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Dot Product (Scalar Product): Formula and Solved Problems

Type
Practice problems
Problems
18
Levels
Easy · Medium · Hard

The dot product, also called the scalar product, multiplies two vectors and gives a number: a⃗⋅b⃗=a1b1+a2b2+a3b3=∣a⃗∣ ∣b⃗∣cos⁡θ\vec a\cdot\vec b = a_1b_1+a_2b_2+a_3b_3 = |\vec a|\,|\vec b|\cos\theta. Below are the formulas you need, then solved problems from easy to hard: angles between vectors, orthogonality, projections and work.

Contents8 sections

Dot product formula

  • Algebraic form (components): a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3
  • Geometric form (θ\theta is the angle between the vectors): a⃗⋅b⃗=∣a⃗∣ ∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a|\,|\vec b|\cos\theta
  • Angle between two vectors: cos⁡θ=a⃗⋅b⃗∣a⃗∣ ∣b⃗∣\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a|\,|\vec b|}.
  • Orthogonal vectors: a⃗⊥b⃗  ⟺  a⃗⋅b⃗=0\vec a\perp\vec b \iff \vec a\cdot\vec b = 0.
  • Projection of a⃗\vec a onto b⃗\vec b: proj⁡b⃗a⃗=a⃗⋅b⃗∣b⃗∣2 b⃗\operatorname{proj}_{\vec b}\vec a = \dfrac{\vec a\cdot\vec b}{|\vec b|^2}\,\vec b.
  • Properties: a⃗⋅b⃗=b⃗⋅a⃗\vec a\cdot\vec b=\vec b\cdot\vec a, a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c, a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a=|\vec a|^2.

Example: a⃗=⟨1,2,2⟩\vec a=\langle 1,2,2\rangle, b⃗=⟨2,−1,2⟩\vec b=\langle 2,-1,2\rangle give a⃗⋅b⃗=2−2+4=4\vec a\cdot\vec b = 2-2+4 = 4. Both vectors have length 3, so cos⁡θ=49\cos\theta = \frac49 and θ≈63.6∘\theta\approx 63.6^\circ.

Computing the Scalar Product from Components

This section develops fluency with the algebraic formula a⃗⋅b⃗=axbx+ayby+azbz\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 u⃗=⟨4,3⟩\vec{u} = \langle 4, 3 \rangle and v⃗=⟨−2,5⟩\vec{v} = \langle -2, 5 \rangle.

  1. Compute the scalar product u⃗⋅v⃗\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: uxvx+uyvyu_x v_x + u_y v_y. The dot product always produces a single number, not a vector.
View Solution
Solution to question 1: u⃗⋅v⃗=(4)(−2)+(3)(5)=−8+15=7 \vec{u} \cdot \vec{v} = (4)(-2) + (3)(5) = -8 + 15 = 7 Solution to question 2: The result 77 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 a⃗=3i−2j+5k\vec{a} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} and b⃗=i+4j−k\vec{b} = \mathbf{i} + 4\mathbf{j} - \mathbf{k}.

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

Problem 3: Scalar Product Using Magnitudes and Angle

Medium

Two vectors p⃗\vec{p} and q⃗\vec{q} satisfy ∣p⃗∣=6|\vec{p}| = 6, ∣q⃗∣=9|\vec{q}| = 9, and the angle between them is θ=π3\theta = \dfrac{\pi}{3}.

  1. Compute p⃗⋅q⃗\vec{p} \cdot \vec{q} using the geometric definition.
  2. A third vector r⃗\vec{r} has the same magnitude as q⃗\vec{q} but the angle between p⃗\vec{p} and r⃗\vec{r} is θ′=2π3\theta' = \dfrac{2\pi}{3}. Find p⃗⋅r⃗\vec{p} \cdot \vec{r} and explain the sign of the result geometrically.
Hint
Use p⃗⋅q⃗=∣p⃗∣∣q⃗∣cos⁡θ\vec{p} \cdot \vec{q} = |\vec{p}||\vec{q}|\cos\theta. For part 2, recall that cos⁡ ⁣(2π3)=−12\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: p⃗⋅q⃗=6×9×cos⁡ ⁣(π3)=54×12=27 \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: p⃗⋅r⃗=6×9×cos⁡ ⁣(2π3)=54×(−12)=−27 \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 r⃗\vec{r} onto p⃗\vec{p} points in the opposite direction to p⃗\vec{p}. Whenever the angle between two vectors is obtuse (90°<θ<180°90° < \theta < 180°), their scalar product is negative.

