The definition of parallel and perpendcicular vectors are presented along with questions and detailed solutions. The questions involve finding vectors given their initial and final points, scalar product of vectors and other concepts that can all be among the formulas for vectors

Two vectors \( \vec{A} \) and \( \vec{B} \) are parallel if and only if they are scalar multiples of one another:

\[ \vec{A} = k \; \vec{B} \]
where \( k \) is a constant not equal to zero.

Let \( \vec{A} = \; < A_x \;,\; A_y > \) and \( \vec{B} = \; < B_x \;,\; B_y > \)

Vectors \( \vec{A} \) and \( \vec{B} \) are parallel if and only if \( \vec{A} = k \vec{B} \)

Using the components, we rewrite the above as
\[ < A_x \;,\; A_y > \; = \; k < B_x \;,\; B_y > \; = \; < k B_x \;,\; k B_y > \]

\( A_x = k B_x \) and \( A_y = k B_y \)

The above can also be written as

\( \dfrac{A_x}{B_x} = k \) and \( \dfrac{A_y}{B_y} = k \)

Conditions under which vectors \( \vec{A} = \; < A_x \;,\; A_y > \) and \( \vec{B} = \; < B_x \;,\; B_y > \) are parallel are given by

\[ \dfrac{A_x}{B_x} = k \; \text{and} \; \dfrac{A_y}{B_y} = k \]
or
\[ \dfrac{A_x}{B_x} = \dfrac{A_y}{B_y} \]
or
\[ A_x \cdot B_y = A_y \cdot B_x \]

Two vectors \( \vec{A} \) and \( \vec{B} \) are perpendicular if and only if their scalar product is equal to zero.

\( \vec{A} = \; < A_x \;,\; A_y > \) and \( \vec{B} = \; < B_x \;,\; B_y > \)

Vectors \( \vec{A} \) and \( \vec{B} \) are perpendicular if and only if \( \vec{A} \cdot \vec{B} \) = 0

Using the components, we write

\( \vec{A} \cdot \vec{B} = A_x \cdot B_x + A_y \cdot B_y\)

Hence vectors \( \vec{A} \) and \( \vec{B} \) are perpendicular if and only if
\[ A_x \cdot B_x + A_y \cdot B_y = 0 \]

Which of the following vectors are parallel?

\( \vec{A} = \; < 2 \;,\; -3 > \) , \( \vec{B} = \; < -6 \;,\; 9 > \) , \( \vec{C} = \; < -1 \;,\; -2 > \) ?

__Solution to Question 1__

The condition for two vectors \( \vec{A} = \; < A_x \;,\; A_y > \) and \( \vec{B} = \; < B_x \;,\; B_y > \) to be parallel is:
\[ A_x \cdot B_y = A_y \cdot B_x \]

Let us test vectors \( \vec{A} \) and \( \vec{B} \) first.

\( A_x \cdot B_y = (2)(9) = 18 \)

\( A_y \cdot B_x = (-6)(-3) = 18 \)

Hence, \( A_x \cdot B_y = 18 = A_y \cdot B_x \) and therefore vectors \( \vec{A} \) and \( \vec{B} \) are parallel.

We now test vectors \( \vec{A} \) and \( \vec{C} \)

The condition for the \( \vec{A} = \; < A_x \;,\; A_y > \) and \( \vec{C} = \; < C_x \;,\; C_y > \) to be parallel is:
\[ A_x \cdot C_y = A_y \cdot C_x \]

\( A_x \cdot C_y = (2)(-2) = -4 \)

\( A_y \cdot C_x = (-3)(-1) = 3 \)

We can see that \( A_x \cdot C_y \ne A_y \cdot C_x \)
and therefore vectors \( \vec{A} \) and \( \vec{C} \) are not parallel.

Vectors \( \vec{B} \) and \( \vec{C} \) are not parallel (there no need to test since \( \vec{A} \) and \( \vec{B} \) are parallel)

__Question 2__

Find the real number \( a \) so that the vectors \( \vec{A} = \; < 2a \;,\; 16> \) and \( \vec{B} = \;< 3 a+2 \;, \;-2 > \) are perpendicular

__Solution to Question 2__

The condition for two vectors \( \vec{A} = \; < A_x \;,\; A_y > \) and \( \vec{B} = \; < B_x \;,\; B_y > \) to be perpendicular is:
\[ A_x \cdot B_x + A_y \cdot B_y = 0 \]

Rewrite the above condition using the components of vectors, we obtain the equation

\( 2a(3a + 2)+ 16(-2) = 0 \)

Expand and rearrange to obtain the quadratic equation

\( 3 a^2 + 2 a - 16 = 0 \)

Solve the equation to find

\( a = 2 \) and \( a = -8 / 3 \)

The vectors \( \vec{A} = < 2a , 16> \) and \( \vec{B} = < 3 a + 2 , -2 > \) are perpendicular for \( a = 2 \) and \( a = - 8 / 3 \).

