Scalar Product (Dot product): Exercises and Solved Problems

The scalar product(also called the dot product or inner product) is a fundamental operation in vector algebra. Given two vectors \(\vec{a}\) and \(\vec{b}\), their scalar product is defined both geometrically as \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\) and algebraically via component-wise multiplication. Mastering the scalar product is essential for computing angles between vectors, testing orthogonality, finding vector projections, and solving physics problems involving work and energy. The exercises below are organized by concept and progress from easy to hard, covering 2D and 3D cases in both \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) notation and component form.

Computing the Scalar Product from Components

This section develops fluency with the algebraic formula \(\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z\). Exercises begin in 2D with integer components and progress to 3D vectors with negative and fractional entries, building the mechanical skill needed for all later topics.

Problem 1: Dot Product of Two 2D Vectors

Easy

Let \(\vec{u} = \langle 4, 3 \rangle\) and \(\vec{v} = \langle -2, 5 \rangle\).

  1. Compute the scalar product \(\vec{u} \cdot \vec{v}\).
  2. State whether the result is a scalar or a vector, and explain why.
Hint
Multiply the corresponding components and add the results: \(u_x v_x + u_y v_y\). The dot product always produces a single number, not a vector.
View Solution
Solution to question 1:
\[ \vec{u} \cdot \vec{v} = (4)(-2) + (3)(5) = -8 + 15 = 7 \]
Solution to question 2: The result \(7\) is a scalar. By definition, the dot product combines two vectors to yield a real number, not a new vector. This is precisely why the operation is called the scalar product.

Problem 2: Dot Product of Two 3D Vectors

Easy

Consider the vectors \(\vec{a} = 3\mathbf{i} – 2\mathbf{j} + 5\mathbf{k}\) and \(\vec{b} = \mathbf{i} + 4\mathbf{j} – \mathbf{k}\).

  1. Calculate \(\vec{a} \cdot \vec{b}\) using the component formula.
  2. Verify that \(\vec{b} \cdot \vec{a}\) gives the same result, illustrating commutativity.
Hint
Apply \(\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z\) with the coefficients of \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) acting as the \(x\), \(y\), \(z\) components respectively.
View Solution
Solution to question 1:
\[ \vec{a} \cdot \vec{b} = (3)(1) + (-2)(4) + (5)(-1) = 3 – 8 – 5 = -10 \]
Solution to question 2:
\[ \vec{b} \cdot \vec{a} = (1)(3) + (4)(-2) + (-1)(5) = 3 – 8 – 5 = -10 \]
Both products equal \(-10\), confirming that the scalar product is commutative: \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\).

Problem 3: Scalar Product Using Magnitudes and Angle

Medium

Two vectors \(\vec{p}\) and \(\vec{q}\) satisfy \(|\vec{p}| = 6\), \(|\vec{q}| = 9\), and the angle between them is \(\theta = \dfrac{\pi}{3}\).

  1. Compute \(\vec{p} \cdot \vec{q}\) using the geometric definition.
  2. A third vector \(\vec{r}\) has the same magnitude as \(\vec{q}\) but the angle between \(\vec{p}\) and \(\vec{r}\) is \(\theta’ = \dfrac{2\pi}{3}\). Find \(\vec{p} \cdot \vec{r}\) and explain the sign of the result geometrically.
Hint
Use \(\vec{p} \cdot \vec{q} = |\vec{p}||\vec{q}|\cos\theta\). For part 2, recall that \(\cos\!\left(\tfrac{2\pi}{3}\right) = -\tfrac{1}{2}\); think about what it means geometrically when the angle between two vectors is obtuse.
View Solution
Solution to question 1:
\[ \vec{p} \cdot \vec{q} = 6 \times 9 \times \cos\!\left(\frac{\pi}{3}\right) = 54 \times \frac{1}{2} = 27 \]
Solution to question 2:
\[ \vec{p} \cdot \vec{r} = 6 \times 9 \times \cos\!\left(\frac{2\pi}{3}\right) = 54 \times \left(-\frac{1}{2}\right) = -27 \]
The negative sign indicates that the projection of \(\vec{r}\) onto \(\vec{p}\) points in the opposite direction to \(\vec{p}\). Whenever the angle between two vectors is obtuse (\(90° < \theta < 180°\)), their scalar product is negative.

