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Geometry and Trigonometry - Scalar (Dot) and Vector (Cross) Products

Grade 12IB

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

🔑Concepts

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The Scalar (Dot) Product measures the extent to which two vectors point in the same direction. Geometrically, it represents the product of the magnitude of one vector and the scalar projection of the second vector onto the first. Visually, if you imagine a light source shining perpendicular to vector a⃗\vec{a}, the dot product relates to the length of the 'shadow' vector b⃗\vec{b} casts upon a⃗\vec{a}.

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Orthogonality and the Dot Product: Two non-zero vectors a⃗\vec{a} and b⃗\vec{b} are perpendicular (orthogonal) if and only if their scalar product is zero, i.e., a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0. Visually, this occurs when the vectors form a 90∘90^{\circ} angle, meaning neither vector has a component in the direction of the other.

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The Vector (Cross) Product results in a third vector that is perpendicular to the plane containing the original two vectors. The direction of this resultant vector is determined by the 'Right-Hand Rule': if you curl the fingers of your right hand from a⃗\vec{a} to b⃗\vec{b}, your thumb points in the direction of a⃗×b⃗\vec{a} \times \vec{b}. Visually, this creates a normal vector n⃗\vec{n} sticking straight out of the surface formed by the two input vectors.

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Geometric Area and the Cross Product: The magnitude of the cross product, ∣a⃗×b⃗∣|\vec{a} \times \vec{b}|, is equal to the area of the parallelogram formed by vectors a⃗\vec{a} and b⃗\vec{b} as adjacent sides. Visually, if you draw these two vectors from a common origin, the 'filling' of the four-sided shape they define has a numerical area equal to the length of the cross product vector.

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Parallelism and the Cross Product: Two non-zero vectors are parallel if and only if their vector product is the zero vector, a⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0}. Visually, this happens when the vectors lie on the same line or are pointing in exactly opposite directions (0∘0^{\circ} or 180∘180^{\circ}), meaning they cannot span a 2D parallelogram (the area is zero).

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Properties of Products: The dot product is commutative (a⃗⋅b⃗=b⃗⋅a⃗\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}), but the cross product is anti-commutative (a⃗×b⃗=−(b⃗×a⃗)\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})). Visually, swapping the order in a cross product flips the resulting vector 180∘180^{\circ} to point in the opposite direction along the same normal line.

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The Scalar Triple Product a⃗⋅(b⃗×c⃗)\vec{a} \cdot (\vec{b} \times \vec{c}) represents the volume of a parallelepiped (a 3D slanted box) defined by the three vectors. Visually, the cross product b⃗×c⃗\vec{b} \times \vec{c} gives the area of the base, and the dot product with a⃗\vec{a} accounts for the vertical height projected onto the base's normal.

📐Formulae

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

a⃗×b⃗=∣ijka1a2a3b1b2b3∣=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}

∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta

\text{Area of a triangle} = \frac{1}{2} |\vec{a} \times \vec{b}|$

\text{Volume of a parallelepiped} = |\vec{a} \cdot (\vec{b} \times \vec{c})|$

💡Examples

Problem 1:

Given vectors u⃗=(2−13)\vec{u} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} and v⃗=(14−2)\vec{v} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}, find the angle θ\theta between them to the nearest degree.

Solution:

Step 1: Calculate the dot product u⃗⋅v⃗\vec{u} \cdot \vec{v}. u⃗⋅v⃗=(2)(1)+(−1)(4)+(3)(−2)=2−4−6=−8\vec{u} \cdot \vec{v} = (2)(1) + (-1)(4) + (3)(-2) = 2 - 4 - 6 = -8 Step 2: Calculate the magnitudes ∣u⃗∣|\vec{u}| and ∣v⃗∣|\vec{v}|. ∣u⃗∣=22+(−1)2+32=4+1+9=14|\vec{u}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14} ∣v⃗∣=12+42+(−2)2=1+16+4=21|\vec{v}| = \sqrt{1^2 + 4^2 + (-2)^2} = \sqrt{1 + 16 + 4} = \sqrt{21} Step 3: Use the cosine formula. cos⁡θ=−81421=−8294≈−0.466\cos \theta = \frac{-8}{\sqrt{14}\sqrt{21}} = \frac{-8}{\sqrt{294}} \approx -0.466 Step 4: Solve for θ\theta. θ=cos⁡−1(−0.466)≈118∘\theta = \cos^{-1}(-0.466) \approx 118^{\circ}

Explanation:

We use the geometric definition of the dot product to isolate cos⁡θ\cos \theta. Since the dot product is negative, we expect an obtuse angle (between 90∘90^{\circ} and 180∘180^{\circ}).

Problem 2:

Find a vector perpendicular to both a⃗=i+2j−k\vec{a} = \mathbf{i} + 2\mathbf{j} - \mathbf{k} and b⃗=2i+3k\vec{b} = 2\mathbf{i} + 3\mathbf{k}.

Solution:

Step 1: Set up the cross product determinant. a⃗×b⃗=∣ijk12−1203∣\vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & 0 & 3 \end{vmatrix} Step 2: Expand the determinant along the top row. a⃗×b⃗=i(2⋅3−(−1)⋅0)−j(1⋅3−(−1)⋅2)+k(1⋅0−2⋅2)\vec{a} \times \vec{b} = \mathbf{i}(2 \cdot 3 - (-1) \cdot 0) - \mathbf{j}(1 \cdot 3 - (-1) \cdot 2) + \mathbf{k}(1 \cdot 0 - 2 \cdot 2) a⃗×b⃗=i(6)−j(3+2)+k(−4)\vec{a} \times \vec{b} = \mathbf{i}(6) - \mathbf{j}(3 + 2) + \mathbf{k}(-4) a⃗×b⃗=6i−5j−4k\vec{a} \times \vec{b} = 6\mathbf{i} - 5\mathbf{j} - 4\mathbf{k} Step 3: Verification (Optional). a⃗⋅(6i−5j−4k)=1(6)+2(−5)+(−1)(−4)=6−10+4=0\vec{a} \cdot (6\mathbf{i} - 5\mathbf{j} - 4\mathbf{k}) = 1(6) + 2(-5) + (-1)(-4) = 6 - 10 + 4 = 0

Explanation:

The cross product of two vectors always produces a vector orthogonal to both. We use the determinant method to calculate the components, ensuring the signs are handled correctly (especially the negative for the j\mathbf{j} component).