The scalar product of two vectors is preseted alog with questions and detailed solutions. Some of the formulas for vectors are also used.

\( \) \( \)\( \) \( \)
The scalar product (also called the dot product and inner product) of vectors \( \vec{A} \) and \( \vec{B} \) is written and defined as follows

For vectors given by their components: \( \vec{A} = < A_x , A_y, A_z > \) and \( \vec{B} = < B_x , B_y, B_z > \), the scalar product is given by \[ \vec{A} \cdot \vec{B} = A_x \cdot B_x + A_y \cdot B_y + A_z \cdot B_z \]

Note that if \( \theta = 90^{\circ} \) , then \( \cos(\theta) = 0 \)

We therefore we can state that:

Two vectors, with magnitudes not equal to zero, are perpendicular if and only if their scalar product is equal to zero.

The scalar product may also be used to find the cosine and therefore the angle between two vectors

\[ \cos \theta = \dfrac{\vec{A} \cdot \vec{B}}{|\vec{A}| \cdot |\vec{B}|} \]

1) \( \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A} \)

2) \( \vec{A} \cdot (\vec{B} + \vec{C} ) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} \)

__Question 1__

Find the real number \( b \) so that vectors \( \vec{A} \) and \( \vec{B} \) given by their components below are perpendicular

\( \vec{A} = < -2 , -b > \) , \( \vec{B} = < -8 , b > \).

__Solution to Question 1__

The condition for two vectors \( \vec{A} = < Ax , Ay > \) and \( \vec{B} = < Bx , By > \) to be perpendicular is that their scalar product is equal to zero:

\[ A_x \cdot B_x + A_y \cdot B_y = 0 \]

Substitute the components by their values and simplify

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

\( 16 - b^2 = 0 \)

\( b^2 = 16 \)

Solve for \( b \) to find the solutions:

\( b = 4 \) and \( b = -4 \)

Two values \( b = 4\) and \( b = - 4\) make the vectors \( \vec{A} = < -2 , -b > \) and \( \vec{B} = < -8 , b > \) perpendicular.

__Question 2__

Find the angle made by the vectors A and B given below

\( \vec{A} = < 2 , 1 , 3 > \) , \( \vec{B} = < 3 , -2 , 1 > \).

__Solution to Question 2__

We first use the components to find the scalar product of the two vectors.

\( \vec{A} \cdot \vec{B} = (2)(3)+(1)(-2)+(3)(1) = 7 \)

We next express the scalar product using the magnitudes and angle θ made by the two vectors.

\( \vec{A} \cdot \vec{B} = |\vec{A}| \cdot |\vec{B}| \cdot \cos \theta = 7 \)

Which gives

\( \cos \theta = \dfrac{7}{|\vec{A}| \cdot |\vec{B}|} \)

Calculate the magnitudes \( |\vec{A}| \) and \( |\vec{B}| \)

\( |\vec{A}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{14} \)

\( |\vec{B}| = = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14} \)

\( \cos \theta = \dfrac{7}{\sqrt{14} \cdot \sqrt{14}} \)

Simplify

\( \cos \theta = \dfrac{7}{14} = 1/2\)

The angle made by the given vector is : \[ \theta = \arccos(1/2) = 60^{\circ} \]

__Question 3__

Given vector \( \vec{U} = < 3 , -7 > \), find the equation of the line through point \( B(2 , 1) \) and perpendicular to vector \( \vec{U} \).

__Solution to Question 3__

A point \( M(x , y) \) is on the line through point \( B(2 , 1) \) and perpendicular to vector \( \vec{U} = < 3 , -7 > \) if and only if the vectors \( \vec{BM} \) and \( \vec{U} \) are perpendicular.

Let us first find the components of vectors BM.

\( \vec{BM} = < x - 2 , y - 1 > \)

Vectors \( \vec{BM} = < x - 2 , y - 1 > \) and \( \vec{U} = < 3 , -7 > \) are perpendiclur if and only if their scalar product is equal to zero. Hence

\( (x - 2) (3) + (y - 1)(-7) = 0 \)

Expand and simplify to obtain the equation of the line through point \( B(2 , 1) and perpendicular to vector \( \vec{U} \)

\[ 3 x - 7 y = - 1 \]

__Question 4__

Given points \( A(1 , 2) \) and \( B(-2 , -2) \), find the equation of the tangent at point \( B \) to the circle with diameter \( AB \).

A tangent to a circle at point \( B \) is perpendicular to segment \( BC \) where \( C \) is the center of the circle (see figure on the right). Any point \( M(x , y) \) on the tangent is such that the scalar product \( \vec{BM} \cdot \vec{BC} \) is equal to zero.

Point \( C \) is the center of the circle and therefore the midpoint of \( A \) and \( B \) . Its coordinates are given by

\( C ( \dfrac{1+(-2)}{2} \; , \; \dfrac{2 + (-2)}{2} ) = C(- \dfrac{1}{2} \; , \; 0) \)

Vectors \( \vec{BM} \) and \( \vec{BC} \) are defined by points \( B \), \( M \) and \( C \) and their components are given by:

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

\( \vec{BC} \; = \; < - \dfrac{1}{2} - (-2) , 0 - (-2)> \; = \; < \dfrac{3}{2} , 2> \)

We now use the fact that the scalar product is equal to zero.

\( \vec{BM} \cdot \vec{BC} \; = \; (x + 2)\dfrac{3}{2} + (y + 2)(2) = 0 \)

Expand and simplify to find the equation of the tangent.

\( \dfrac{3}{2} x + 2 y = - 7 \)

__Question 5__

Find the angle between the lines given by the equations: \( y = 2 x + 4 \) and \( y = x + 3\).

Let L1 be the line with equation \( y = 2 x + 4 \) and line L2 the line with equation \( y = x + 3\).

First find the point of intersection by solving the system of equations: \[ y = 2 x + 4 \text{ and } y = x + 3 \]

The point of intersection is at \( (-1 , 2) \)

We now find the y-intercepts of the two lines

For line L1 the y-intercept is \( (0 , 4) \) and for L2 the y-intercept is \( (0 , 3) \)

We now find two vectors V1 and V2 parallel to L1 and L2 respectively.

\( \vec{V1} = < 0 - (-1) , 4 - 2 > = < 1 , 2 > \)

\( \vec{V2} = < 0 - (-1) , 3 - 2 > = < 1 , 1 > \)

We now calculate the angle \( \theta \) between the lines given by their equations.

\( \theta = \arccos (\dfrac{\vec{V1} \cdot \vec{V2}}{\vec{V1} \cdot \vec{V2}}) \)

The dot product \( \vec{V1} \cdot \vec{V2} \) is calculate using the coordinates

\( \vec{V1} \cdot \vec{V2} = < 1 , 2 > \cdot < 1 , 1 > = 3 \)

The magnitudes of \( \vec{V1} \) and \( \vec{V2} \)

\( |\vec{V1}| = \sqrt{1^2 + 2^2} = \sqrt{5} \)

\( |\vec{V2}| = \sqrt{1^2 + 1^2} = \sqrt{2} \)

\( \theta = \arccos \left(\dfrac{3}{\sqrt{5} \sqrt{2}}\right) \approx 18.43^{\circ}\)

- Introduction to Vectors
- Formulas for Vectors
- Parallel and Perpendicular Vectors with questions (some of which may be challenging) and detailed solutions.