Parallel Vectors
Two vectors A and B are parallel if and only if they are scalar multiples of one another.
A = kB , k is a constant not equal to zero.
Let A = (Ax , Ay) and B = (Bx , By)
A and B are parallel if and only if A = kB
(Ax , Ay) = k (Bx , By) = (k Ax , k By)
Ax = k Bx and Ay = k By or Ax / Bx = k and Ay / By = k
Condition under which vectors A = (Ax , Ay) and B = (Bx , By) are parallel is given by
Ax / Bx = Ay / By or Ax By = Bx Ay
Perpendicular Vectors
Two vectors A and B are perpendicular if and only if their scalar product is equal to zero.
Let A = (Ax , Ay) and B = (Bx , By)
Vectors A and B are perpendicular if and only if A·B = 0
(Ax , Ay) · (Bx , By) = Ax Bx + Ay By
Hence vectors A and B are perpendicular if and only if
Ax Bx + Ay By = 0
Question 1
Which of the following vectors are parallel?
A = (2 , -3) , B = (-6 , 9) , C = (-1 , -2)?
Solution to Question 1
The condition for two vectors A = (Ax , Ay) and B = (Bx , By) to be parallel is: Ax By = Bx Ay.
Let us test vectors A and B first.
Ax By = (2)(9) = 18
Bx Ay = (-3)(-6) = 18
Vectors A and B are parallel.
We now test vectors A and C
Ax Cy = (2)(-2) = -4
Cx Ay = (-1)(-3) = 3
Vectors A and C are not parallel.
Vectors B and C are not parallel (there no need to test since A and B are parallel)
Question 2
Find the real number a so that the vectors A = (2a , 16) and B = (3a+2 , -2) are perpendicular
Solution to Question 2
The condition for two vectors A = (Ax , Ay) and B = (Bx , By) to be perpendicular is: Ax Bx + Ay By=0
Rewrite the above condition using the components of vectors, we obtain the equation
2a(3a + 2)+ 16(-3) = 0
Expand and rearrange to obtain the quadratic equation
3 a 2 + 2 a - 16 = 0
Solve the equation to find
a = 2 and a = -8 / 3
Question 3
Find the real number k so that the points A(-2 , k), B(2 , 3) and C(2k , -4) are the vertices of a right triangle with right angle at B.
Solution to Question 3
ABC is a right triangle at B if and only if vectors BA and BC are perpendicular. And two vectors are perpendicular if and only if their scalar product is equal to zero.
Let us first find the components of vectors BA and BC given the coordinates of the three points.
BA = (-2 - 2 , k - 3 ) = (-4 , k - 3)
BC = (2 k - 2 , -4 - 3 ) = (2 k - 2 , -7)
The scalar product BA·BC = 0 is written using the components of the two vectors
(-4)(2k - 2) + (k - 3)(-7) = 0
Expand, simplify and solve for k
k = - 13
Question 4
Find the equation of the circle with diameter the points A(2 , -2) and B(4 , -3).
Solution to Question 4
For a point M(x , y) to be on the circle defined by its diameter, triangle AMC must be a right triangle with the right angle at M. Triangle AMC is right at point M if and only if the scalar product MA·MB
is equal to zero.
Let us first find the components of vectors MA and MB given the coordinates of the three points.
MA = (2 - x , -2 - y )
MB = (4 - x , -3 - y)
The scalar product MA·MB = 0 is written using the components of the two vectors
(2 - x)(4 - x) + (-2 - y)(-3 - y) = 0
Expand and simplify to obtain the equation of the circle
x 2 - 6 x + y 2 + 5y + 14 = 0
Question 5
Given vector U = (2 , -5), find.
a) the equation of the line through point A(1 , 1) and parallel to vector U.
b) the equation of the line through point B(-2 , -3) and perpendicular to vector U.
Solution to Question 5
a) A point M(x , y) is on the line through point A(1 , 1) and parallel to vector U = (2 , -5) if and only if the vectors AM and U are parallel.
Let us first find the components of vectors AM.
AM = (x - 1 , y - 1)
Vectors AM = (x - 1 , y - 1) and U = (2 , -5) are parallel if and only if
(x - 1) (-5) = (2)(y - 1)
Expand and simplify to obtain the equation of the line
5 x + 2 y = 3
b) A point M(x , y) is on the line through point B(-2 , -3) and perpendicular to vector U = (2 , -5) if and only if the vectors BM and U are perpendicular.
Let us first find the components of vectors BM.
BM = (x - (-2) , y - (-3)) = (x + 2 , y + 3)
Vectors BM = (x + 2 , y + 3) and U = (2 , -5) are perpendiclur if and only if
(x + 2) (2) + (y + 3)(-5) = 0
Expand and simplify to obtain the equation of the line
2 x - 5 y = 11
Question 6
Find the equation of the tangents through the point D(2 , 4) to the circle of center C(0 , 0) and radius 2.
Solution to Question 6
From point D outside the circle, two tangent through D to the the circle of center C may be found(see figure 1 below).
Fig1. - Tangent through a point to a circle.
Two conditions for point T to be the point of tangency:
1) Vector TD and vector TC are perpendicular
2) The magnitude (or length) of vector TC is equal to the radius.
If a and b are the x and y coordinates of point T, then vectors TD and TC are given by their components as follows:
TD = (2 - a ,4 - b)
TC = (0 - a ,0 - b)
We now use condition 1) above: Two vectors are perpendicular if and only if their scalar product is equal to zero. Hence
(2 - a ,4 - b) · (0 - a ,0 - b) = 0
Expand and simply and rewrite equation as
a 2 + b 2 - 2 a - 4 b = 0
We next use condition 2) above using the square of the magnitude.
(0 - a) 2 + (0 - b) 2 = 2 2
Which may be simplified to
a 2 + b 2 = 4
We now need to solve the two equations, given by the two conditions, a 2 + b 2 - 2 a - 4 b = 0 and a 2 + b 2 = 4 simultaneously. Substitute a 2 + b 2 by 4 in the first equation to obtain a new equation
4 - 2 a - 4 b = 0
which may be written (dividing all terms by 2) as
2 - a - 2 b = 0
Solve the above for a to get
a = 2 - 2 b
We now substitute a by 2 - 2 b in the equation a 2 + b 2 = 4 to obtain
(2 - 2 b) 2 + b 2 = 4
Expand and simplify
4 - 8b + 4 b 2 + b 2 = 4
- 8 b + 5 b 2 = 0
Solve for b, two solutions:
b = 0 and b = 8 / 5
Use the equation a = 2 - 2 b to find the corresponding values of a
for b = 0 , a = 2
for b = 8 / 5 , a = -6 / 5
The two points of tangency are
T1 = (2 , 0) and T2 = (8 / 5 , - 6 / 5)
The equations of the tangents are: (find equation of line through two points)
