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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

infoNote

A 3D3D vector v\mathbf{v} is represented as:

v=(vxvyvz)=vxi+vyj+vzk\mathbf{v} = \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = v_x \mathbf{i} + v_y \mathbf{j} + v_z \mathbf{k}

where:

  • vx,vy,vzv_x, v_y, v_z are the components along the xx -, y y -, and zz -axes, respectively.
  • i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} are the unit vectors along the xx -, yy -, and z z -axes.

2. Basic Operations with 3D Vectors

a) Vector Addition and Subtraction

If a=(axayaz)\mathbf{a} = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} and b=(bxbybz)\mathbf{b} = \begin{pmatrix} b_x \\ b_y \\ b_z \end{pmatrix}, then:

a+b=(ax+bxay+byaz+bz)\mathbf{a} + \mathbf{b} = \begin{pmatrix} a_x + b_x \\ a_y + b_y \\ a_z + b_z \end{pmatrix}

ab=(axbxaybyazbz)\mathbf{a} - \mathbf{b} = \begin{pmatrix} a_x - b_x \\ a_y - b_y \\ a_z - b_z \end{pmatrix}

b) Magnitude of a Vector

The magnitude (length) of a vector v=(vxvyvz)\mathbf{v} = \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} is given by:

v=vx2+vy2+vz2|\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}

c) Dot Product

The dot product of two vectors a=(axayaz)\mathbf{a} = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} and b=(bxbybz)\mathbf{b} = \begin{pmatrix} b_x \\ b_y \\ b_z \end{pmatrix} is:

ab=axbx+ayby+azbz\mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z

This is also equal to:

ab=abcosθ\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta

where θ\theta is the angle between the vectors.

d) Cross Product

The cross product of two vectors a=(axayaz)\mathbf{a} = \begin{pmatrix} a_x \\ a_y \\ a_z \end{pmatrix} and b=(bxbybz)\mathbf{b} = \begin{pmatrix} b_x \\ b_y \\ b_z \end{pmatrix} is:

a×b=ijkaxayazbxbybz=(aybzazbyazbxaxbzaxbyaybx)\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}

The cross product gives a vector that is perpendicular to both a\mathbf{a} and b\mathbf{b}, with a magnitude equal to the area of the parallelogram formed by a\mathbf{a} and b\mathbf{b}.


3. Problem-Solving Examples

infoNote

Question : a) Finding the Angle Between Two Vectors

Problem: Find the angle between the vectors a=(324)\mathbf{a} = \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix} and b=(142)\mathbf{b} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}.

Solution:

  1. Dot Product:

ab=3×1+(2)×4+4×(2)=388=:highlight[13]\mathbf{a} \cdot \mathbf{b} = 3 \times 1 + (-2) \times 4 + 4 \times (-2) = 3 - 8 - 8 = :highlight[-13]

  1. Magnitudes:

a=32+(2)2+42=9+4+16=:highlight[29]|\mathbf{a}| = \sqrt{3^2 + (-2)^2 + 4^2} = \sqrt{9 + 4 + 16} = :highlight[\sqrt{29}]

b=12+42+(2)2=1+16+4=:highlight[21]|\mathbf{b}| = \sqrt{1^2 + 4^2 + (-2)^2} = \sqrt{1 + 16 + 4} = :highlight[\sqrt{21}]

  1. Angle θ\theta :

cosθ=abab=132921=13609\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{-13}{\sqrt{29} \cdot \sqrt{21}} = \frac{-13}{\sqrt{609}}

θ=cos1(13609):highlight[126.87]\theta = \cos^{-1}\left(\frac{-13}{\sqrt{609}}\right) \approx :highlight[126.87^\circ]

The angle between the vectors is approximately 126.87°.

infoNote

Question : b) Finding the Area of a Parallelogram Formed by Two Vectors

Problem: Find the area of the parallelogram formed by vectors a=(231)\mathbf{a} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} and b=(114)\mathbf{b} = \begin{pmatrix} 1 \\ -1 \\ 4 \end{pmatrix}.

Solution:

  1. Cross Product:

a×b=(3×41×(1)1×12×42×(1)3×1)=(12+11823)=(:highlight[13]:highlight[7]:highlight[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} :highlight[13] \\ :highlight[-7] \\ :highlight[-5] \end{pmatrix}

  1. Magnitude of the Cross Product:

a×b=132+(7)2+(5)2=169+49+25=243:highlight[15.59]|\mathbf{a} \times \mathbf{b}| = \sqrt{13^2 + (-7)^2 + (-5)^2} = \sqrt{169 + 49 + 25} = \sqrt{243} \approx :highlight[15.59]

The area of the parallelogram is approximately 15.59 square units.

infoNote

Question : c) Finding the Volume of a Parallelepiped Formed by Three Vectors

Problem: Find the volume of the parallelepiped formed by the vectors a=(123)\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, b=(401)\mathbf{b} = \begin{pmatrix} 4 \\ 0 \\ 1 \end{pmatrix}, and c=(212)\mathbf{c} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}.

Solution:

  1. Cross Product b×c\mathbf{b} \times \mathbf{c} :

b×c=(0×21×(1)1×24×24×(1)0×2)=(0+12840)=(:highlight[1]:highlight[6]:highlight[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} :highlight[1] \\ :highlight[-6] \\ :highlight[-4] \end{pmatrix}

  1. Dot Product a(b×c)\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) :

a(b×c)=1×1+2×(6)+3×(4)=11212=:highlight[23]\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 1 \times 1 + 2 \times (-6) + 3 \times (-4) = 1 - 12 - 12 = :highlight[-23]

  1. Volume (absolute value):

V=a(b×c)=23=:highlight[23] cubic unitsV = | \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) | = |-23| = :highlight[23] \text{ cubic units}

The volume of the parallelepiped is 23 cubic units.


Summary

infoNote
  • 3D vectors allow us to solve complex spatial problems involving magnitude and direction in three dimensions.
  • Vector operations such as addition, dot products, and cross products provide
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