__Question 3__

Find the real number \( k \) so that the points \( A(-2 \;,\; k), B(2 \;,\; 3) \text{ and } C(2k \;,\; -4) \) are the vertices of a right triangle with right angle at point \( B \).

__Solution to Question 3__

\( ABC \) is a right triangle at \( B \) if and only if vectors \( \vec{BA} \) and \( \vec{BC} \) 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 \( \vec{BA} \) and \( \vec{BC} \) given the coordinates of the three points.

\( \vec{BA} = < -2 - 2 ,\;,\; k - 3 > = < -4 \;,\; k - 3 > \)

\( \vec{BC} = < 2 k - 2 \;,\; -4 - 3 > = < 2 k - 2 ,\;,\; -7 > \)

The scalar product \( \vec{BA} \cdot \vec{BC} = 0 \) 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 \( AB \) where \( A(2 \;,\; -2) \) and \( B(4 \;,\; -3) \).

__Solution to Question 4__

According to Thales theorem , for point \( M(x \;,\; y) \) to be on the circle defined by its diameter \( AB \), 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 \( \vec{MA} \cdot \vec{MB} \) is equal to zero.

Let us first find the components of vectors \( \vec{MA} \) and \( \vec{MB} given the coordinates of the three points \( A(2 \;,\; -2) \), \( B(4 \;,\; -3) \) and \( M(x \;,\; y) \).

\( \vec{MA} = < 2 - x \;,\; -2 - y > \)

\( \vec{MB} = < 4 - x \;,\; -3 - y > \)

The scalar product \( \vec{MA} \cdot \vec{MB} = 0 \) 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 \( \vec{u} = < 2 \;,\; -5> \), find.

a) the equation of the line through point \( A(1 \;,\; 1) \) and parallel to vector \( \vec{u} \).

b) the equation of the line through point \( B(-2 \;,\; -3) \) and perpendicular to vector \( \vec{u} \).

__Solution to Question 5__

a)

Point \( M(x \;,\; y) \) is on the line through point \( A(1 \;,\; 1) \) and parallel to vector \( \vec{u} = < 2 \;,\; -5> \) if and only if the vectors \( \vec{AM} \) and \( \vec{u} \) are parallel.

Let us first find the components of vectors AM.

\( \vec{AM} = \; < x - 1 \;,\; y - 1 > \)

Vectors \( \vec{AM} = \; < x - 1 \;,\; y - 1 > \) and \( \vec{u} = < 2 \;,\; -5> \) 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 \( \vec{u} = < 2 , -5> \) if and only if the vectors \( \vec{BM} \) and \( \vec{u} \) are perpendicular.

Let us first find the components of vectors \( \vec{BM} \).

\( \vec{BM} = < x - (-2) , y - (-3) > = < x + 2 , y + 3 > \)

Vectors \( \vec{BM} = < x + 2 , y + 3 > \) and \( \vec{u} = < 2 , -5> \) 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).

1) Vectors \( \vec{TD} \) and \( \vec{TC} \) are perpendicular

2) The magnitude (or length) of vector \( \vec{TC} \) is equal to the radius.

Let \( a \) and \( b \) be the x and y coordinates of point \( T \).

Vectors \( \vec{TD} \) and \( \vec{TC} \) are given by their components as follows:

\( \vec{TD} = < 2 - a \;,\; 4 - b > \)

\( \vec{TC} = < 0 - a \;,\; 0 - b > \)

We now use condition 1) above: Two vectors are perpendicular if and only if their scalar product is equal to zero. Hence

\( < 2 - a \;,\; 4 - b > \cdot < 0 - a \;,\; 0 - b > = 0 \)

Calculate the dot (scalar) product using the coordinates

\( (2 - a) ( - a) + (4 - b) (-b) = 0 \)

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 of vector \( \vec{TC} \) to be equal to the square of the radius.

\( (0 - a)^2 + (0 - b)^2 = 2^2 \)

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

\( a^2 + b^2 = 4 \)

found above 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

\( 4 - 8b + 4 b^2 + b^2 = 4 \)

\( - 8 b + 5 b^2 = 0 \)

Factor: \( b(- 8 + 5 b) = 0 \)

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

\( T_1 = (2 \; , \; 0) \) and \( T_2 = ( - 6 / 5 \; , \; 8 / 5 ) \)

The equations of the tangents are: (find equation of line through two points \( T \) and \( D \))

Tangent T1:

tangent through the points \( T_1 = (2 \; , \; 0) \) and \( D(2 \;,\; 4) \) is given by

\( x = 2 \)

Tangent T2:

tangent through the points \( T_2 = ( - 6 / 5 \; , \; 8 / 5 ) \) and \( D(2 \;,\; 4) \) is given by

\( - 3 x + 4 y = 10 \)