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
Two vectors, with magnitudes not equal to zero, are perpendicular if and only if their scalar product is equal to zero.
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 \( AB \).
Question 5
Find the angle between the lines given by the equations: \( y = 2 x + 4 \) and \( y = x + 3\).