Problem 4: Combining Both Definitions

Hard

Let \(\vec{a} = 2\mathbf{i} – \mathbf{j} + 2\mathbf{k}\) and \(\vec{b} = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).

  1. Compute \(\vec{a} \cdot \vec{b}\) using components.
  2. Compute \(|\vec{a}|\) and \(|\vec{b}|\).
  3. Use both results to find the cosine of the angle between \(\vec{a}\) and \(\vec{b}\), then verify consistency between the two formulas.
Hint
After finding the dot product and both magnitudes, isolate \(\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\). Check that this value lies in \([-1,1]\).
View Solution
Solution to question 1:
\[ \vec{a} \cdot \vec{b} = (2)(-1) + (-1)(2) + (2)(2) = -2 – 2 + 4 = 0 \]
Solution to question 2:
\[ |\vec{a}| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3, \qquad |\vec{b}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
Solution to question 3:
\[ \cos\theta = \frac{0}{3 \times 3} = 0 \implies \theta = 90° \]
Both definitions are consistent: the component formula gives \(\vec{a}\cdot\vec{b} = 0\), and the geometric formula confirms \(\cos\theta = 0\), meaning the vectors are perpendicular.

Angle Between Vectors Using the Scalar Product

Finding the angle between two vectors is one of the most common applications of the scalar product. By rearranging the geometric definition, one obtains \(\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\). The exercises in this section cover acute, right, and obtuse angles in both 2D and 3D, and introduce the role of the arccosine function.

Problem 5: Angle Between Two 2D Vectors

Easy

Find the angle \(\theta\) between the vectors \(\vec{u} = \langle 1, \sqrt{3} \rangle\) and \(\vec{v} = \langle \sqrt{3}, 1 \rangle\).

  1. Compute the scalar product \(\vec{u} \cdot \vec{v}\).
  2. Compute \(|\vec{u}|\) and \(|\vec{v}|\), then determine \(\theta\) in degrees.
Hint
Both vectors have magnitude 2. Once you have \(\cos\theta\), look for an exact value among the standard angles \(0°, 30°, 45°, 60°, 90°\).
View Solution
Solution to question 1:
\[ \vec{u} \cdot \vec{v} = (1)(\sqrt{3}) + (\sqrt{3})(1) = 2\sqrt{3} \]
Solution to question 2:
\[ |\vec{u}| = \sqrt{1 + 3} = 2, \qquad |\vec{v}| = \sqrt{3 + 1} = 2 \]
\[ \cos\theta = \frac{2\sqrt{3}}{2 \times 2} = \frac{\sqrt{3}}{2} \implies \theta = \arccos\!\left(\frac{\sqrt{3}}{2}\right) = 30° \]

Problem 6: Angle Between Two 3D Vectors

Medium

Given \(\vec{a} = \langle 2, 1, 3 \rangle\) and \(\vec{b} = \langle 3, -2, 1 \rangle\):

  1. Find \(\vec{a} \cdot \vec{b}\), \(|\vec{a}|\), and \(|\vec{b}|\).
  2. Calculate the angle \(\theta\) between \(\vec{a}\) and \(\vec{b}\), giving your answer in degrees to two decimal places.
Hint
Compute each magnitude using \(|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}\), then apply \(\theta = \arccos\!\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\right)\).
View Solution
Solution to question 1:
\[ \vec{a} \cdot \vec{b} = (2)(3) + (1)(-2) + (3)(1) = 6 – 2 + 3 = 7 \]
\[ |\vec{a}| = \sqrt{4 + 1 + 9} = \sqrt{14}, \qquad |\vec{b}| = \sqrt{9 + 4 + 1} = \sqrt{14} \]
Solution to question 2:
\[ \cos\theta = \frac{7}{\sqrt{14} \cdot \sqrt{14}} = \frac{7}{14} = \frac{1}{2} \implies \theta = \arccos\!\left(\frac{1}{2}\right) = 60.00° \]

Problem 7: Angle in a Triangle via the Scalar Product

Hard

Points \(A(1, 2, -1)\), \(B(3, 0, 4)\), and \(C(-1, 3, 2)\) form a triangle.

