Scalar Product of Vectors

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

    Scalar Product of Vectors

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

    angle between vectors and scalar product
    Fig1. - Angle between vectors and scalar product.
    \[ \vec{A} \cdot \vec{B} = |\vec{A}| \cdot |\vec{B}| \cdot \cos \theta \] where \( |\vec{A}| \) is the magnitude of vector \( \vec{A} \), \( |\vec{B}| \) is the magnitude of vector \( \vec{B} \) and \( \theta \) is the angle made by the two vectors. The result of a scalar product of two vectors is a scalar quantity.
    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
    \[ \cos \theta = \dfrac{\vec{A} \cdot \vec{B}}{|\vec{A}| \cdot |\vec{B}|} \]


    Properties of the Scalar Product

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


    Questions and Applications the Scalar Product

    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 \( AB \).

    tangent to circle problem
    Fig1. - Tangent to circle.

     Solution to Question 4
    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 \( y = x + 3\).

    Angle Between two Lines
    Figure 1.Angle Between two Lines .

     Solution to Question 5
    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 \( \vec{V2} \)
    \( |\vec{V1}| = \sqrt{1^2 + 2^2} = \sqrt{5} \)
    \( |\vec{V2}| = \sqrt{1^2 + 1^2} = \sqrt{2} \)
    \( \theta = \arccos \left(\dfrac{3}{\sqrt{5} \sqrt{2}}\right) \approx 18.43^{\circ}\)

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