Problem 4: Combining Both Definitions

Hard

Let a⃗=2i−j+2k\vec{a} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k} and b⃗=−i+2j+2k\vec{b} = -\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}.

  1. Compute a⃗⋅b⃗\vec{a} \cdot \vec{b} using components.
  2. Compute ∣a⃗∣|\vec{a}| and ∣b⃗∣|\vec{b}|.
  3. Use both results to find the cosine of the angle between a⃗\vec{a} and b⃗\vec{b}, then verify consistency between the two formulas.
Hint
After finding the dot product and both magnitudes, isolate cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}. Check that this value lies in [−1,1][-1,1].
View Solution
Solution to question 1: a⃗⋅b⃗=(2)(−1)+(−1)(2)+(2)(2)=−2−2+4=0 \vec{a} \cdot \vec{b} = (2)(-1) + (-1)(2) + (2)(2) = -2 - 2 + 4 = 0 Solution to question 2: ∣a⃗∣=4+1+4=9=3,∣b⃗∣=1+4+4=9=3 |\vec{a}| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3, \qquad |\vec{b}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 Solution to question 3: cos⁡θ=03×3=0  ⟹  θ=90° \cos\theta = \frac{0}{3 \times 3} = 0 \implies \theta = 90° Both definitions are consistent: the component formula gives a⃗⋅b⃗=0\vec{a}\cdot\vec{b} = 0, and the geometric formula confirms cos⁡θ=0\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⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\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 θ\theta between the vectors u⃗=⟨1,3⟩\vec{u} = \langle 1, \sqrt{3} \rangle and v⃗=⟨3,1⟩\vec{v} = \langle \sqrt{3}, 1 \rangle.

  1. Compute the scalar product u⃗⋅v⃗\vec{u} \cdot \vec{v}.
  2. Compute ∣u⃗∣|\vec{u}| and ∣v⃗∣|\vec{v}|, then determine θ\theta in degrees.
Hint
Both vectors have magnitude 2. Once you have cos⁡θ\cos\theta, look for an exact value among the standard angles 0°,30°,45°,60°,90°0°, 30°, 45°, 60°, 90°.
View Solution
Solution to question 1: u⃗⋅v⃗=(1)(3)+(3)(1)=23 \vec{u} \cdot \vec{v} = (1)(\sqrt{3}) + (\sqrt{3})(1) = 2\sqrt{3} Solution to question 2: ∣u⃗∣=1+3=2,∣v⃗∣=3+1=2 |\vec{u}| = \sqrt{1 + 3} = 2, \qquad |\vec{v}| = \sqrt{3 + 1} = 2 cos⁡θ=232×2=32  ⟹  θ=arccos⁡ ⁣(32)=30° \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 a⃗=⟨2,1,3⟩\vec{a} = \langle 2, 1, 3 \rangle and b⃗=⟨3,−2,1⟩\vec{b} = \langle 3, -2, 1 \rangle:

  1. Find a⃗⋅b⃗\vec{a} \cdot \vec{b}, ∣a⃗∣|\vec{a}|, and ∣b⃗∣|\vec{b}|.
  2. Calculate the angle θ\theta between a⃗\vec{a} and b⃗\vec{b}, giving your answer in degrees to two decimal places.
Hint
Compute each magnitude using ∣v⃗∣=vx2+vy2+vz2|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}, then apply θ=arccos⁡ ⁣(a⃗⋅b⃗∣a⃗∣∣b⃗∣)\theta = \arccos\!\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\right).
View Solution
Solution to question 1: a⃗⋅b⃗=(2)(3)+(1)(−2)+(3)(1)=6−2+3=7 \vec{a} \cdot \vec{b} = (2)(3) + (1)(-2) + (3)(1) = 6 - 2 + 3 = 7 ∣a⃗∣=4+1+9=14,∣b⃗∣=9+4+1=14 |\vec{a}| = \sqrt{4 + 1 + 9} = \sqrt{14}, \qquad |\vec{b}| = \sqrt{9 + 4 + 1} = \sqrt{14} Solution to question 2: cos⁡θ=714⋅14=714=12  ⟹  θ=arccos⁡ ⁣(12)=60.00° \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)A(1, 2, -1), B(3,0,4)B(3, 0, 4), and C(−1,3,2)C(-1, 3, 2) form a triangle.

