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IB Mathematics AA Vector product of two vectors Study Notes

IB Mathematics AA Vector product of two vectors Study Notes

IB Mathematics AA vector product of two vectors Study Notes

IB Mathematics AA Vector product of two vectors Notes Offer a clear explanation of Vector product of two vectors, including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Vector product of two vectors.

Vector Product of Two Vectors (Cross Product)

Introduction

The vector product, also known as the cross product, is an operation that takes two vectors and returns another vector. This resultant vector is perpendicular to the plane containing the two input vectors. The vector product is widely used in physics and engineering to calculate torque, angular momentum, and areas of geometric shapes.

Definition of the Vector Product

The vector product of two vectors \( \vec{v} \) and \( \vec{w} \) is defined as:

\[ \vec{v} \times \vec{w} = |\vec{v}||\vec{w}|\sin \theta \, \vec{n} \]

Here:

  • \( |\vec{v}| \) and \( |\vec{w}| \) are the magnitudes of \( \vec{v} \) and \( \vec{w} \).
  • \( \theta \) is the angle between \( \vec{v} \) and \( \vec{w} \) (measured in the plane containing both vectors).
  • \( \vec{n} \) is the unit vector perpendicular to the plane containing \( \vec{v} \) and \( \vec{w} \), with its direction determined by the right-hand screw rule.

Key Properties of the Vector Product

The vector product has several important properties:

  • \( \vec{v} \times \vec{w} = -\vec{w} \times \vec{v} \) (anti-commutative property).
  • \( \vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w} \) (distributive property).
  • \( (k\vec{v}) \times \vec{w} = k (\vec{v} \times \vec{w}) \), where \( k \) is a scalar.
  • \( \vec{v} \times \vec{v} = 0 \) (the vector product of any vector with itself is zero).

Geometric Interpretation

The magnitude of the vector product, \( |\vec{v} \times \vec{w}| \), represents the area of the parallelogram formed by \( \vec{v} \) and \( \vec{w} \). If the two vectors are:

  • Parallel: \( \vec{v} \times \vec{w} = 0 \), since \( \sin \theta = 0 \).
  • Perpendicular: \( |\vec{v} \times \vec{w}| = |\vec{v}||\vec{w}| \), since \( \sin \theta = 1 \).
Area of a Triangle Using the Vector Product

The area of a parallelogram formed by two vectors \( \vec{v} \) and \( \vec{w} \) is:

\[ \text{Area of parallelogram} = |\vec{v} \times \vec{w}| \]

The area of the triangle is half the area of the parallelogram:

\[ \text{Area of triangle} = \frac{1}{2} |\vec{v} \times \vec{w}| \]

Right-Hand Screw Rule

The direction of the resultant vector \( \vec{u} \times \vec{v} \) is determined by the right-hand screw rule. To use this rule:

  • Point the fingers of your right hand in the direction of \( \vec{u} \).
  • Rotate your fingers toward \( \vec{v} \).
  • Your thumb points in the direction of the resultant vector \( \vec{u} \times \vec{v} \).

Example Calculations

Example 1: Basic Vector Product

Given \( \vec{v} = [2, 3, 4] \) and \( \vec{w} = [1, 0, -1] \):

The vector product is calculated as:

\[ \vec{v} \times \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & 0 & -1 \end{vmatrix} \]

Expanding the determinant:

\[ \vec{v} \times \vec{w} = \hat{i} (3 \cdot -1 – 4 \cdot 0) – \hat{j} (2 \cdot -1 – 4 \cdot 1) + \hat{k} (2 \cdot 0 – 3 \cdot 1) \]

\[ \vec{v} \times \vec{w} = -3\hat{i} + 6\hat{j} – 3\hat{k} \]

Thus, \( \vec{v} \times \vec{w} = [-3, 6, -3] \).

Example 2: Area of a Parallelogram

Given \( \vec{v} = [3, 0, 0] \) and \( \vec{w} = [0, 4, 0] \):

  • The magnitude of the vector product is: \[ |\vec{v} \times \vec{w}| = |[3, 0, 0] \times [0, 4, 0]| = |[0, 0, 12]| = 12 \]
  • The area of the parallelogram is \( 12 \), and the area of the triangle is \( 6 \).

