitudepassing through the point \latex{ C }.
Let us add this vector to the position vector of the point \latex{ C }.
Based on the construction it can be seen that the point designated bythe vector \latex{(\overrightarrow{a}+\overrightarrow{b} )+\overrightarrow{c} =\overrightarrow{a} +\overrightarrow{b}+\overrightarrow{c}} lies on the straight line of the altitudepassing through the point \latex{ C }.
If we add the vectors up in a different order, then it can similarly be shown that this end-point will lie also on the straight line of the altitude passing through the vertex \latex{ B }. Since there is only one point (the orthocentre \latex{ M } of the triangle), which lies on all altitudes, the point designated by the vector \latex{\overrightarrow{a} +\overrightarrow{b} +\overrightarrow{c}} is actually the orthocentre.
Based on the relation relating to the centroid, which was proven earlier, and based on the solution of example \latex{ 3 } we can formulate the following theorem:
Based on the relation relating to the centroid, which was proven earlier, and based on the solution of example \latex{ 3 } we can formulate the following theorem:
\latex{\overrightarrow{c}}
\latex{\overrightarrow{a}}
\latex{\overrightarrow{b}}
\latex{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}
\latex{\overrightarrow{a}+\overrightarrow{b}}
\latex{ O }
\latex{ C }
\latex{ B }
\latex{ A }
Figure 7
THEOREM: The centre \latex{ O } of the circumscribed circle, the centroid \latex{ S } and the orthocentre \latex{ M } of a triangle are always collinear, where the point \latex{ S } is the trisection point of the line segment \latex{ OM }: \latex{\frac{OS}{SM}=\frac{1}{2} .}
\latex{ S }
\latex{ O }
\latex{ M }
\latex{ C }
\latex{ B }
\latex{ A }
\latex{ F_c }
Figure 8
Proof
Let us choose the centre \latex{ O } of the circumscribed circle of the triangle as the reference centre, and let us give the position vectors pointing from this point to the centroid \latex{ S } and the orthocentre \latex{ M }:
\latex{\overrightarrow{OS}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c} }{3}} , \latex{\overrightarrow{OM}= \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c.} }
The two relations imply that
\latex{\overrightarrow{OM}=3\times \overrightarrow{OS,} } i.e. the two vectors point in the same direction, and the relation between their lengths means that the point \latex{ S } will be a trisection point of the line segment \latex{ OM }. (Figure 8)
This straight line is called the Euler line of the triangle.
The theorem implies the statement below.
Consequence: The orthocentre \latex{ M } is twice the distance from the vertex \latex{ C } as the centre \latex{ O } of the circumscribed circle from the side \latex{ c }, i.e. \latex{MC=2\times OF_{C}}.
In Figure 8 it can be seen that \latex{OSF_{C} \Delta \sim MSC\Delta} since their angles are equal.
\latex{(OSF_{c}\nless = MSC \nless and OF_{C} \parallel MC.)} The ratio of similarity is \latex{\frac{1}{2}} (because of the previous theorem).
Obviously this statement can be formulated for any vertex and any pair of sides.
The theorem implies the statement below.
Consequence: The orthocentre \latex{ M } is twice the distance from the vertex \latex{ C } as the centre \latex{ O } of the circumscribed circle from the side \latex{ c }, i.e. \latex{MC=2\times OF_{C}}.
In Figure 8 it can be seen that \latex{OSF_{C} \Delta \sim MSC\Delta} since their angles are equal.
\latex{(OSF_{c}\nless = MSC \nless and OF_{C} \parallel MC.)} The ratio of similarity is \latex{\frac{1}{2}} (because of the previous theorem).
Obviously this statement can be formulated for any vertex and any pair of sides.
LEONHARD EULER
(\latex{ 1707–1783 }),
a Swiss mathematician. One of the most versatile philosophersof all times who gave usmore theorems thanany other. He started his research in Berlin, but hespent most of his life in the court of autocratrix Catherine in Saint Petersburg, where at the end of his life he dictated his mathematical papersto his servant completely blind. In Switzerland his works started to bepublished: they totalled more than \latex{ 70 } large-formatvolumes.
(\latex{ 1707–1783 }),
a Swiss mathematician. One of the most versatile philosophersof all times who gave usmore theorems thanany other. He started his research in Berlin, but hespent most of his life in the court of autocratrix Catherine in Saint Petersburg, where at the end of his life he dictated his mathematical papersto his servant completely blind. In Switzerland his works started to bepublished: they totalled more than \latex{ 70 } large-formatvolumes.