  1. Find the vectors AB→\overrightarrow{AB} and AC→\overrightarrow{AC}.
  2. Use the scalar product to calculate the angle ∠BAC\angle BAC at vertex AA, in degrees to one decimal place.
  3. Determine the angle ∠ABC\angle ABC at vertex BB using the same method.
Hint
For ∠BAC\angle BAC, the two relevant vectors both start at AA: they are AB→\overrightarrow{AB} and AC→\overrightarrow{AC}. For ∠ABC\angle ABC, the vectors BA→\overrightarrow{BA} and BC→\overrightarrow{BC} both start at BB. Subtract coordinates carefully.
View Solution
Solution to question 1: AB→=B−A=⟨2,−2,5⟩,AC→=C−A=⟨−2,1,3⟩ \overrightarrow{AB} = B - A = \langle 2, -2, 5 \rangle, \qquad \overrightarrow{AC} = C - A = \langle -2, 1, 3 \rangle Solution to question 2: AB→⋅AC→=(2)(−2)+(−2)(1)+(5)(3)=−4−2+15=9 \overrightarrow{AB} \cdot \overrightarrow{AC} = (2)(-2) + (-2)(1) + (5)(3) = -4 - 2 + 15 = 9 ∣AB→∣=4+4+25=33,∣AC→∣=4+1+9=14 |\overrightarrow{AB}| = \sqrt{4 + 4 + 25} = \sqrt{33}, \qquad |\overrightarrow{AC}| = \sqrt{4 + 1 + 9} = \sqrt{14} cos⁡(∠BAC)=933 14=9462≈921.49≈0.4188 \cos(\angle BAC) = \frac{9}{\sqrt{33}\,\sqrt{14}} = \frac{9}{\sqrt{462}} \approx \frac{9}{21.49} \approx 0.4188 ∠BAC=arccos⁡(0.4188)≈65.3° \angle BAC = \arccos(0.4188) \approx 65.3° Solution to question 3: BA→=A−B=⟨−2,2,−5⟩,BC→=C−B=⟨−4,3,−2⟩ \overrightarrow{BA} = A - B = \langle -2, 2, -5 \rangle, \qquad \overrightarrow{BC} = C - B = \langle -4, 3, -2 \rangle BA→⋅BC→=(−2)(−4)+(2)(3)+(−5)(−2)=8+6+10=24 \overrightarrow{BA} \cdot \overrightarrow{BC} = (-2)(-4) + (2)(3) + (-5)(-2) = 8 + 6 + 10 = 24 ∣BA→∣=33,∣BC→∣=16+9+4=29 |\overrightarrow{BA}| = \sqrt{33}, \qquad |\overrightarrow{BC}| = \sqrt{16 + 9 + 4} = \sqrt{29} cos⁡(∠ABC)=2433 29=24957≈2430.94≈0.7758 \cos(\angle ABC) = \frac{24}{\sqrt{33}\,\sqrt{29}} = \frac{24}{\sqrt{957}} \approx \frac{24}{30.94} \approx 0.7758 ∠ABC=arccos⁡(0.7758)≈39.0° \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. a⃗=⟨3,−4⟩\vec{a} = \langle 3, -4 \rangle and b⃗=⟨8,6⟩\vec{b} = \langle 8, 6 \rangle
  2. c⃗=⟨2,5⟩\vec{c} = \langle 2, 5 \rangle and d⃗=⟨5,−2⟩\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: a⃗⋅b⃗=(3)(8)+(−4)(6)=24−24=0 \vec{a} \cdot \vec{b} = (3)(8) + (-4)(6) = 24 - 24 = 0 Since a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0, vectors a⃗\vec{a} and b⃗\vec{b} are orthogonal. Solution to question 2: c⃗⋅d⃗=(2)(5)+(5)(−2)=10−10=0 \vec{c} \cdot \vec{d} = (2)(5) + (5)(-2) = 10 - 10 = 0 Since c⃗⋅d⃗=0\vec{c} \cdot \vec{d} = 0, vectors c⃗\vec{c} and d⃗\vec{d} are also orthogonal.

