Given The Vertices { A(-1,3), B(2,1), $}$ And { C(5,3) $}$ Of Triangle { \triangle ABC $}$:1. Express In Column Notation The Unit Vectors Parallel To { \overrightarrow{AB} $}$ And [$ \overrightarrow{AC}

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Introduction

In geometry and trigonometry, vectors play a crucial role in describing the relationships between points in a plane or in space. Given the vertices of a triangle, we can express the vectors representing the sides of the triangle using column notation. In this article, we will explore how to express the unit vectors parallel to AB→\overrightarrow{AB} and AC→\overrightarrow{AC} in triangle △ABC\triangle ABC, where the vertices are A(−1,3)A(-1,3), B(2,1)B(2,1), and C(5,3)C(5,3).

Column Notation of Vectors

Column notation is a way of representing vectors as columns of numbers. For a vector v\mathbf{v} with components v1,v2,…,vnv_1, v_2, \ldots, v_n, the column notation is given by:

v=(v1v2â‹®vn)\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix}

Expressing Vectors AB→\overrightarrow{AB} and AC→\overrightarrow{AC}

To express the vectors AB→\overrightarrow{AB} and AC→\overrightarrow{AC} in column notation, we need to find the components of these vectors. The vector AB→\overrightarrow{AB} is given by the difference between the coordinates of points BB and AA, while the vector AC→\overrightarrow{AC} is given by the difference between the coordinates of points CC and AA.

Let's calculate the components of AB→\overrightarrow{AB} and AC→\overrightarrow{AC}:

  • For AB→\overrightarrow{AB}:
    • xx-component: 2−(−1)=32 - (-1) = 3
    • yy-component: 1−3=−21 - 3 = -2
    • Therefore, AB→=(3−2)\overrightarrow{AB} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}
  • For AC→\overrightarrow{AC}:
    • xx-component: 5−(−1)=65 - (-1) = 6
    • yy-component: 3−3=03 - 3 = 0
    • Therefore, AC→=(60)\overrightarrow{AC} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}

Unit Vectors

A unit vector is a vector with a magnitude of 1. To find the unit vector parallel to a given vector, we need to divide the vector by its magnitude. The magnitude of a vector v=(v1v2)\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} is given by:

∥v∥=v12+v22\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2}

Finding the Unit Vectors Parallel to AB→\overrightarrow{AB} and AC→\overrightarrow{AC}

Now that we have the components of AB→\overrightarrow{AB} and AC→\overrightarrow{AC}, we can find their magnitudes and then calculate the unit vectors parallel to these vectors.

  • For AB→\overrightarrow{AB}:
    • Magnitude: ∥AB→∥=32+(−2)2=9+4=13\|\overrightarrow{AB}\| = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
    • Unit vector: uAB=AB→∥AB→∥=113(3−2)=(3/13−2/13)\mathbf{u}_{AB} = \frac{\overrightarrow{AB}}{\|\overrightarrow{AB}\|} = \frac{1}{\sqrt{13}} \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 3/\sqrt{13} \\ -2/\sqrt{13} \end{pmatrix}
  • For AC→\overrightarrow{AC}:
    • Magnitude: ∥AC→∥=62+02=36=6\|\overrightarrow{AC}\| = \sqrt{6^2 + 0^2} = \sqrt{36} = 6
    • Unit vector: uAC=AC→∥AC→∥=16(60)=(10)\mathbf{u}_{AC} = \frac{\overrightarrow{AC}}{\|\overrightarrow{AC}\|} = \frac{1}{6} \begin{pmatrix} 6 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}

Conclusion

In this article, we have expressed the vectors AB→\overrightarrow{AB} and AC→\overrightarrow{AC} in column notation and found the unit vectors parallel to these vectors. We have also calculated the magnitudes of these vectors and used them to find the unit vectors. The unit vectors are essential in many applications, including physics, engineering, and computer graphics.

References

  • [1] "Vector Notation" by Math Open Reference
  • [2] "Unit Vectors" by Khan Academy
  • [3] "Column Notation" by Wolfram MathWorld

Further Reading

  • "Vector Calculus" by Michael Corral
  • "Linear Algebra and Its Applications" by Gilbert Strang
  • "Calculus: Early Transcendentals" by James Stewart
    Q&A: Vector Notation and Unit Vectors in Triangle ABC =====================================================

Introduction

In our previous article, we explored how to express the vectors AB→\overrightarrow{AB} and AC→\overrightarrow{AC} in column notation and found the unit vectors parallel to these vectors. In this article, we will answer some frequently asked questions related to vector notation and unit vectors in triangle △ABC\triangle ABC.

Q: What is the difference between a vector and a unit vector?

A: A vector is a quantity with both magnitude and direction, while a unit vector is a vector with a magnitude of 1. Unit vectors are often used to represent the direction of a vector.

Q: How do I find the magnitude of a vector?

A: To find the magnitude of a vector v=(v1v2)\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}, you can use the formula:

∥v∥=v12+v22\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2}

Q: What is the purpose of unit vectors?

A: Unit vectors are used to represent the direction of a vector. They are often used in physics, engineering, and computer graphics to simplify calculations and represent the direction of a vector.

Q: How do I find the unit vector parallel to a given vector?

A: To find the unit vector parallel to a given vector v\mathbf{v}, you can divide the vector by its magnitude:

u=v∥v∥\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}

Q: Can I use unit vectors to represent the direction of a vector in 3D space?

A: Yes, you can use unit vectors to represent the direction of a vector in 3D space. However, you will need to use a 3D vector representation, such as v=(v1v2v3)\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}.

Q: How do I find the unit vector parallel to the vector AB→\overrightarrow{AB} in triangle △ABC\triangle ABC?

A: To find the unit vector parallel to the vector AB→\overrightarrow{AB}, you can use the formula:

uAB=AB→∥AB→∥=113(3−2)=(3/13−2/13)\mathbf{u}_{AB} = \frac{\overrightarrow{AB}}{\|\overrightarrow{AB}\|} = \frac{1}{\sqrt{13}} \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 3/\sqrt{13} \\ -2/\sqrt{13} \end{pmatrix}

Q: How do I find the unit vector parallel to the vector AC→\overrightarrow{AC} in triangle △ABC\triangle ABC?

A: To find the unit vector parallel to the vector AC→\overrightarrow{AC}, you can use the formula:

uAC=AC→∥AC→∥=16(60)=(10)\mathbf{u}_{AC} = \frac{\overrightarrow{AC}}{\|\overrightarrow{AC}\|} = \frac{1}{6} \begin{pmatrix} 6 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}

Conclusion

In this article, we have answered some frequently asked questions related to vector notation and unit vectors in triangle â–³ABC\triangle ABC. We hope that this article has provided you with a better understanding of these concepts and how to apply them in different situations.

References

  • [1] "Vector Notation" by Math Open Reference
  • [2] "Unit Vectors" by Khan Academy
  • [3] "Column Notation" by Wolfram MathWorld

Further Reading

  • "Vector Calculus" by Michael Corral
  • "Linear Algebra and Its Applications" by Gilbert Strang
  • "Calculus: Early Transcendentals" by James Stewart