Exercise 1

Given the vectors \vec { u } = (1, 2, 3), \vec { v } = (2, 0, 1) and \vec { w } = (-1, 3, 0), calculate the following:

1. \vec { u } . \vec { v }, \vec { v } . \vec { w }

2. \vec { u } \times \vec { v }, \vec { u } \times \vec { w }

3. (\vec { u } \times \vec { v }) . \vec { w }

4. \left | \vec { u } \right | \left | \vec { v } \right |

5. \cos { \overline{ \vec { u }, \vec { v } } }

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Exercise 2

For what values of a do the vectors \vec { u } = (1, 1, 1), \vec { v } = (1, a, 1), and \vec { w } = (1, 1, a) form a basis?

Exercise 3

Determining the value of the coefficient k for the vectors \vec { x } = k \vec { u } - 2 \vec { v } + 3 \vec { w } = - \vec { u } + k \vec { v } + \vec { w } if the vectors are:

1. Orthogonal.

2. Parallel.

Exercise 4

Find the direction cosines of the vector \vec { u } = (2, 2, 1).

Exercise 5

Calculate the angle between the vectors \vec { u } = (1, 1, -1) and \vec { v } = (2, 2, 1).

Exercise 6

Given the vectors \vec { u } = (3, 1, -1) and \vec { v } = (2, 3, 4), calculate:

1 The magnitudes of \vec { u } and \vec { v }·

2 The cross product of \vec { u } and \vec { v }·

3 The unit vector orthogonal to \vec { u } and \vec { v }·

4 The area of the parallelogram whose sides are the vectors \vec { u } and \vec { v }·

Exercise 7

Calculate the triple product of: [\vec { u } \times \vec { v }, \vec { v } \times \vec { w }, \vec { w } \times \vec { u }] if \vec { u } = (1, 0, 1) \quad \vec { v } = (0, 1, 1) \quad \vec { w } = (1, 1, 0).

Exercise 8

Given the vectors \vec { u } = (2, 1, 3), \vec { v } = (1, 2, 3),  and \vec { w } = (-1, -1, 0), calculate the triple product [\vec { u }, \vec { v }, \vec { w }]. Also, what is the volume of the parallelepiped whose edges are formed by these vectors?

 

Solution of exercise 1

Given the vectors \vec { u } = (1, 2, 3), \vec { v } = (2, 0, 1) and \vec { w } = (-1, 3, 0), calculate the following:

1. \vec { u } . \vec { v }, \vec { v } . \vec { w }

\vec { u } . \vec { v } = (1, 2, 3) . (2, 0, 1) = 1 . 2 + 2 . 0 + 3 . 1 = 5

\vec { v } . \vec { w } = (2, 0, 1) . (-1, 3, 0) = 2 . (-1) + 0 . 3 + 1 . 0 = -2

 

2. \vec { u } \times \vec { v }, \vec { u } \times \vec { w }

\vec { u } \times \vec { v } = \begin{vmatrix} \vec { i } & \vec { j } & \vec { k } \\ 1 & 2 & 3 \\ 2 & 0 & 1 \end{vmatrix} = \begin{vmatrix} 2 & 3 \\ 0 & 1 \end{vmatrix} \vec { i } - \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} \vec { j } + \begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} \vec { k } = 2 \vec { i } + 5 \vec { j } - 4 \vec { k }

\vec { u } \times \vec { w } = \begin{vmatrix} \vec { i } & \vec { j } & \vec { k } \\ 1 & 2 & 3 \\ -1 & 3 & 0 \end{vmatrix} = \begin{vmatrix} 2 & 3 \\ 3 & 0 \end{vmatrix} \vec { i } - \begin{vmatrix} 1 & 3 \\ -1 & 0 \end{vmatrix} \vec { j } + \begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} \vec { k } = -9 \vec { i } - 3 \vec { j } + 5 \vec { k }

 

