Emma

June 26, 2019

Chapters

Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 5
Exercise 6
Solution of exercise 1
Solution of exercise 2
Solution of exercise 3
Solution of exercise 4
Solution of exercise 5
Solution of exercise 6

Exercise 1

Find two unit vectors for $(2, −2, 3)$ and $(3, −3, 2)$ and determine the orthogonal vector for the two.

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Exercise 2

Find a unit vector that is perpendicular to $\overrightarrow { u } = (2, 3, 4)$ and $\overrightarrow { v } = (-1, 3, -5)$ .

Exercise 3

Given the vectors $\overrightarrow { u } = 3i - j + k$ and $\overrightarrow { v } = 2i - 3j + k$ , find the product $\overrightarrow { u } \times \overrightarrow { v }$ and verify that this vector is orthogonal to $\overrightarrow { u }$ and $\overrightarrow { v }$ . Also, find the vector $\overrightarrow { v } \times \overrightarrow { u }$ and compare it with $\overrightarrow { u } \times \overrightarrow { v }$ .

Exercise 4

Consider the following figure:

Determine:

1 The coordinates of D if ABCD is a parallelogram.

2 The area of the parallelogram.

Exercise 5

Given the points $A = (1, 0, 1), B = (1, 1, 1)$ and $C = (1, 6, a)$ , determine:

1 What values of a are collinear.

2 Determine if values exist for a so that A, B, and C are three vertices of a parallelogram of area $3$ . If values do exist, determine the coordinates of C:

Exercise 6

$A = (−3, 4, 0), B = (3, 6, 3)$ and $C = (−1, 2, 1)$ are the three vertices of a triangle.

1. Calculate the cosine of each of the three angles in the triangle.

2. Calculate the area of the triangle.

Solution of exercise 1

Find two unit vectors for $(2, −2, 3)$ and $(3, −3, 2)$ and determine the orthogonal vector for the two.

$\overrightarrow { w } = \begin{vmatrix} i & j & k \\ 2 & -2 & 3 \\ 3 & -3 & 2 \end{vmatrix} = i \begin{vmatrix} -2 & 3 \\ -3 & 2 \end{vmatrix} - j \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} + k \begin{vmatrix} 2 & -2 \\ 3 & -3 \end{vmatrix}$

$\overrightarrow { w } = i(-4 + 9) - j(4 - 9) + k(-6 + 6)$

$\overrightarrow { w } = 5i + 5j$

$\left| \overrightarrow { w } \right| = \sqrt { { 5 }^{ 2 } + { 5 }^{ 2 } } = 5 \sqrt { 2 }$

$\hat { u } = (\frac { 5 }{ 5 \sqrt { 2 } }, \frac { 5 }{ 5 \sqrt { 2 } }, 0 ) = (\frac { 1 }{ \sqrt { 2 } }, \frac { 1 }{ \sqrt { 2 }}, 0)$

$\hat { v } = (\frac { -5 }{ 5 \sqrt { 2 } }, \frac { -5 }{ 5 \sqrt { 2 } }, 0 ) = (\frac { -1 }{ \sqrt { 2 } }, \frac { -1 }{ \sqrt { 2 }}, 0)$

Solution of exercise 2

Find a unit vector that is perpendicular to $\overrightarrow { u } = (2, 3, 4)$ and $\overrightarrow { v } = (-1, 3, -5)$ .

$\overrightarrow { u } \times \overrightarrow { v } = \begin{vmatrix} i & j & k \\ 2 & 3 & 4 \\ -1 & 3 & -5 \end{vmatrix} = i \begin{vmatrix} 3 & 4 \\ 3 & -5 \end{vmatrix} - j \begin{vmatrix} 2 & 4 \\ -1 & -5 \end{vmatrix} + k \begin{vmatrix} 2 & 3 \\ -1 & 3 \end{vmatrix} = -27i + 6j + 9k$

$\left| \overrightarrow { u } \times \overrightarrow { v } \right| = \sqrt { { (-27) }^{ 2 }+{ 6 }^{ 2 }+{ 9 }^{ 2 } } = \sqrt { 846 }$

$\overrightarrow { w } = ( \frac { -27 }{ \sqrt { 846 } }, \frac { 6 }{ \sqrt { 846 } }, \frac { 9 }{ \sqrt { 846 }} )$

Solution of exercise 3

$\overrightarrow { u } \times \overrightarrow { v } = \begin{vmatrix} i & j & k \\ 3 & -1 & 1 \\ 2 & -3 & 1 \end{vmatrix} = i \begin{vmatrix} -1 & 1 \\ -3 & 1 \end{vmatrix} - j \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} + k \begin{vmatrix} 3 & -1 \\ 2 & -3 \end{vmatrix}$

