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Vectors - Scalar (Dot) Product of Vectors

Grade 12ICSE

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

🔑Concepts

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Geometric Definition: The dot product (or scalar product) of two vectors a⃗\vec{a} and b⃗\vec{b} is defined as a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta, where θ\theta is the angle between them (0≤θ≤π0 \le \theta \le \pi). Visually, if you draw both vectors from the same origin, the dot product represents the length of the projection of one vector onto the other, multiplied by the magnitude of that other vector.

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Algebraic Definition in Components: In a 3D Cartesian coordinate system, if a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} and b⃗=b1i^+b2j^+b3k^\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}, their dot product is the sum of the products of their corresponding components: a1b1+a2b2+a3b3a_1b_1 + a_2b_2 + a_3b_3. This result is always a scalar quantity, not a vector.

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Condition for Perpendicularity: Two non-zero vectors a⃗\vec{a} and b⃗\vec{b} are perpendicular (orthogonal) if and only if their dot product is zero (00). Geometrically, this occurs because the angle θ=90∘\theta = 90^\circ and cos⁡90∘=0\cos 90^\circ = 0. This is a fundamental tool for solving problems involving right angles in 3D space.

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Projection of a Vector: The scalar projection of a⃗\vec{a} on b⃗\vec{b} is the magnitude of the 'shadow' that a⃗\vec{a} casts onto the line of b⃗\vec{b}. It is calculated as a⃗⋅b⃗∣b⃗∣\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}. If we need the vector projection, we multiply this scalar by the unit vector in the direction of b⃗\vec{b}, which is b⃗∣b⃗∣\frac{\vec{b}}{|\vec{b}|}.

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Properties of Dot Product: The dot product is commutative, meaning a⃗⋅b⃗=b⃗⋅a⃗\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}. It is also distributive over addition: a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}. Additionally, the dot product of a vector with itself gives the square of its magnitude: a⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2.

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Angle Between Two Vectors: The angle θ\theta between two vectors can be determined by rearranging the dot product formula: cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}. If the dot product is positive, the angle is acute; if negative, the angle is obtuse; and if zero, the angle is exactly 90∘90^\circ.

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Dot Products of Unit Vectors: For the standard basis vectors i^,j^,k^\hat{i}, \hat{j}, \hat{k}, the dot product of any unit vector with itself is 11 (e.g., i^⋅i^=1\hat{i} \cdot \hat{i} = 1) because they are parallel, and the dot product between any two different unit vectors is 00 (e.g., i^⋅j^=0\hat{i} \cdot \hat{j} = 0) because they are mutually perpendicular.

📐Formulae

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

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

cos⁡θ=a1b1+a2b2+a3b3a12+a22+a32b12+b22+b32\cos \theta = \frac{a_1b_1 + a_2b_2 + a_3b_3}{\sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}}

\text{Scalar Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

\text{Vector Projection of } \vec{a} \text{ on } \vec{b} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$

∣a⃗∣=a⃗⋅a⃗|\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}

\vec{a} \perp \vec{b} \iff \vec{a} \cdot \vec{b} = 0$

💡Examples

Problem 1:

Find the angle between the vectors a⃗=i^+2j^−k^\vec{a} = \hat{i} + 2\hat{j} - \hat{k} and b⃗=−i^+j^−2k^\vec{b} = -\hat{i} + \hat{j} - 2\hat{k}.

Solution:

  1. Calculate the dot product: a⃗⋅b⃗=(1)(−1)+(2)(1)+(−1)(−2)=−1+2+2=3\vec{a} \cdot \vec{b} = (1)(-1) + (2)(1) + (-1)(-2) = -1 + 2 + 2 = 3. \ 2. Calculate the magnitude of a⃗\vec{a}: ∣a⃗∣=12+22+(−1)2=1+4+1=6|\vec{a}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}. \ 3. Calculate the magnitude of b⃗\vec{b}: ∣b⃗∣=(−1)2+12+(−2)2=1+1+4=6|\vec{b}| = \sqrt{(-1)^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}. \ 4. Use the cosine formula: cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=366=36=12\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{3}{\sqrt{6}\sqrt{6}} = \frac{3}{6} = \frac{1}{2}. \ 5. Since cos⁡θ=12\cos \theta = \frac{1}{2}, θ=cos⁡−1(12)=60∘\theta = \cos^{-1}(\frac{1}{2}) = 60^\circ or π3\frac{\pi}{3} radians.

Explanation:

This solution uses the algebraic component method to find the dot product and magnitudes, then substitutes them into the geometric definition to solve for the angle.

Problem 2:

Find the value of λ\lambda such that the vectors a⃗=2i^+λj^+k^\vec{a} = 2\hat{i} + \lambda\hat{j} + \hat{k} and b⃗=i^−2j^+3k^\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k} are perpendicular.

Solution:

  1. For vectors to be perpendicular, their dot product must be zero: a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0. \ 2. Write the dot product in terms of components: (2)(1)+(λ)(−2)+(1)(3)=0(2)(1) + (\lambda)(-2) + (1)(3) = 0. \ 3. Simplify the equation: 2−2λ+3=02 - 2\lambda + 3 = 0. \ 4. Solve for λ\lambda: 5−2λ=0  ⟹  2λ=5  ⟹  λ=52=2.55 - 2\lambda = 0 \implies 2\lambda = 5 \implies \lambda = \frac{5}{2} = 2.5.

Explanation:

The problem applies the orthogonality condition where the dot product of two perpendicular vectors is zero, creating a linear equation to solve for the unknown scalar.