  1. Find the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
  2. Use the scalar product to calculate the angle \(\angle BAC\) at vertex \(A\), in degrees to one decimal place.
  3. Determine the angle \(\angle ABC\) at vertex \(B\) using the same method.
Hint
For \(\angle BAC\), the two relevant vectors both start at \(A\): they are \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\). For \(\angle ABC\), the vectors \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) both start at \(B\). Subtract coordinates carefully.
View Solution
Solution to question 1:
\[ \overrightarrow{AB} = B – A = \langle 2, -2, 5 \rangle, \qquad \overrightarrow{AC} = C – A = \langle -2, 1, 3 \rangle \]
Solution to question 2:
\[ \overrightarrow{AB} \cdot \overrightarrow{AC} = (2)(-2) + (-2)(1) + (5)(3) = -4 – 2 + 15 = 9 \]
\[ |\overrightarrow{AB}| = \sqrt{4 + 4 + 25} = \sqrt{33}, \qquad |\overrightarrow{AC}| = \sqrt{4 + 1 + 9} = \sqrt{14} \]
\[ \cos(\angle BAC) = \frac{9}{\sqrt{33}\,\sqrt{14}} = \frac{9}{\sqrt{462}} \approx \frac{9}{21.49} \approx 0.4188 \]
\[ \angle BAC = \arccos(0.4188) \approx 65.3° \]
Solution to question 3:
\[ \overrightarrow{BA} = A – B = \langle -2, 2, -5 \rangle, \qquad \overrightarrow{BC} = C – B = \langle -4, 3, -2 \rangle \]
\[ \overrightarrow{BA} \cdot \overrightarrow{BC} = (-2)(-4) + (2)(3) + (-5)(-2) = 8 + 6 + 10 = 24 \]
\[ |\overrightarrow{BA}| = \sqrt{33}, \qquad |\overrightarrow{BC}| = \sqrt{16 + 9 + 4} = \sqrt{29} \]
\[ \cos(\angle ABC) = \frac{24}{\sqrt{33}\,\sqrt{29}} = \frac{24}{\sqrt{957}} \approx \frac{24}{30.94} \approx 0.7758 \]
\[ \angle ABC = \arccos(0.7758) \approx 39.0° \]

Orthogonality and Perpendicular Vectors

Two non-zero vectors are orthogonal (perpendicular) if and only if their scalar product equals zero. This section explores orthogonality conditions, including how to find unknown components that make vectors perpendicular, and how to decompose a vector into parallel and orthogonal parts relative to another.

Problem 8: Testing Orthogonality in 2D

Easy

Determine which of the following pairs of vectors are orthogonal.

  1. \(\vec{a} = \langle 3, -4 \rangle\) and \(\vec{b} = \langle 8, 6 \rangle\)
  2. \(\vec{c} = \langle 2, 5 \rangle\) and \(\vec{d} = \langle 5, -2 \rangle\)
Hint
Compute the scalar product for each pair. A zero result means orthogonal; a non-zero result means not orthogonal.
View Solution
Solution to question 1:
\[ \vec{a} \cdot \vec{b} = (3)(8) + (-4)(6) = 24 – 24 = 0 \]
Since \(\vec{a} \cdot \vec{b} = 0\), vectors \(\vec{a}\) and \(\vec{b}\) are orthogonal.
Solution to question 2:
\[ \vec{c} \cdot \vec{d} = (2)(5) + (5)(-2) = 10 – 10 = 0 \]
Since \(\vec{c} \cdot \vec{d} = 0\), vectors \(\vec{c}\) and \(\vec{d}\) are also orthogonal.