$\overrightarrow { u } \times \overrightarrow { v } = i(-1 + 3) - j(3 - 2) + k(-9 + 2) = 2i - j - 7k$

$(\overrightarrow { u } \times \overrightarrow { v }) \bot \overrightarrow { u } \qquad (\overrightarrow { u } \times \overrightarrow { v }) . \overrightarrow { u } = 0$

$(2, -1, -7) . (3, -1, 1) = 6 + 1 - 7 = 0$

$(\overrightarrow { u } \times \overrightarrow { v }) \bot \overrightarrow { v } \qquad (\overrightarrow { u } \times \overrightarrow { v }) . \overrightarrow { v } = 0$

$(2, -1, -7) . (2, -3, 1) = 4 + 3 - 7 = 0$

$\overrightarrow { v } \times \overrightarrow { u } = \begin{vmatrix} i & j & k \\ 2 & -3 & 1 \\ 3 & -1 & 1 \end{vmatrix} = i \begin{vmatrix} -3 & 1 \\ -1 & 1 \end{vmatrix} - j \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} + k \begin{vmatrix} 2 & -3 \\ 3 & -1 \end{vmatrix}$

$\overrightarrow { v } \times \overrightarrow { u } = i(-3 + 1) - j(2 - 3) + k(-2 + 9) = -2i + j + 7k$

$\overrightarrow { v } \times \overrightarrow { u } = - (\overrightarrow { u } \times \overrightarrow { v })$

Solution of exercise 4

Consider the following figure:

Determine:

1 The coordinates of D if ABCD is a parallelogram.

$(x - 1, y - 1, z) = (2+1, 2+1, 1)$

$x - 1 = 3 \qquad x = 4$

$y - 1 = 3 \qquad y = 4$

$z = 1$

$D = (4, 4, 1)$

2 The area of the parallelogram.

$A = \left| \overrightarrow { BC } \times \overrightarrow { BA } \right|$

$\overrightarrow { BC } = (3, 3, 1) \qquad \overrightarrow { BA } = (2, 2, 1)$

$\overrightarrow { BC } \times \overrightarrow { BA } = \begin{vmatrix} i & j & k \\ 3 & 3 & 1 \\ 2 & 2 & 1 \end{vmatrix} = i \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} - j \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} + k \begin{vmatrix} 3 & 3 \\ 2 & 2 \end{vmatrix}$

$\overrightarrow { BC } \times \overrightarrow { BA } = i (3 - 2) - j (3 - 2) + k (6 - 6) = i - j$

$A = \left| \overrightarrow { BC } \times \overrightarrow { BA } \right| = \sqrt { { 1 }^{ 2 } + { (-1) }^{ 2 } } = \sqrt { 2 } { u }^{ 2 }$

Solution of exercise 5

Given the points $A = (1, 0, 1), B = (1, 1, 1)$ and $C = (1, 6, a)$ , determine:

1 What values of a are collinear.

If A, B, and C are collinear, the vectors $\overrightarrow { AB }$ and $\overrightarrow { AC }$ are linearly dependent and have proportional components.

$\overrightarrow { AB } = (1 - 1, 1 - 0, 1 - 1) = (0, 1, 0)$

$\overrightarrow { AC } = (1 - 1, 6 - 0, a - 1) = (0, 6, a - 1)$

$(0, 6, a - 1) = k (0, 1, 0) \qquad a - 1 =0 \qquad a = 1$

2 Determine if values exist for a so that A, B, and C are three vertices of a parallelogram of area $3$ . If values do exist, determine the coordinates of C:

$(\overrightarrow { AB } \times \overrightarrow { AC } ) = \begin{vmatrix} i & j & k \\ 0 & 1 & 0 \\ 0 & 6 & a - 1 \end{vmatrix} = i \begin{vmatrix} 1 & 0 \\ 6 & a-1 \end{vmatrix} - j \begin{vmatrix} 0 & 0 \\ 0 & a-1 \end{vmatrix} + k \begin{vmatrix} 0 & 1 \\ 0 & 6 \end{vmatrix}$

$(\overrightarrow { AB } \times \overrightarrow { AC } ) = i (a - 1) - j (0 - 0) + k (0 - 0) = i(a - 1)$

$A = \left| \overrightarrow { AB } \times \overrightarrow { AC } \right| = 3$

$3 = \sqrt { { (a - 1) }^{ 2 } + { 0 }^{ 2 } + { 0 }^{ 2 } }$

$a - 1 = 3 \qquad or \qquad a -1 = -3$

$a = 4 \qquad or \qquad a = -2$

$C (1, 6, 4) \qquad or \qquad C (1, 6, -2)$

Solution of exercise 6

$A = (−3, 4, 0), B = (3, 6, 3)$ and $C = (−1, 2, 1)$ are the three vertices of a triangle.

