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Geometry and Trigonometry - Vectors in 2D and 3D

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Vectors in 2D and 3D: A vector is a quantity with both magnitude and direction, represented by a directed line segment (an arrow). In 3D space, a vector v\mathbf{v} can be expressed in component form as (xyz)\begin{pmatrix} x \\ y \\ z \end{pmatrix} or in unit vector notation as xi+yj+zkx\mathbf{i} + y\mathbf{j} + z\mathbf{k}, where i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} are unit vectors along the x,y,zx, y, z axes respectively.

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Magnitude of a Vector: The magnitude represents the length of the vector arrow. For a 3D vector v=(v1v2v3)\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}, the magnitude is denoted by ∣v∣|\mathbf{v}| and is calculated using the 3D version of Pythagoras' Theorem: ∣v∣=v12+v22+v32|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}. Visually, this is the distance from the tail to the tip of the vector.

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Vector Addition and Scalar Multiplication: Adding two vectors a+b\mathbf{a} + \mathbf{b} is performed by adding their corresponding components. Visually, this follows the 'tip-to-tail' rule, where the resultant vector is the diagonal of the parallelogram formed by a\mathbf{a} and b\mathbf{b}. Scalar multiplication kvk\mathbf{v} changes the length of the vector by factor kk; if k<0k < 0, the vector's direction is reversed.

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The Scalar (Dot) Product: The dot product a⋅b\mathbf{a} \cdot \mathbf{b} results in a scalar value. It is calculated as a1b1+a2b2+a3b3a_1b_1 + a_2b_2 + a_3b_3 or as ∣a∣∣b∣cos⁡θ|\mathbf{a}||\mathbf{b}| \cos \theta, where θ\theta is the angle between the vectors. If a⋅b=0\mathbf{a} \cdot \mathbf{b} = 0, the vectors are perpendicular (90∘90^\circ), which means they are visually at a right angle to each other.

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Vector Equation of a Line: A line in 2D or 3D can be defined by a fixed point a\mathbf{a} (position vector) and a direction vector b\mathbf{b}. The equation is r=a+tb\mathbf{r} = \mathbf{a} + t\mathbf{b}, where tt is a scalar parameter. Visually, as tt changes, the point r\mathbf{r} moves along a straight path that passes through a\mathbf{a} and stays parallel to b\mathbf{b}.

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Unit Vectors: A unit vector is a vector with a magnitude of exactly 11. Any non-zero vector v\mathbf{v} can be converted into a unit vector v^\hat{\mathbf{v}} in the same direction by dividing the vector by its magnitude: v^=v∣v∣\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|}.

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Parallel and Collinear Vectors: Two vectors are parallel if one is a scalar multiple of the other, i.e., a=kb\mathbf{a} = k\mathbf{b}. Visually, they point in the same or opposite directions but have the same 'slope' in space. Three points A,B,CA, B, C are collinear if the vectors AB⃗\vec{AB} and BC⃗\vec{BC} are parallel and share a common point BB.

📐Formulae

Magnitude: ∣v∣=x2+y2+z2|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}

Dot Product (Algebraic): a⋅b=a1b1+a2b2+a3b3\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3

Dot Product (Geometric): a⋅b=∣a∣∣b∣cos⁡θ\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos \theta

Angle between vectors: cos⁡θ=a⋅b∣a∣∣b∣\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}

Vector Line Equation: r=a+tb\mathbf{r} = \mathbf{a} + t\mathbf{b}

Unit Vector: v^=1∣v∣v\hat{\mathbf{v}} = \frac{1}{|\mathbf{v}|}\mathbf{v}

Distance between two points: d=(x2−x1)2+(y2−y1)2+(z2−z1)2d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

💡Examples

Problem 1:

Find the angle between the vectors a=(21−2)\mathbf{a} = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix} and b=(034)\mathbf{b} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}.

Solution:

  1. Calculate the dot product: a⋅b=(2)(0)+(1)(3)+(−2)(4)=0+3−8=−5\mathbf{a} \cdot \mathbf{b} = (2)(0) + (1)(3) + (-2)(4) = 0 + 3 - 8 = -5.
  2. Calculate the magnitude of a\mathbf{a}: ∣a∣=22+12+(−2)2=4+1+4=9=3|\mathbf{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3.
  3. Calculate the magnitude of b\mathbf{b}: ∣b∣=02+32+42=0+9+16=25=5|\mathbf{b}| = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{0+9+16} = \sqrt{25} = 5.
  4. Use the cosine formula: cos⁡θ=−5(3)(5)=−515=−13\cos \theta = \frac{-5}{(3)(5)} = \frac{-5}{15} = -\frac{1}{3}.
  5. Find the angle: θ=arccos⁡(−13)≈109.5∘\theta = \arccos(-\frac{1}{3}) \approx 109.5^\circ.

Explanation:

To find the angle between two vectors, we use the dot product formula. First, compute the algebraic dot product and the lengths of both vectors. Then, solve for the cosine of the angle and use the inverse cosine function.

Problem 2:

Determine the vector equation of the line passing through the points A(1,4,2)A(1, 4, 2) and B(3,−1,5)B(3, -1, 5).

Solution:

  1. Identify a position vector a\mathbf{a} for a point on the line. Using point AA, a=(142)\mathbf{a} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}.
  2. Find the direction vector b\mathbf{b} by calculating AB⃗=b−a=(3−1−1−45−2)=(2−53)\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 3-1 \\ -1-4 \\ 5-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \\ 3 \end{pmatrix}.
  3. Write the equation in the form r=a+tb\mathbf{r} = \mathbf{a} + t\mathbf{b}.
  4. Result: r=(142)+t(2−53)\mathbf{r} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} + t \begin{pmatrix} 2 \\ -5 \\ 3 \end{pmatrix}.

Explanation:

A vector line requires a starting point (position vector) and a direction. The direction is found by subtracting the coordinates of the first point from the second point. The final equation combines these using a parameter tt.