Problem 9: Finding the Unknown Component for Perpendicularity

Medium

Let \(\vec{A} = \langle -2, -t \rangle\) and \(\vec{B} = \langle -8, t \rangle\), where \(t\) is a real parameter.

  1. Find all values of \(t\) such that \(\vec{A}\) and \(\vec{B}\) are perpendicular.
  2. For each value of \(t\) found, write out the two vectors explicitly.
Hint
Set \(\vec{A} \cdot \vec{B} = 0\) and solve the resulting equation in \(t\). You should obtain a quadratic.
View Solution
Solution to question 1:
\[ \vec{A} \cdot \vec{B} = (-2)(-8) + (-t)(t) = 16 – t^2 = 0 \]
\[ t^2 = 16 \implies t = 4 \quad \text{or} \quad t = -4 \]
Solution to question 2: For \(t = 4\): \(\vec{A} = \langle -2, -4 \rangle\) and \(\vec{B} = \langle -8, 4 \rangle\). For \(t = -4\): \(\vec{A} = \langle -2, 4 \rangle\) and \(\vec{B} = \langle -8, -4 \rangle\). In both cases one can verify \(\vec{A}\cdot\vec{B} = 0\).

Problem 10: Orthogonal Decomposition of a Vector

Hard

Given \(\vec{u} = \langle 3, -1, 2 \rangle\) and \(\vec{v} = \langle 1, 4, -1 \rangle\):

  1. Decompose \(\vec{v}\) into \(\vec{v} = \vec{v}_{\parallel} + \vec{v}_{\perp}\), where \(\vec{v}_{\parallel}\) is parallel to \(\vec{u}\) and \(\vec{v}_{\perp}\) is orthogonal to \(\vec{u}\).
  2. Verify that \(\vec{v}_{\perp} \cdot \vec{u} = 0\).
Hint
The component of \(\vec{v}\) parallel to \(\vec{u}\) is \(\vec{v}_{\parallel} = \dfrac{\vec{v}\cdot\vec{u}}{|\vec{u}|^2}\,\vec{u}\). Then \(\vec{v}_{\perp} = \vec{v} – \vec{v}_{\parallel}\).
View Solution
Solution to question 1:
\[ \vec{v} \cdot \vec{u} = (1)(3) + (4)(-1) + (-1)(2) = 3 – 4 – 2 = -3 \]
\[ |\vec{u}|^2 = 9 + 1 + 4 = 14 \]
\[ \vec{v}_{\parallel} = \frac{-3}{14}\langle 3, -1, 2 \rangle = \left\langle -\frac{9}{14},\; \frac{3}{14},\; -\frac{6}{14} \right\rangle \]
\[ \vec{v}_{\perp} = \vec{v} – \vec{v}_{\parallel} = \left\langle 1 + \frac{9}{14},\; 4 – \frac{3}{14},\; -1 + \frac{6}{14} \right\rangle = \left\langle \frac{23}{14},\; \frac{53}{14},\; -\frac{8}{14} \right\rangle \]
Solution to question 2:
\[ \vec{v}_{\perp} \cdot \vec{u} = \frac{23}{14}(3) + \frac{53}{14}(-1) + \left(-\frac{8}{14}\right)(2) = \frac{69 – 53 – 16}{14} = \frac{0}{14} = 0 \checkmark \]

Vector Projection and Scalar Component

The vector projection of \(\vec{b}\) onto \(\vec{a}\) and the associated scalar projection (component) are central tools in applied mathematics and physics. This section develops both concepts, starting with straightforward cases and building toward problems requiring full projection decompositions.

