where \( || \vec{u} || \) is the magnitude of vector \( \vec{u} \), \( || \vec{v} || \) is the magnitude of vector \( \vec{v} \) and \( \theta \) is the angle between the vectors \( \vec{u} \) and \( \vec{v} \).
If the components of vectors \( \vec{u} \) and \( \vec{v} \) are known: \( \vec{u} = (u_x , u_y ,u_z)\) and \( \vec{v} = (v_x , v_y , v_z) \), it can be shown that the scalar product may be expressed as follows:
Find \( a \) so that the vectors \( \lt a,-6,3 \gt \) and \( <1,0,-2> \) are perpendicular.
solution
For two vectors to be perpendicular, their scalar product must be equal to zero.
\( \lt a,-6,3 \gt \cdot <1,0,-2> = a(1) + (-6)(0)+(3)(-2) = a - 6 = 0 \)
Solve for a
\( a = 6\)
Scalar and Vector Projection of a Vector onto Another
In many applications, it is important to find the component of a vector in the direction of another vector. As shown below, vector \( \vec{u}\) is projected onto vector \( \vec{v}\) by dropping a perpendicular from the terminal point of \( \vec{u}\) to the line through \( \vec{v}\). The component of \( \vec{u}\) along \( \vec{v}\) is a scalar quantity called the scalar projection and is given by
.
The vector projection of \( \vec{u}\) on \( \vec{v}\) is a vector quantity obtained by multiplying the component \( \text{comp}_{\vec{v}}\vec{u} \) by the unit vector in the direction of vector \( \vec{v}\) and is given by
The cross product \( \vec{u} \times \vec{v} \) is perpendicular to both \( \vec{v} \) and \( \vec{u} \)
The right hand rule, to find the direction of the cross product, is as follows: point the index in the direction of \( \vec {u} \), the middle finger in the direction of \( \vec{v} \) and the direction of the cross product \( \vec {u} \times \vec {v} \) is in the same direction as that of the thumb.
Theorems on Cross Products
If \( \vec{u} \), \( \vec{v} \) and \( \vec{w} \) are vectors and k is a scalar, then
1) The cross product \( \vec{u} \times \vec{v} \) is perpendicular to both \( \vec{v} \) and \( \vec{u} \)
2) \( \vec{u} \times \vec{v} = - \vec{v} \times \vec{u} \)
3) \( \vec{u} \times \vec{v} = 0 \) if and only if \( \vec{u} \) and \( \vec{v} \) are parallel , if both \( \vec{u} \) and \( \vec{v} \) are non zero vectors.
4) \( \vec{u} \times (\ \vec{v} + \vec{w} ) = \vec{v} \times \vec{u} + \vec{u} \times \vec{w} \)
5) \( (k \vec{u}) \times \vec{v} = \vec{u} \times ( k\vec{v}) = k ( \vec{u} \times \vec{v} ) \)
6) \( ||\vec{u} \times \vec{v} || = ||\vec{u}|| ||\vec{v}|| sin \theta\) , where \( \theta \) is the angle between \( \vec{u}\) and \( \vec{v} \).
Area of a Parallelogram
A parallelogram is a quadrilateral (4 sides) with opposite sides parallel. In the figure below is shown the parallelogram A, B, C and D. Hence, we have equality between the vectors.
\( \vec{AB} = \vec{DC} \) and \( \vec{AD} = \vec{BC} \)
The area of the parallelagram is given by \( || \vec{AB} \times \vec{AD} || \)
The area of a triangle may be calculated as half the area the corresponding parallelogram.
Volume of a Parallelepiped
A parallelepiped is a 3d figure formed by 6 parallelograms as shown in the figure below. We have equality between several vectors.
\( \vec{AE} = \vec{DH} = \vec{CG} = \vec{BF} = \vec{u} \)
\( \vec{AD} = \vec{BC} = \vec{EH} = \vec{FG} = \vec{v}\)
\( \vec{AB} = \vec{DC} = \vec{EF} = \vec{HG} = \vec{w}\)
The volume V of the parallelepiped is given by
V \( = |\vec{u}\cdot (\vec{v} \times \vec{w})| = | \vec{v}\cdot (\vec{w} \times \vec{u})| = | \vec{w}\cdot (\vec{v} \times \vec{u})| \)
Example 3
Calculate the cross product of the vectors \(\vec{u} = \lt 1,1,3 >\) and \(\vec{v} = \lt 1,0,2 > \).
