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Vector Algebra - Scalar (dot) product of vectors

Grade 12CBSE

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

🔑Concepts

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Definition of Scalar Product: The scalar (or dot) product of two non-zero vectors a⃗\vec{a} and b⃗\vec{b} is a real number 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 the two vectors. Visually, imagine two arrows originating from the same point; the dot product measures how much one vector extends in the direction of the other, scaled by their magnitudes.

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Angle between Vectors: The angle θ\theta between vectors is restricted to the interval 0≤θ≤π0 \leq \theta \leq \pi. If a⃗⋅b⃗>0\vec{a} \cdot \vec{b} > 0, the angle is acute (0≤θ<π/20 \leq \theta < \pi/2). If a⃗⋅b⃗<0\vec{a} \cdot \vec{b} < 0, the angle is obtuse (π/2<θ≤π\pi/2 < \theta \leq \pi). Visually, an acute angle shows both vectors pointing generally in the same direction, while an obtuse angle shows them pointing away from each other.

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Orthogonality Condition: 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. In a coordinate system, this corresponds to the vectors forming a perfect 90∘90^{\circ} angle, such as the relationship between the X and Y axes.

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Dot Product of Unit Vectors: For the standard mutually perpendicular unit vectors i^\hat{i}, j^\hat{j}, and k^\hat{k} along the axes, the dot product of a vector with itself is 1 (i^⋅i^=j^⋅j^=k^⋅k^=1\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1), while the dot product between different unit vectors is 0 (i^⋅j^=j^⋅k^=k^⋅i^=0\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0).

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Scalar Product in Component Form: 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}, then a⃗⋅b⃗=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3. This algebraic method allows for calculation without knowing the angle θ\theta explicitly.

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Projection of a Vector: The projection of vector a⃗\vec{a} on vector b⃗\vec{b} is given by a⃗⋅b⃗∣b⃗∣\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}. Visually, this is the length of the 'shadow' that vector a⃗\vec{a} casts on the line containing vector b⃗\vec{b} when light is shining perpendicular to b⃗\vec{b}. It forms the base of a right-angled triangle where a⃗\vec{a} is the hypotenuse.

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Distributive Property: The scalar product is distributive over addition, meaning a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}. Geometrically, the sum of the projections of two vectors onto a third vector equals the projection of their resultant vector onto that same third vector.

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Self-Dot Product and Magnitude: The dot product of a vector with itself is equal to the square of its magnitude: a⃗⋅a⃗=∣a⃗∣2\vec{a} \cdot \vec{a} = |\vec{a}|^2. This provides a crucial link between the algebraic dot product and the geometric length of the vector arrow.

📐Formulae

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

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=a1b1+a2b2+a3b3a12+a22+a32b12+b22+b32\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{a_1 b_1 + a_2 b_2 + a_3 b_3}{\sqrt{a_1^2 + a_2^2 + a_3^2} \sqrt{b_1^2 + b_2^2 + b_3^2}}

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3

Projection of a⃗ on b⃗=1∣b⃗∣(a⃗⋅b⃗)\text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{1}{|\vec{b}|} (\vec{a} \cdot \vec{b})

Projection vector of a⃗ on b⃗=(a⃗⋅b⃗∣b⃗∣2)b⃗\text{Projection vector of } \vec{a} \text{ on } \vec{b} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}

∣a⃗+b⃗∣2=∣a⃗∣2+∣b⃗∣2+2(a⃗⋅b⃗)|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})

∣a⃗−b⃗∣2=∣a⃗∣2+∣b⃗∣2−2(a⃗⋅b⃗)|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})

💡Examples

Problem 1:

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

Solution:

Step 1: Calculate the dot product a⃗⋅b⃗\vec{a} \cdot \vec{b}. a⃗⋅b⃗=(1)(1)+(1)(−1)+(−1)(1)=1−1−1=−1\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1 Step 2: Calculate the magnitudes ∣a⃗∣|\vec{a}| and ∣b⃗∣|\vec{b}|. ∣a⃗∣=12+12+(−1)2=3|\vec{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} ∣b⃗∣=12+(−1)2+12=3|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} Step 3: Use the formula cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}. cos⁡θ=−13⋅3=−13\cos \theta = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = \frac{-1}{3} Step 4: Solve for θ\theta. θ=cos⁡−1(−13)\theta = \cos^{-1}\left(-\frac{1}{3}\right)

Explanation:

To find the angle, we first calculate the scalar product using components and then divide it by the product of the magnitudes of the two vectors to find cos⁡θ\cos \theta.

Problem 2:

Find the projection of the vector a⃗=2i^+3j^+2k^\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k} on the vector b⃗=i^+2j^+k^\vec{b} = \hat{i} + 2\hat{j} + \hat{k}.

Solution:

Step 1: Calculate the dot product a⃗⋅b⃗\vec{a} \cdot \vec{b}. a⃗⋅b⃗=(2)(1)+(3)(2)+(2)(1)=2+6+2=10\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10 Step 2: Calculate the magnitude of vector b⃗\vec{b}. ∣b⃗∣=12+22+12=1+4+1=6|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} Step 3: Calculate the projection using the formula a⃗⋅b⃗∣b⃗∣\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}. Projection=106\text{Projection} = \frac{10}{\sqrt{6}} Step 4: Rationalize the denominator (optional). Projection=1066=563\text{Projection} = \frac{10\sqrt{6}}{6} = \frac{5\sqrt{6}}{3}

Explanation:

The projection is a scalar value representing the magnitude of the component of a⃗\vec{a} along the direction of b⃗\vec{b}. It is found by dividing the dot product of the vectors by the magnitude of the vector on which the projection is being taken.