Problem 11: Scalar Projection in 2D

Easy

Given \(\vec{a} = \langle 4, 3 \rangle\) and \(\vec{b} = \langle 2, -1 \rangle\):

  1. Find the scalar projection of \(\vec{b}\) onto \(\vec{a}\), defined as \(\text{comp}_{\vec{a}}\vec{b} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|}\).
  2. Interpret the sign of the result geometrically.
Hint
Compute the dot product first, then divide by \(|\vec{a}| = \sqrt{a_x^2 + a_y^2}\). A positive value means the projection points in the same direction as \(\vec{a}\); a negative value means it points opposite.
View Solution
Solution to question 1:
\[ \vec{a}\cdot\vec{b} = (4)(2) + (3)(-1) = 8 – 3 = 5 \]
\[ |\vec{a}| = \sqrt{16 + 9} = 5 \]
\[ \text{comp}_{\vec{a}}\vec{b} = \frac{5}{5} = 1 \]
Solution to question 2: The scalar projection equals \(1 > 0\), so the component of \(\vec{b}\) in the direction of \(\vec{a}\) is positive, meaning the angle between the two vectors is acute and \(\vec{b}\) has a net forward component along \(\vec{a}\).

Problem 12: Vector Projection in 3D

Medium

Let \(\vec{a} = \langle 1, -1, 1 \rangle\) and \(\vec{b} = \langle 3, 1, 4 \rangle\).

  1. Find the vector projection \(\text{proj}_{\vec{a}}\vec{b} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\,\vec{a}\).
  2. Find the component of \(\vec{b}\) orthogonal to \(\vec{a}\), denoted \(\vec{b}_{\perp}\).
  3. Verify that \(\vec{b}_{\perp} \perp \vec{a}\).
Hint
Compute \(\vec{a}\cdot\vec{b}\) and \(|\vec{a}|^2\) first. Then \(\vec{b}_{\perp} = \vec{b} – \text{proj}_{\vec{a}}\vec{b}\). To verify perpendicularity, check that the dot product with \(\vec{a}\) is zero.
View Solution
Solution to question 1:
\[ \vec{a}\cdot\vec{b} = (1)(3) + (-1)(1) + (1)(4) = 3 – 1 + 4 = 6 \]
\[ |\vec{a}|^2 = 1 + 1 + 1 = 3 \]
\[ \text{proj}_{\vec{a}}\vec{b} = \frac{6}{3}\langle 1,-1,1\rangle = 2\langle 1,-1,1\rangle = \langle 2,-2,2\rangle \]
Solution to question 2:
\[ \vec{b}_{\perp} = \vec{b} – \text{proj}_{\vec{a}}\vec{b} = \langle 3-2,\; 1-(-2),\; 4-2\rangle = \langle 1, 3, 2\rangle \]
Solution to question 3:
\[ \vec{b}_{\perp}\cdot\vec{a} = (1)(1) + (3)(-1) + (2)(1) = 1 – 3 + 2 = 0 \checkmark \]

Problem 13: Component of One Vector in the Direction of Another

Hard

Find the component of \(\vec{b} = 3\mathbf{i} + \mathbf{j} + 4\mathbf{k}\) in the direction of \(\vec{a} = \mathbf{i} – \mathbf{j} + \mathbf{k}\), and express the result both as a scalar and as a vector.

  1. Compute the unit vector \(\hat{a}\) in the direction of \(\vec{a}\).
  2. Find the scalar component \(\text{comp}_{\vec{a}}\vec{b} = \vec{b}\cdot\hat{a}\).
  3. Express the vector component (projection) of \(\vec{b}\) along \(\vec{a}\).
Hint
The unit vector is \(\hat{a} = \vec{a}/|\vec{a}|\). Multiplying the scalar component by \(\hat{a}\) gives the vector component. All three sub-questions build on each other.
View Solution
Solution to question 1:
\[ |\vec{a}| = \sqrt{1 + 1 + 1} = \sqrt{3}, \qquad \hat{a} = \frac{1}{\sqrt{3}}\langle 1,-1,1\rangle \]
Solution to question 2:
\[ \text{comp}_{\vec{a}}\vec{b} = \vec{b}\cdot\hat{a} = \frac{1}{\sqrt{3}}\bigl[(3)(1)+(1)(-1)+(4)(1)\bigr] = \frac{3-1+4}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]
Solution to question 3:
\[ \text{proj}_{\vec{a}}\vec{b} = \left(\text{comp}_{\vec{a}}\vec{b}\right)\hat{a} = 2\sqrt{3}\cdot\frac{1}{\sqrt{3}}\langle 1,-1,1\rangle = 2\langle 1,-1,1\rangle = \langle 2,-2,2\rangle \]