3. (\vec { u } \times \vec { v }) . \vec { w }

(\vec { u } \times \vec { v }) . \vec { w } = (2, 5, -4) . (-1, 3, 0) = 13

4. \left | \vec { u } \right | \left | \vec { v } \right |

\left | \vec { u } \right | = \sqrt { { 1 }^{ 2 } + { 2 }^{ 2 } + { 3 }^{ 2 } } = \sqrt { 14 }

\left | \vec { v } \right | = \sqrt { { 2 }^{ 2 } + { 0 }^{ 2 } + { 1 }^{ 2 } } = \sqrt { 5 }

5. \cos { \overline{ \vec { u }, \vec { v } } }

\cos { \overline{ \vec { u }, \vec { v } } } = \frac { 5 }{ \sqrt { 14 } . \sqrt { 5 } } = 0.5976

 

Solution of exercise 2

For what values of a do the vectors \vec { u } = (1, 1, 1), \vec { v } = (1, a, 1), and \vec { w } = (1, 1, a) form a basis?

\begin{vmatrix} 1 & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \end{vmatrix} \neq 0

{ a }^{ 2 } + 1 + 1 - a - a - 1 \neq 0

{ a }^{ 2 } - 2a + 1 \neq 0

{ (a - 1) }^{ 2 } \neq 0 \qquad a \neq 1

For a \neq 1, the vectors form a basis.

 

Solution of exercise 3

Determining the value of the coefficient k for the vectors \vec { x } = k \vec { u } - 2 \vec { v } + 3 \vec { w } = - \vec { u } + k \vec { v } + \vec { w } if the vectors are:

1. Orthogonal.

\vec { x } . \vec { y } = 0

\vec { x } . \vec { y } = (k \vec { u } - 2 \vec { v } + 3 \vec { w })(- \vec { u } + k \vec { v } + \vec { w }) = - k - 2k + 3

-3 \vec { k } + 3 = 0 \qquad k = 1

2. Parallel.

\frac { k }{ - 1 } = \frac { -2 }{ k } = \frac { 3 }{ 1 }

\left\{\begin{matrix} { k }^{ 2 } = 2 \\ k = -3 \end{matrix}\right

The system does not have a solution.

 

Solution of exercise 4

Find the direction cosines of the vector \vec { u } = (2, 2, 1).

\left | \vec { u } \right | = \sqrt { { 2 }^{ 2 } + { 2 }^{ 2 } + { 1 }^{ 2 } }

\cos { \alpha } = \frac { 2 }{ 3 } \qquad \cos { \beta } = \frac { 2 }{ 3 } \qquad \cos { \gamma } = \frac { 1 }{ 3 }

 

Solution of exercise 5

Calculate the angle between the vectors \vec { u } = (1, 1, -1) and \vec { v } = (2, 2, 1).

\cos { \alpha } = \frac { 2 + 2 - 1 }{ \sqrt { 1 + 1 +1 } \sqrt { 4 + 4 + 1 }} = \frac { 3 }{ 3 \sqrt { 3 } }

\alpha = \arccos { (\frac { 1 }{ \sqrt { 3 } }) } = { 54.74 }^{ \circ }

 

Solution of exercise 6

Given the vectors \vec { u } = (3, 1, -1) and \vec { v } = (2, 3, 4), calculate:

1 The magnitudes of \vec { u } and \vec { v }·

\left | \vec { u } \right | = \sqrt { { 3 }^{ 2 } + { 1 }^{ 2 } + { (-1) }^{ 2 } } = \sqrt { 11 }

\left | \vec { v } \right | = \sqrt { { 2 }^{ 2 } + { 3 }^{ 2 } + { (4) }^{ 2 } } = \sqrt { 29 }

