11.2.2 Problem Solving using 3D Vectors
Solving problems using 3D vectors involves extending the concepts of vector addition, scalar (dot) products, vector (cross) products, and vector components from 2D to three dimensions. These techniques are useful in physics, engineering, and other fields where quantities like force, velocity, and displacement operate in three-dimensional space.
1. Understanding 3D Vectors
2. Basic Operations with 3D Vectors
a) Vector Addition and Subtraction
If a = ( a x a y a z ) \mathbf{a} = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} a = a x a y a z and b = ( b x b y b z ) \mathbf{b} = \begin{pmatrix} b_x \\ b_y \\ b_z \end{pmatrix} b = b x b y b z , then:
a + b = ( a x + b x a y + b y a z + b z ) \mathbf{a} + \mathbf{b} = \begin{pmatrix} a_x + b_x \\ a_y + b_y \\ a_z + b_z \end{pmatrix} a + b = a x + b x a y + b y a z + b z
a − b = ( a x − b x a y − b y a z − b z ) \mathbf{a} - \mathbf{b} = \begin{pmatrix} a_x - b_x \\ a_y - b_y \\ a_z - b_z \end{pmatrix} a − b = a x − b x a y − b y a z − b z
b) Magnitude of a Vector
The magnitude (length) of a vector v = ( v x v y v z ) \mathbf{v} = \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} v = v x v y v z is given by:
∣ v ∣ = v x 2 + v y 2 + v z 2 |\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} ∣ v ∣ = v x 2 + v y 2 + v z 2
c) Dot Product
The dot product of two vectors a = ( a x a y a z ) \mathbf{a} = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} a = a x a y a z and b = ( b x b y b z ) \mathbf{b} = \begin{pmatrix} b_x \\ b_y \\ b_z \end{pmatrix} b = b x b y b z is:
a ⋅ b = a x b x + a y b y + a z b z \mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z a ⋅ b = a x b x + a y b y + a z b z
This is also equal to:
a ⋅ b = ∣ a ∣ ∣ b ∣ cos θ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta a ⋅ b = ∣ a ∣∣ b ∣ cos θ
where θ \theta θ is the angle between the vectors .
d) Cross Product
The cross product of two vectors a = ( a x a y a z ) \mathbf{a} = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} a = a x a y a z and b = ( b x b y b z ) \mathbf{b} = \begin{pmatrix} b_x \\ b_y \\ b_z \end{pmatrix} b = b x b y b z is:
a × b = ∣ i j k a x a y a z b x b y b z ∣ = ( a y b z − a z b y a z b x − a x b z a x b y − a y b x ) \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} = \begin{pmatrix} a_y b_z - a_z b_y \\ a_z b_x - a_x b_z \\ a_x b_y - a_y b_x \end{pmatrix} a × b = i a x b x j a y b y k a z b z = a y b z − a z b y a z b x − a x b z a x b y − a y b x
The cross product gives a vector that is perpendicular to both a \mathbf{a} a and b \mathbf{b} b , with a magnitude equal to the area of the parallelogram formed by a \mathbf{a} a and b \mathbf{b} b .
3. Problem-Solving Examples
Problem:
Find the angle between the vectors a = ( 3 − 2 4 ) \mathbf{a} = \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix} a = 3 − 2 4 and b = ( 1 4 − 2 ) \mathbf{b} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix} b = 1 4 − 2 .
Solution:
Dot Product:
a ⋅ b = 3 × 1 + ( − 2 ) × 4 + 4 × ( − 2 ) = 3 − 8 − 8 = − 13 \mathbf{a} \cdot \mathbf{b} = 3 \times 1 + (-2) \times 4 + 4 \times (-2) = 3 - 8 - 8 = -13 a ⋅ b = 3 × 1 + ( − 2 ) × 4 + 4 × ( − 2 ) = 3 − 8 − 8 = − 13
Magnitudes:
∣ a ∣ = 3 2 + ( − 2 ) 2 + 4 2 = 9 + 4 + 16 = 29 |\mathbf{a}| = \sqrt{3^2 + (-2)^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29} ∣ a ∣ = 3 2 + ( − 2 ) 2 + 4 2 = 9 + 4 + 16 = 29
∣ b ∣ = 1 2 + 4 2 + ( − 2 ) 2 = 1 + 16 + 4 = 21 |\mathbf{b}| = \sqrt{1^2 + 4^2 + (-2)^2} = \sqrt{1 + 16 + 4} = \sqrt{21} ∣ b ∣ = 1 2 + 4 2 + ( − 2 ) 2 = 1 + 16 + 4 = 21
Angle θ \theta θ :
cos θ = a ⋅ b ∣ a ∣ ∣ b ∣ = − 13 29 ⋅ 21 = − 13 609 \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{-13}{\sqrt{29} \cdot \sqrt{21}} = \frac{-13}{\sqrt{609}} cos θ = ∣ a ∣∣ b ∣ a ⋅ b = 29 ⋅ 21 − 13 = 609 − 13
θ = cos − 1 ( − 13 609 ) ≈ 126.8 7 ∘ \theta = \cos^{-1}\left(\frac{-13}{\sqrt{609}}\right) \approx 126.87^\circ θ = cos − 1 ( 609 − 13 ) ≈ 126.8 7 ∘
The angle between the vectors is approximately 126.87° .