Exercises
{{exercise_number}}. Let the edge vectors starting from one vertex of a cube be \latex{\overrightarrow{a} ,\overrightarrow{b}} and \latex{\overrightarrow{c}}. Express the position vectors pointing to the other vertices of the cube in terms of these vectors.
{{exercise_number}}. Let the position vectors starting from the vertex \latex{ D } of a regular tetrahedron \latex{ ABCD } be the vectors \latex{\overrightarrow{a} ,\overrightarrow{b}} and \latex{\overrightarrow{c}} , denoted according to the end-points. Express the following vectorsin terms of these:
- \latex{\overrightarrow{AB}+\overrightarrow{CD;}}
- \latex{\overrightarrow{AC}+ \overrightarrow{CD}+\overrightarrow{DB};}
- \latex{\overrightarrow{AB}-\overrightarrow{CD}};
- \latex{\overrightarrow{AD}+\overrightarrow{CD}-\overrightarrow{BC}.}
{{exercise_number}}. The vectors \latex{\overrightarrow{a}} and \latex{\overrightarrow{b}} are perpendicular to each other, and \latex{|\overrightarrow{a} |=3, |\overrightarrow{b}|=4} units. Constructthe following vectors and give their magnitudes:
- \latex{\overrightarrow{a}+\overrightarrow{b};}
- \latex{\overrightarrow{a} -\overrightarrow{b};}
- \latex{2\times \overrightarrow{a} +3\times \overrightarrow{b} ;}
- \latex{2\times \overrightarrow{a} -5\times \overrightarrow{b} .}
{{exercise_number}}. The vector \latex{\overrightarrow{a}} is given. Construct the following vectors:
- \latex{\sqrt{2}\times \overrightarrow{a};}
- \latex{\frac{\sqrt{3} }{2} \times \overrightarrow{a} ;}
- \latex{-\frac{3}{4} \times \overrightarrow{a} ;}
- \latex{-\sqrt{5} \times \overrightarrow{a} .}
{{exercise_number}}. Let \latex{\overrightarrow{m}=\overrightarrow{a}+\overrightarrow{b}} and \latex{\overrightarrow{n}=\overrightarrow{a} -\overrightarrow{b}.} Express the following vectors in terms of the vectors \latex{\overrightarrow{a}} and \latex{\overrightarrow{b}}:
- \latex{2\times \overrightarrow{m} +3\times \overrightarrow{n};}
- \latex{2\times \overrightarrow{m} -4\times \overrightarrow{n} ;}
- \latex{\frac{1}{2} \times \overrightarrow{m} +\frac{2}{3} \times \overrightarrow{n} ;}
- \latex{-2\times \overrightarrow{m} +3\times (\overrightarrow{m}-\overrightarrow{n} ).}
{{exercise_number}}. The lengths of the sides of the triangle \latex{ ABC } are \latex{a=6, b=7, c= 8.\,cm}. Let us choose the point \latex{ A } as the centre of the reference system, and let the position vectors pointing to the points \latex{ B } and \latex{ C } be \latex{\overrightarrow{b}} and \latex{\overrightarrow{c}} respectively. The angle bisector of the angle at the vertex \latex{ A } intersects the side \latex{ BC } at the point \latex{ D }. Express the position vector pointing to the point \latex{ D } with the help of the vectors \latex{\overrightarrow{b}} and \latex{\overrightarrow{c}} .
{{exercise_number}}. The vectors \latex{\overrightarrow{a} } and \latex{\overrightarrow{b} } with a common starting point are given so that \latex{\|\overrightarrow{a} |=3, |\overrightarrow{b} |=4.} Construct a vector in their plane that includes angles of equal measure with these vectors.
{{exercise_number}}. The position vectors of the four vertices of a parallelogram are \latex{\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}} and \latex{\overrightarrow{d}} respectively.Prove that \latex{\overrightarrow{a} +\overrightarrow{c}=\overrightarrow{b}+\overrightarrow{d}. }
{{exercise_number}}. Extend the lengths of the sides of the triangle\latex{ ABC } to their double according to the figure. Show that the centroid of the \latex{A_{1} B_{1} C_{1}} obtained is also the centroid of the triangle ABC.
\latex{ C }
\latex{ B }
\latex{ B_1 }
\latex{ A }
\latex{ A_1 }
\latex{ C_1 }