Algebraic Properties: Commutativity, Distributivity, and Bilinearity

The scalar product satisfies key algebraic properties: it is commutative (\(\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}\)), distributive over addition (\(\vec{a}\cdot(\vec{b}+\vec{c}) = \vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}\)), and compatible with scalar multiplication. These exercises develop algebraic fluency and the ability to manipulate dot product expressions symbolically, which is essential for proofs and derivations.

Problem 14: Distributivity of the Scalar Product

Medium

Let \(\vec{a} = \langle 1, 2, -1 \rangle\), \(\vec{b} = \langle 3, 0, 4 \rangle\), and \(\vec{c} = \langle -2, 1, 1 \rangle\).

  1. Compute \(\vec{a} \cdot (\vec{b} + \vec{c})\) directly by first adding \(\vec{b}\) and \(\vec{c}\).
  2. Compute \(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}\) separately, then verify both results are equal.
Hint
Add vectors component-wise: \(\vec{b}+\vec{c} = \langle b_x+c_x, b_y+c_y, b_z+c_z\rangle\). Then apply the dot product formula to each expression.
View Solution
Solution to question 1:
\[ \vec{b}+\vec{c} = \langle 1, 1, 5\rangle \]
\[ \vec{a}\cdot(\vec{b}+\vec{c}) = (1)(1)+(2)(1)+(-1)(5) = 1+2-5 = -2 \]
Solution to question 2:
\[ \vec{a}\cdot\vec{b} = (1)(3)+(2)(0)+(-1)(4) = 3+0-4 = -1 \]
\[ \vec{a}\cdot\vec{c} = (1)(-2)+(2)(1)+(-1)(1) = -2+2-1 = -1 \]
\[ \vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c} = -1 + (-1) = -2 \]
Both methods give \(-2\), confirming the distributive property.

Problem 15: Expanding \(|\vec{a}+\vec{b}|^2\) Using the Scalar Product

Hard

Prove the following identity using properties of the scalar product, then verify it numerically with \(\vec{a} = \langle 2, -1, 3 \rangle\) and \(\vec{b} = \langle 1, 4, -2 \rangle\):

\[ |\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2 \]

  1. Prove the identity algebraically using the definition \(|\vec{v}|^2 = \vec{v}\cdot\vec{v}\) and distributivity.
  2. Verify both sides numerically with the given vectors.
Hint
Write \(|\vec{a}+\vec{b}|^2 = (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})\) and expand using the distributive property. Use commutativity to simplify the middle terms.
View Solution
Solution to question 1 (proof):
\[ |\vec{a}+\vec{b}|^2 = (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) = \vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{a} + \vec{b}\cdot\vec{b} \]
Since \(\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}\):
\[ = |\vec{a}|^2 + 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2 \qquad \square \]
Solution to question 2 (numerical verification):
\[ \vec{a}+\vec{b} = \langle 3, 3, 1\rangle \implies |\vec{a}+\vec{b}|^2 = 9+9+1 = 19 \]
\[ |\vec{a}|^2 = 4+1+9 = 14, \quad |\vec{b}|^2 = 1+16+4 = 21 \]
\[ \vec{a}\cdot\vec{b} = (2)(1)+(-1)(4)+(3)(-2) = 2-4-6 = -8 \]
\[ |\vec{a}|^2 + 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2 = 14 + 2(-8) + 21 = 14 – 16 + 21 = 19 \checkmark \]

Applications: Work Done by a Force

In physics, the work done by a constant force \(\vec{F}\) over a displacement \(\vec{d}\) is defined as the scalar product \(W = \vec{F}\cdot\vec{d}\). This section applies the scalar product to mechanical work, covering cases where force and displacement are parallel, at an angle, and in 3D space.