Problem:
Find the area of the parallelogram formed by vectors a = ( 2 3 1 ) \mathbf{a} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} a = 2 3 1 and b = ( 1 − 1 4 ) \mathbf{b} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix} b = 1 − 1 4 .
Solution:
Cross Product:
a × b = ( 3 × 4 − 1 × ( − 1 ) 1 × 1 − 2 × 4 2 × ( − 1 ) − 3 × 1 ) = ( 12 + 1 1 − 8 − 2 − 3 ) = ( 13 − 7 − 5 ) \mathbf{a} \times \mathbf{b} = \begin{pmatrix} 3 \times 4 - 1 \times (-1) \\ 1 \times 1 - 2 \times 4 \\ 2 \times (-1) - 3 \times 1 \end{pmatrix} = \begin{pmatrix} 12 + 1 \\ 1 - 8 \\ -2 - 3 \end{pmatrix} = \begin{pmatrix} 13 \\ -7 \\ -5 \end{pmatrix} a × b = 3 × 4 − 1 × ( − 1 ) 1 × 1 − 2 × 4 2 × ( − 1 ) − 3 × 1 = 12 + 1 1 − 8 − 2 − 3 = 13 − 7 − 5
Magnitude of the Cross Product:
∣ a × b ∣ = 1 3 2 + ( − 7 ) 2 + ( − 5 ) 2 = 169 + 49 + 25 = 243 ≈ 15.59 |\mathbf{a} \times \mathbf{b}| = \sqrt{13^2 + (-7)^2 + (-5)^2} = \sqrt{169 + 49 + 25} = \sqrt{243} \approx 15.59 ∣ a × b ∣ = 1 3 2 + ( − 7 ) 2 + ( − 5 ) 2 = 169 + 49 + 25 = 243 ≈ 15.59
The area of the parallelogram is approximately 15.59 square units .
Problem:
Find the volume of the parallelepiped formed by the vectors a = ( 1 2 3 ) \mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} a = 1 2 3 , b = ( 4 0 1 ) \mathbf{b} = \begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix} b = 4 0 1 , and c = ( 2 − 1 2 ) \mathbf{c} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} c = 2 − 1 2 .
Solution:
Cross Product b × c \mathbf{b} \times \mathbf{c} b × c :
b × c = ( 0 × 2 − 1 × ( − 1 ) 1 × 2 − 4 × 2 4 × ( − 1 ) − 0 × 2 ) = ( 0 + 1 2 − 8 − 4 − 0 ) = ( 1 − 6 − 4 ) \mathbf{b} \times \mathbf{c} = \begin{pmatrix} 0 \times 2 - 1 \times (-1) \\ 1 \times 2 - 4 \times 2 \\ 4 \times (-1) - 0 \times 2 \end{pmatrix} = \begin{pmatrix} 0 + 1 \\ 2 - 8 \\ -4 - 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -6 \\ -4 \end{pmatrix} b × c = 0 × 2 − 1 × ( − 1 ) 1 × 2 − 4 × 2 4 × ( − 1 ) − 0 × 2 = 0 + 1 2 − 8 − 4 − 0 = 1 − 6 − 4
Dot Product a ⋅ ( b × c ) \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) a ⋅ ( b × c ) :
a ⋅ ( b × c ) = 1 × 1 + 2 × ( − 6 ) + 3 × ( − 4 ) = 1 − 12 − 12 = − 23 \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 1 \times 1 + 2 \times (-6) + 3 \times (-4) = 1 - 12 - 12 = -23 a ⋅ ( b × c ) = 1 × 1 + 2 × ( − 6 ) + 3 × ( − 4 ) = 1 − 12 − 12 = − 23
Volume (absolute value):
V = ∣ a ⋅ ( b × c ) ∣ = ∣ − 23 ∣ = 23 cubic units V = | \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) | = |-23| = 23 \text{ cubic units} V = ∣ a ⋅ ( b × c ) ∣ = ∣ − 23∣ = 23 cubic units
The volume of the parallelepiped is 23 cubic units .
Summary