Applications of the Vector Product

The vector product has practical applications in physics, geometry, and engineering:

  • Torque: \( \vec{\tau} = \vec{r} \times \vec{F} \), where \( \vec{r} \) is the position vector and \( \vec{F} \) is the force.
  • Angular Momentum: \( \vec{L} = \vec{r} \times \vec{p} \), where \( \vec{p} \) is the linear momentum.
  • Area Calculation: Finding areas of parallelograms and triangles.

IB Mathematics AA SL Vector product of two vectors Exam Style Worked Out Questions

Question

The points A(1, 2, 1) , B(−3, 1, 4) , C(5, −1, 2) and D(5, 3, 7) are the vertices of a tetrahedron.

a. Find the vectors \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \).[2]

b.Find the Cartesian equation of the plane \(\prod \) that contains the face ABC.[4]

 
▶️Answer/Explanation

Markscheme

\(\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}}
  { – 4} \\
  { – 1} \\
  3
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}}  = \left( {\begin{array}{*{20}{c}}
  4 \\
  { – 3} \\
  1
\end{array}} \right)\)     A1A1

Note: Accept row vectors.

 

[2 marks]

a.

\(\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}}  = \left| {\begin{array}{*{20}{c}}
  {\boldsymbol{i}}&{\boldsymbol{j}}&{\boldsymbol{k}} \\
  { – 4}&{ – 1}&3 \\
  4&{ – 3}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
  8 \\
  {16} \\
  {16}
\end{array}} \right)\)     M1A1

normal \({\boldsymbol{n}} = \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  2
\end{array}} \right)\) so \({\boldsymbol{r}} \cdot \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  1
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  2
\end{array}} \right)\)     (M1)

\(x + 2y + 2z = 7\)     A1

Note: If attempt to solve by a system of equations:
Award A1 for 3 correct equations, A1 for eliminating a variable and A2 for the correct answer.

 

[4 marks]

b.

Examiners report

Most candidates attempted this question and scored at least a few marks in (a) and (b). Part (c) was more challenging to many candidates who were unsure how to find the required distance. Part (d) was attempted by many candidates some of whom benefited from follow through marks due to errors in previous parts. However, many candidates failed to give the correct answer to this question due to the use of the simplified vector found in (b) showing little understanding of the role of the magnitude of this vector. Part (e) was poorly answered. Overall, this question was not answered to the expected level, showing that many candidates have difficulties with vectors and are unable to answer even standard questions on this topic.

a.

Most candidates attempted this question and scored at least a few marks in (a) and (b). Part (c) was more challenging to many candidates who were unsure how to find the required distance. Part (d) was attempted by many candidates some of whom benefited from follow through marks due to errors in previous parts. However, many candidates failed to give the correct answer to this question due to the use of the simplified vector found in (b) showing little understanding of the role of the magnitude of this vector. Part (e) was poorly answered. Overall, this question was not answered to the expected level, showing that many candidates have difficulties with vectors and are unable to answer even standard questions on this topic.

b.

Question

Consider the points A(1, 2, 3), B(1, 0, 5) and C(2, −1, 4).

Find \(\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} \).

[4]
a.

Hence find the area of the triangle ABC.

[2]
b.
▶️Answer/Explanation

Markscheme

\(\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}}
  1 \\
  0 \\
  5
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  3
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0 \\
  { – 2} \\
  2
\end{array}} \right)\), \(\overrightarrow {{\text{AC}}}  = \left( {\begin{array}{*{20}{c}}
  2 \\
  { – 1} \\
  4
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
  1 \\
  2 \\
  3
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  1 \\
  { – 3} \\
  1
\end{array}} \right)\)     A1A1

Note: Award the above marks if the components are seen in the line below.

 

\(\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}}  = \left| {\begin{array}{*{20}{c}}
  i&j&k \\
  0&{ – 2}&2 \\
  1&{ – 3}&1
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
  4 \\
  2 \\
  2
\end{array}} \right)\)     (M1)A1

[4 marks]

a.

area \( = \frac{1}{2}\left| {\left( {\overrightarrow {{\text{AB}}}  \times \overrightarrow {{\text{AC}}} } \right)} \right|\)     (M1)

\( = \frac{1}{2}\sqrt {{4^2} + {2^2} + {2^2}}  = \frac{1}{2}\sqrt {24} {\text{ }}\left( { = \sqrt 6 } \right)\)     A1

Note: Award M0A0 for attempts that do not involve the answer to (a).

[2 marks]

b.

Examiners report

Candidates showed a good understanding of the vector techniques required in this question.

a.

Candidates showed a good understanding of the vector techniques required in this question.

b.

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