2 The cross product of \vec { u } and \vec { v }·

\vec { u } \times \vec { v } = \begin{vmatrix} \vec { i } & \vec { j } & \vec { k } \\ 3 & 1 & -1 \\ 2 & 3 & 4 \end{vmatrix} = \vec { i } \begin{vmatrix} 1 & -1 \\ 3 & 4  \end{vmatrix} - \vec { j } \begin{vmatrix} 3 & -1 \\ 2 & 4 \end{vmatrix} + \vec { k } \begin{vmatrix} 3 & 1 \\ 2  & 3 \end{vmatrix} = 7 \vec { i } - 14 \vec { j } + 7 \vec { k }

3 The unit vector orthogonal to \vec { u } and \vec { v }·

\left | \vec { u } \times \vec { v } \right | = \sqrt { { 7 }^{ 2 } + { 14 }^{ 2 } + { 7 }^{ 2 } } = \sqrt { 294 }

\vec { w } = ( \frac { 7 }{ \sqrt { 294 } }, \frac { -14 }{ \sqrt { 294 } }, \frac { 7 }{ \sqrt { 294 } })

4 The area of the parallelogram whose sides are the vectors \vec { u } and \vec { v }·

A = \left | \vec { u } \times \vec { v } \right | = \sqrt { 294 } { u }^{ 2 }

 

Solution of exercise 7

Calculate the triple product of: [\vec { u } \times \vec { v }, \vec { v } \times \vec { w }, \vec { w } \times \vec { u }] if \vec { u } = (1, 0, 1) \quad \vec { v } = (0, 1, 1) \quad \vec { w } = (1, 1, 0).

\vec { u } = (1, 0, 1) \quad \vec { v } = (0, 1, 1) \quad \vec { w } = (1, 1, 0)

\vec { u } \times \vec { v } = \begin{vmatrix} \vec { i } & \vec { j } & \vec { k } \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix} = \vec { i } \begin{vmatrix} 0 & 1 \\ 1 & 1  \end{vmatrix} - \vec { j } \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} + \vec { k } \begin{vmatrix} 1 & 0 \\ 0  & 1 \end{vmatrix} = - \vec { i } - \vec { j } + \vec { k }

\vec { v } \times \vec { w } = \begin{vmatrix} \vec { i } & \vec { j } & \vec { k } \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \vec { i } \begin{vmatrix} 1 & 1 \\ 1 & 0  \end{vmatrix} - \vec { j } \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} + \vec { k } \begin{vmatrix} 0 & 1 \\ 1  & 1 \end{vmatrix} = - \vec { i } + \vec { j } - \vec { k }

\vec { w } \times \vec { u } = \begin{vmatrix} \vec { i } & \vec { j } & \vec { k } \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{vmatrix} = \vec { i } \begin{vmatrix} 1 & 0 \\ 0 & 1  \end{vmatrix} - \vec { j } \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} + \vec { k } \begin{vmatrix} 1 & 1 \\ 1  & 0 \end{vmatrix} = \vec { i } - \vec { j } - \vec { k }

(\vec { v } \times \vec { w }) \times (\vec { w } \times \vec { u }) = \begin{vmatrix} \vec { i } & \vec { j } & \vec { k } \\ -1 & 1 & -1 \\ 1 & -1 & -1 \end{vmatrix} = - 2 \vec { i } - 2 \vec { j }

[\vec { u } \times \vec { v }, \vec { v } \times \vec { w }, \vec { w } \times \vec { u }] = (- \vec { i } - \vec { j } + \vec { k }) (-2 \vec { i } - 2 \vec { j }) = 4

 

Solution of exercise 8

Given the vectors \vec { u } = (2, 1, 3), \vec { v } = (1, 2, 3),  and \vec { w } = (-1, -1, 0), calculate the triple product [\vec { u }, \vec { v }, \vec { w }]. Also, what is the volume of the parallelepiped whose edges are formed by these vectors?

[ \vec { u }, \vec { v }, \vec { w }] = \begin{vmatrix} 2 & 1 & 3 \\ 1 & 2 & 3 \\ -1 & -1 & 0 \end{vmatrix} = 6

V = 6 { u }^{ 3 }

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.