Vector: A ector is a matrix that has only one row then we call the matrix a row ector or only one column then we call it a column ector. A row ector is of the form: a a a... A column ector is of the form: b b b b m an A quantity such as force displacement or elocity is called a ector and is represented by a directed line segment A ector in the plane is directed line segment. The directed line segment AB has initial point A and terminal point B; its length is denoted by AB. Two ectors are equal if they hae the same length and direction. Component form If is a two dimensional ector in the plane equal to the ector with initial point at the origin and terminal point ) then the Component form of is: ) If is a three dimensional ector in the plane equal to the ector with initial point at the origin and terminal point ) then the Component form of is: )
The numbers and are called the components of. Gien the points P x y ) and Q x y ) the standard position ector ) equal to PQ is x x y y z z) z z The magnitude or length of the ector PQ is the nonnegatie number x x ) y y) z z) The only ector with length 0 is the zero ector 0 00) or 0 000). This ector is also the only ector with no specific direction. Find a) component form and b) length of the ector with initial point P 4 ) and terminal point Q 5 ) Solution: a) 5-4 -) The component form of PQ is - - ) b) The length or magnitude of PQ is ) ) ) 9 Vector Addition and Multiplication of a ector by a scalar Addition: Let u u u ) and ) be ectors with k a scalar. u u u u u )
Scalar multiplication: ku ku ku ku ) If the length of ku is the absolute alue of the scalar k times the length of u. The ector ) u u has the same length as u but points in the opposite direction. If u u u ) and ) u u u u ) u Note that u ) u and the difference u as the sum u ) Properties of ector operations: Let u and w be ectors and a and b be scalars. ) u u ) u ) w u w) ) u 0 u 4) u u) 0 5) 0u 0 6) u u 7) a bu) ab) u 8) a u ) au a 9) a b) u au bu Unit ectors A ector of length is called unit ector. The standard unit ectors are: i 00) j 00) k 00) ) 00) 0 0) 00 ) 00) 00) 00) i j k
We call the scalar or number) the i-component of the ector the j-component of the ector and the k-component. In component form P x y ) and P x y ) is z z P P x x )i y y )j z z ) k If 0 then u is a unit ector in the direction of called the direction of the nonzero ector. Solution Find a unit ector u in the direction of the ector P 0 ) and P 0). P P )i - 0)j 0 -) k i j- k P P ) ) -) 9 u PP i j k i j PP The unit ector u is the direction of P P. k Midpoint of a line segment The Midpoint M of a line segment joining points P x y ) and P x y ) is the point z x x) y y) z z ) z 4
The midpoint of the segment joining P 0) and P 744) is 7 4 0 4 5) Product of ectors u & are ectors There are two kinds of multiplication of two ectors: - The scalar product dot product) u.. The result is a scalar. - The ector product cross product) u. The result is a ector. ) The dot product In this section we show how to calculate easily the angle between two ectors directly from their components. The dot product is also called inner or scalar products because the product results in scalar not a ector. Def.: The dot product Note: u u u of ectors u u u ) and ) is: u u i i i j j j. j k 0 k k k j u Angle between two ectors The angle between two nonzero ectors u u u ) and ) is gien by u u cos u cos where 0 ) u u 5
Find the angle in the triangle ACB determined by the ertices A 00) B5) and C5) B5) C5) CA 5 ) and CB -) CA CB 5) ) -)) 4 CA 5) ) 9 A CB ) ) cos 4 9 Orthogonal ectors Vectors u u u ) and ) are orthogonal or perpendicular) if and only if u 0 u a) u ) and 46) are orthogonal because u 0 b) u i j k and j 4k are orthogonal because u 0 c) 0 is orthogonal to eery ector u since 0u 000) u u u) 0 Properties of the Dot product If u and w are any ectors and c is a scalar then ) u u ) cu) u c) c u ) ) u w) u u w 4) u u u 5) 0u 0 6
Vector projection Vector projection of u onto proj u u proj u ) "The ector projection of u onto ") Find the ector projection of u 6i j k onto i j k and the scalar component of u in the direction of. Solution: We find proj u from eq.): proj u u u 6 6 4-4 - 4 8 8 i j k) i j k) i j k 4 4 9 9 9 9 We find the scalar component of u in the direction of from eq.): Problems: ) Let u ) and 5). Find the a) component form and b) magnitude length) of the ector.. u 5. 4 u 5 5 ) Find the component form of the ector: a. The ector PQ where P ) and Q-). b. The ector OP where O is the origin and P is the midpoint of segment RS where R -) and S 4). c. The ector from the point A ) to the origin. d. The sum of AB and CD where A -) B 0) C -) and D ) 7
) Let u and w as in the figure: find a) u b) u w c) u and d) u w w u 4) Find the ectors whose lengths and directions are gien. Try to do the calculation without writing: Length a. i Direction b. - k 4 c. j k 5 5 6 d. 7 i j k 7 7 7 5) Find a) the direction of P P and b) the midpoint of line segment P P. a. P 5 ) and P 50) b. P 000) and P ) 6) Find u u the cosine of the angle between and u and the ector proj u. a) i 4 j 5 k u i 4 j 5 k 4 b) ) i ) k u 5i j 5 5 c) i j u i j k d) 5 i j u i 7 j e) u 8
7) Find the angles between the ectors: a) u i j k i 4k b) u i 7 j i j k c) u i j k i j k 8) Find the measures of the angles between the diagonals of the rectangle whose ertices are A 0) B0) C4) and D4) References: - Adanced Engineering Mathematics Erwin Kreyszic)- 8 th Edition. - Calculus Haward Anton). أ). رياض احمد عزت ) Studies - Adanced Mathematics for Engineering 9