Problem 16: Work Done by a 2D Force

Easy

A force \(\vec{F} = \langle 5, 3 \rangle\) N acts on an object that undergoes a displacement \(\vec{d} = \langle 4, -2 \rangle\) m.

  1. Calculate the work done \(W = \vec{F}\cdot\vec{d}\), giving your answer in joules.
  2. If the same force were applied perpendicular to the displacement, state the work done and explain why.
Hint
Work is simply the dot product of force and displacement vectors. Recall that perpendicular vectors have a dot product of zero — no energy is transferred when force and motion are at right angles.
View Solution
Solution to question 1:
\[ W = \vec{F}\cdot\vec{d} = (5)(4) + (3)(-2) = 20 – 6 = 14 \text{ J} \]
Solution to question 2: If the force is perpendicular to the displacement, \(W = |\vec{F}||\vec{d}|\cos 90° = 0\). No work is done. A physical example is a centripetal force acting perpendicular to circular motion — it changes the direction of velocity but transfers no energy to the object.

Problem 17: Work Done by a 3D Force

Medium

A particle moves from point \(P(2, 1, 2)\) to point \(Q(3, 2, 5)\) (in metres) under a constant force \(\vec{F} = 7\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\) N.

  1. Find the displacement vector \(\overrightarrow{PQ}\).
  2. Calculate the work done by the force during this displacement.
Hint
The displacement vector is \(\overrightarrow{PQ} = Q – P\). Work is then \(W = \vec{F}\cdot\overrightarrow{PQ}\).
View Solution
Solution to question 1:
\[ \overrightarrow{PQ} = \langle 3-2,\; 2-1,\; 5-2\rangle = \langle 1, 1, 3\rangle \text{ m} \]
Solution to question 2:
\[ W = \vec{F}\cdot\overrightarrow{PQ} = (7)(1)+(5)(1)+(2)(3) = 7+5+6 = 18 \text{ J} \]

Problem 18: Work Done When Force Acts at an Angle

Hard

A force of magnitude \(|\vec{F}| = 20\) N acts on a crate, making an angle of \(40°\) with the horizontal floor. The crate is dragged horizontally a distance of \(8\) m.

  1. Model the displacement as \(\vec{d} = \langle 8, 0 \rangle\) m and the force as \(\vec{F} = \langle 20\cos 40°, 20\sin 40° \rangle\) N. Compute the work done in component form.
  2. Apply the geometric definition \(W = |\vec{F}||\vec{d}|\cos\theta\) and verify that both approaches agree.
  3. Determine the percentage of the total force magnitude that contributes to useful work along the direction of motion.
Hint
The angle between \(\vec{F}\) and \(\vec{d}\) (horizontal) is exactly \(40°\). The fraction of force doing useful work is \(\cos 40°\).
View Solution
Solution to question 1:
\[ \vec{F} \approx \langle 15.32,\; 12.86\rangle \text{ N} \]
\[ W = \vec{F}\cdot\vec{d} = (15.32)(8) + (12.86)(0) \approx 122.6 \text{ J} \]
Solution to question 2:
\[ W = 20 \times 8 \times \cos 40° = 160 \times 0.7660 \approx 122.6 \text{ J} \checkmark \]
Both methods agree.
Solution to question 3:
\[ \frac{|\vec{F}|\cos 40°}{|\vec{F}|} = \cos 40° \approx 0.766 = 76.6\% \]
Only \(76.6\%\) of the applied force contributes to horizontal motion; the remaining \(23.4\%\) acts vertically against the floor’s normal reaction.