Problem 9: Finding the Unknown Component for Perpendicularity

Medium

Let A⃗=⟨−2,−t⟩\vec{A} = \langle -2, -t \rangle and B⃗=⟨−8,t⟩\vec{B} = \langle -8, t \rangle, where tt is a real parameter.

  1. Find all values of tt such that A⃗\vec{A} and B⃗\vec{B} are perpendicular.
  2. For each value of tt found, write out the two vectors explicitly.
Hint
Set A⃗⋅B⃗=0\vec{A} \cdot \vec{B} = 0 and solve the resulting equation in tt. You should obtain a quadratic.
View Solution
Solution to question 1: A⃗⋅B⃗=(−2)(−8)+(−t)(t)=16−t2=0 \vec{A} \cdot \vec{B} = (-2)(-8) + (-t)(t) = 16 - t^2 = 0 t2=16  ⟹  t=4ort=−4 t^2 = 16 \implies t = 4 \quad \text{or} \quad t = -4 Solution to question 2: For t=4t = 4: A⃗=⟨−2,−4⟩\vec{A} = \langle -2, -4 \rangle and B⃗=⟨−8,4⟩\vec{B} = \langle -8, 4 \rangle. For t=−4t = -4: A⃗=⟨−2,4⟩\vec{A} = \langle -2, 4 \rangle and B⃗=⟨−8,−4⟩\vec{B} = \langle -8, -4 \rangle. In both cases one can verify A⃗⋅B⃗=0\vec{A}\cdot\vec{B} = 0.

Problem 10: Orthogonal Decomposition of a Vector

Hard

Given u⃗=⟨3,−1,2⟩\vec{u} = \langle 3, -1, 2 \rangle and v⃗=⟨1,4,−1⟩\vec{v} = \langle 1, 4, -1 \rangle:

  1. Decompose v⃗\vec{v} into v⃗=v⃗∥+v⃗⊥\vec{v} = \vec{v}_{\parallel} + \vec{v}_{\perp}, where v⃗∥\vec{v}_{\parallel} is parallel to u⃗\vec{u} and v⃗⊥\vec{v}_{\perp} is orthogonal to u⃗\vec{u}.
  2. Verify that v⃗⊥⋅u⃗=0\vec{v}_{\perp} \cdot \vec{u} = 0.
Hint
The component of v⃗\vec{v} parallel to u⃗\vec{u} is v⃗∥=v⃗⋅u⃗∣u⃗∣2 u⃗\vec{v}_{\parallel} = \dfrac{\vec{v}\cdot\vec{u}}{|\vec{u}|^2}\,\vec{u}. Then v⃗⊥=v⃗−v⃗∥\vec{v}_{\perp} = \vec{v} - \vec{v}_{\parallel}.
View Solution
Solution to question 1: v⃗⋅u⃗=(1)(3)+(4)(−1)+(−1)(2)=3−4−2=−3 \vec{v} \cdot \vec{u} = (1)(3) + (4)(-1) + (-1)(2) = 3 - 4 - 2 = -3 ∣u⃗∣2=9+1+4=14 |\vec{u}|^2 = 9 + 1 + 4 = 14 v⃗∥=−314⟨3,−1,2⟩=⟨−914,  314,  −614⟩ \vec{v}_{\parallel} = \frac{-3}{14}\langle 3, -1, 2 \rangle = \left\langle -\frac{9}{14},\; \frac{3}{14},\; -\frac{6}{14} \right\rangle v⃗⊥=v⃗−v⃗∥=⟨1+914,  4−314,  −1+614⟩=⟨2314,  5314,  −814⟩ \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: v⃗⊥⋅u⃗=2314(3)+5314(−1)+(−814)(2)=69−53−1614=014=0✓ \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 b⃗\vec{b} onto a⃗\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 a⃗=⟨4,3⟩\vec{a} = \langle 4, 3 \rangle and b⃗=⟨2,−1⟩\vec{b} = \langle 2, -1 \rangle:

  1. Find the scalar projection of b⃗\vec{b} onto a⃗\vec{a}, defined as compa⃗b⃗=a⃗⋅b⃗∣a⃗∣\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 ∣a⃗∣=ax2+ay2|\vec{a}| = \sqrt{a_x^2 + a_y^2}. A positive value means the projection points in the same direction as a⃗\vec{a}; a negative value means it points opposite.
View Solution
Solution to question 1: a⃗⋅b⃗=(4)(2)+(3)(−1)=8−3=5 \vec{a}\cdot\vec{b} = (4)(2) + (3)(-1) = 8 - 3 = 5 ∣a⃗∣=16+9=5 |\vec{a}| = \sqrt{16 + 9} = 5 compa⃗b⃗=55=1 \text{comp}_{\vec{a}}\vec{b} = \frac{5}{5} = 1 Solution to question 2: The scalar projection equals 1>01 > 0, so the component of b⃗\vec{b} in the direction of a⃗\vec{a} is positive, meaning the angle between the two vectors is acute and b⃗\vec{b} has a net forward component along a⃗\vec{a}.

Problem 12: Vector Projection in 3D

Medium

Let a⃗=⟨1,−1,1⟩\vec{a} = \langle 1, -1, 1 \rangle and b⃗=⟨3,1,4⟩\vec{b} = \langle 3, 1, 4 \rangle.

  1. Find the vector projection proja⃗b⃗=a⃗⋅b⃗∣a⃗∣2 a⃗\text{proj}_{\vec{a}}\vec{b} = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\,\vec{a}.
  2. Find the component of b⃗\vec{b} orthogonal to a⃗\vec{a}, denoted b⃗⊥\vec{b}_{\perp}.
  3. Verify that b⃗⊥⊥a⃗\vec{b}_{\perp} \perp \vec{a}.
Hint
Compute a⃗⋅b⃗\vec{a}\cdot\vec{b} and ∣a⃗∣2|\vec{a}|^2 first. Then b⃗⊥=b⃗−proja⃗b⃗\vec{b}_{\perp} = \vec{b} - \text{proj}_{\vec{a}}\vec{b}. To verify perpendicularity, check that the dot product with a⃗\vec{a} is zero.
View Solution
Solution to question 1: a⃗⋅b⃗=(1)(3)+(−1)(1)+(1)(4)=3−1+4=6 \vec{a}\cdot\vec{b} = (1)(3) + (-1)(1) + (1)(4) = 3 - 1 + 4 = 6 ∣a⃗∣2=1+1+1=3 |\vec{a}|^2 = 1 + 1 + 1 = 3 proja⃗b⃗=63⟨1,−1,1⟩=2⟨1,−1,1⟩=⟨2,−2,2⟩ \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: b⃗⊥=b⃗−proja⃗b⃗=⟨3−2,  1−(−2),  4−2⟩=⟨1,3,2⟩ \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: b⃗⊥⋅a⃗=(1)(1)+(3)(−1)+(2)(1)=1−3+2=0✓ \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 b⃗=3i+j+4k\vec{b} = 3\mathbf{i} + \mathbf{j} + 4\mathbf{k} in the direction of a⃗=i−j+k\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 a^\hat{a} in the direction of a⃗\vec{a}.
  2. Find the scalar component compa⃗b⃗=b⃗⋅a^\text{comp}_{\vec{a}}\vec{b} = \vec{b}\cdot\hat{a}.
  3. Express the vector component (projection) of b⃗\vec{b} along a⃗\vec{a}.
Hint
The unit vector is a^=a⃗/∣a⃗∣\hat{a} = \vec{a}/|\vec{a}|. Multiplying the scalar component by a^\hat{a} gives the vector component. All three sub-questions build on each other.
View Solution
Solution to question 1: ∣a⃗∣=1+1+1=3,a^=13⟨1,−1,1⟩ |\vec{a}| = \sqrt{1 + 1 + 1} = \sqrt{3}, \qquad \hat{a} = \frac{1}{\sqrt{3}}\langle 1,-1,1\rangle Solution to question 2: compa⃗b⃗=b⃗⋅a^=13[(3)(1)+(1)(−1)+(4)(1)]=3−1+43=63=23 \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: proja⃗b⃗=(compa⃗b⃗)a^=23⋅13⟨1,−1,1⟩=2⟨1,−1,1⟩=⟨2,−2,2⟩ \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 (a⃗⋅b⃗=b⃗⋅a⃗\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}), 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}), 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 a⃗=⟨1,2,−1⟩\vec{a} = \langle 1, 2, -1 \rangle, b⃗=⟨3,0,4⟩\vec{b} = \langle 3, 0, 4 \rangle, and c⃗=⟨−2,1,1⟩\vec{c} = \langle -2, 1, 1 \rangle.