1. Calculate the cosine of each of the three angles in the triangle.

$\overrightarrow { AB } = (3 + 3, 6 - 4, 3 - 0) = (6, 2, 3) \qquad \qquad \overrightarrow { BA } = (-6, -2, -3)$

$\overrightarrow { AC } = (-1 + 3, 2 - 4, 1 - 0) = (2, -2, 1) \qquad \qquad \overrightarrow { CA } = (-2, 2, -1)$

$\overrightarrow { BC } = (-1 - 3, 2 - 6, 1 - 3) = (-4, -4, -2) \qquad \qquad \overrightarrow { CB } = (4, 4, 2)$

$\cos { \overrightarrow { AB }, \overrightarrow { AC } } = \frac { 12 - 4 +3 }{ \sqrt { 36 + 4 + 9 } \sqrt { 4 + 4 + 1}} = \frac { 11 }{ 21 }$

$\cos { \overrightarrow { BA }, \overrightarrow { BC } } = \frac { 24 + 8 + 6 }{ \sqrt { 36 + 4 + 9 } \sqrt { 16 + 16 + 4 }} = \frac { 38 }{ 21 }$

$\cos { \overrightarrow { CA }, \overrightarrow { CB } } = \frac { -8 + 8 - 2 }{ \sqrt { 4 + 4 + 1 } \sqrt { 16 + 16 + 4 }} = \frac { -1 }{ 9 }$

2. Calculate the area of the triangle.

$A = \frac { 1 }{ 2 } \left| \overrightarrow { AB } \times \overrightarrow { AC } \right|$

$\overrightarrow { AB } \times \overrightarrow { AC } = \begin{vmatrix} i & j & k \\ 6 & 2 & 3 \\ 2 & -2 & 1 \end{vmatrix} = i \begin{vmatrix} 2 & 3 \\ -2 & 1 \end{vmatrix} - j \begin{vmatrix} 6 & 3 \\ 2 & 1 \end{vmatrix} + k \begin{vmatrix} 6 & 2 \\ 2 & -2 \end{vmatrix} = 8i - 16k$

$\overrightarrow { AB } \times \overrightarrow { AC } = \sqrt { { 8 }^{ 2 } + { 16 }^{ 2 } } = \sqrt { 64 + 256 } = 8 \sqrt { 5 }$

$A = \frac { 1 }{ 2 } \left| \overrightarrow { AB } \times \overrightarrow { AC } \right| = \frac { 1 }{ 2 } (8 \sqrt { 5 }) = 4 \sqrt { 5 }$

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

Cross Product Problems

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Solution of exercise 1

Solution of exercise 2

Solution of exercise 3

Solution of exercise 4

Solution of exercise 5

Solution of exercise 6

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Linearly Independent Vectors

Unit Vector

Angle Between Two Vectors

Vector Projection

Scalar Multiplication

Vector Addition

Cross Product

Dot Product

Adding Vectors

Direction Cosine

Linearly Dependent Vectors

Basis Vector

3D Vectors

Linear Dependence and Independence

Scalar Triple Product

Standard Basis

Orthogonal and Orthonormal Vectors

Vectors in the Plane

Linear Combination of Vectors

Cartesian and Polar Coordinates

Scalar Product of Vectors

Vector Formulas

Dot and Triple Product Problems

Vector Problems

Cross Product Problems

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Solution of exercise 1

Solution of exercise 2

Solution of exercise 3

Solution of exercise 4

Solution of exercise 5

Solution of exercise 6

Cancel reply

Theory

Linearly Independent Vectors

Unit Vector

Angle Between Two Vectors

Vector Projection

Scalar Multiplication

Vector Addition

Cross Product

Dot Product

Adding Vectors

Direction Cosine

Linearly Dependent Vectors

Basis Vector

3D Vectors

Linear Dependence and Independence

Scalar Triple Product

Standard Basis

Orthogonal and Orthonormal Vectors

Vectors in the Plane

Linear Combination of Vectors

Cartesian and Polar Coordinates

Scalar Product of Vectors

Formulas

Vector Formulas

Exercises

Dot and Triple Product Problems

Vector Problems