Find two unit vectors perpendicular to the vectors \( \vec{u} = \lt 1,-2,1 \gt \) and \( \vec{v} = \lt -2,0,4> \).
solution
The cross product \( \vec{w} = \vec{u} \times \vec{v} \) is a vector perpendicular to both vectors \( \vec{u} \; \text{and} \; \vec{v} \).
Let us calculte \( \vec{u} \times \vec{v} \) as follows:
We now need to find a unit vector \( \vec{u_1} \) in the same direction as \( \vec{w} \) and is given by
\( \vec{u_1} = \dfrac{1}{||\vec{w}||} \vec{w}\)
and a second unit vector \( \vec{u_2} \) in the opposite direction of \( \vec{w} \) and is given by
\( \vec{u_2} = -\vec{u_1}\)
\( ||\vec{w}|| = \sqrt{(-8)^2+(- 6)^2+(- 4)^2 } = 2\sqrt{29}\)
\( \vec{u_1} = \dfrac{1}{2\sqrt{29}} (-8\vec{i} - 6 \vec{j} - 4 \vec{k}) = -\dfrac{4}{\sqrt{29}}\vec{i} -\dfrac{3}{\sqrt{29}}\vec{j}-\dfrac{2}{\sqrt{29}}\vec{k}\)
\( \vec{u_2} = \dfrac{4}{\sqrt{29}}\vec{i} + \dfrac{3}{\sqrt{29}}\vec{j}+ \dfrac{2}{\sqrt{29}}\vec{k}\)
Example 5
Explain why the following statement are not true.
a) \( \vec{u} \times \vec{u} = ||\vec{u}||^2\)
b) \( \vec{u} \cdot (\vec{u} \times \vec{w} )= (\vec{u} \cdot \vec{u}) \times \vec{w} \)
solution
a) The left side \( \vec{u} \times \vec{u} \) is a cross product and the result is a vector. The right side \( ||\vec{u}||^2\) is a scalar quantity. A vector and a scalar cannot be compared.
b) The left side \( \vec{u} \cdot (\vec{u} \times \vec{w} ) \) is a scalar product of \( \vec{u} \) and \( (\vec{u} \times \vec{w} ) \) and the result is a scalar. The right side is the product of a scalar quantity \( \vec{u} \cdot \vec{u} \) and vector \( \vec{w} \) and the result is a vector. A scalar and a vector cannot be compared.
Answer the following Questions
Detailed Solutions and explanations to these questions.
1) Calculate \( \vec{u} \cdot (\vec{u} \times \vec{v}) \) given that \( \vec{u} = \lt a,b,c \gt \) and \( \vec{v} = \lt d,e,f \gt \).
2) Find \( k \) so that vectors \( \vec{u} = \lt -2,-k,1 \gt \) and \( \vec{v} = \lt 8,-2,-3 > \) are perpendicular
3)Find \( k \) so that the vectors \( \vec{u} = \lt -3,2,-2 \gt \), \( \vec{v} = \lt 2,1,k> \) and \( \vec{w} = \lt -1,3,-5> \) are on the same plane (or coplanar)?
4) Find angle \( \theta \) between the vectors \( \vec{u} = \lt 2,0,1 \gt \) and \( \vec{v} = \lt 8,-2,-3 > \).
5) Find the vector projection of \( \vec{u} = \lt -1,-1,1 \gt \) onto \( \vec{v} = \lt 2,1,1 > \).
6) Find that \( k \) so that the points \( A(-1,2,k) \), \( B(-3,6,3) \) and \( C(1,3,6) \) are the vertices of a right triangle with a right angle at \( A \).
7) Given vector \( \vec{v} = \lt 3,-1,-2 \gt \), find the vector \( \vec{u} \) such that \( \vec{v} \times \vec{u} = \lt 4,2,5 > \) and \( ||\vec{u}|| = 3\))
8) Points A, B, C and D forms a parallelogram.
a)Find the coordinates of point D.
b)Find the area of a parallelogram.
9) In the cube below find the angle between the diagonals AG and BH.
10) Find a vector that is orthogonal to the plane containing the points A(1,2,-3), B(0,-2,1) and C(-2,0,1).
11) Find the area of the triangle whose vertices are the points A(1,0,-3), B(1,-2,0) and C(0,2,1).
12)Find the volume of the parallelepiped shown below.