  1. Compute a⃗⋅(b⃗+c⃗)\vec{a} \cdot (\vec{b} + \vec{c}) directly by first adding b⃗\vec{b} and c⃗\vec{c}.
  2. Compute a⃗⋅b⃗+a⃗⋅c⃗\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} separately, then verify both results are equal.
Hint
Add vectors component-wise: b⃗+c⃗=⟨bx+cx,by+cy,bz+cz⟩\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: b⃗+c⃗=⟨1,1,5⟩ \vec{b}+\vec{c} = \langle 1, 1, 5\rangle a⃗⋅(b⃗+c⃗)=(1)(1)+(2)(1)+(−1)(5)=1+2−5=−2 \vec{a}\cdot(\vec{b}+\vec{c}) = (1)(1)+(2)(1)+(-1)(5) = 1+2-5 = -2 Solution to question 2: a⃗⋅b⃗=(1)(3)+(2)(0)+(−1)(4)=3+0−4=−1 \vec{a}\cdot\vec{b} = (1)(3)+(2)(0)+(-1)(4) = 3+0-4 = -1 a⃗⋅c⃗=(1)(−2)+(2)(1)+(−1)(1)=−2+2−1=−1 \vec{a}\cdot\vec{c} = (1)(-2)+(2)(1)+(-1)(1) = -2+2-1 = -1 a⃗⋅b⃗+a⃗⋅c⃗=−1+(−1)=−2 \vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c} = -1 + (-1) = -2 Both methods give −2-2, confirming the distributive property.

Problem 15: Expanding ∣a⃗+b⃗∣2|\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 a⃗=⟨2,−1,3⟩\vec{a} = \langle 2, -1, 3 \rangle and b⃗=⟨1,4,−2⟩\vec{b} = \langle 1, 4, -2 \rangle:

∣a⃗+b⃗∣2=∣a⃗∣2+2 a⃗⋅b⃗+∣b⃗∣2 |\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 ∣v⃗∣2=v⃗⋅v⃗|\vec{v}|^2 = \vec{v}\cdot\vec{v} and distributivity.
  2. Verify both sides numerically with the given vectors.
Hint
Write ∣a⃗+b⃗∣2=(a⃗+b⃗)⋅(a⃗+b⃗)|\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): ∣a⃗+b⃗∣2=(a⃗+b⃗)⋅(a⃗+b⃗)=a⃗⋅a⃗+a⃗⋅b⃗+b⃗⋅a⃗+b⃗⋅b⃗ |\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 a⃗⋅b⃗=b⃗⋅a⃗\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}: =∣a⃗∣2+2 a⃗⋅b⃗+∣b⃗∣2□ = |\vec{a}|^2 + 2\,\vec{a}\cdot\vec{b} + |\vec{b}|^2 \qquad \square Solution to question 2 (numerical verification): a⃗+b⃗=⟨3,3,1⟩  ⟹  ∣a⃗+b⃗∣2=9+9+1=19 \vec{a}+\vec{b} = \langle 3, 3, 1\rangle \implies |\vec{a}+\vec{b}|^2 = 9+9+1 = 19 ∣a⃗∣2=4+1+9=14,∣b⃗∣2=1+16+4=21 |\vec{a}|^2 = 4+1+9 = 14, \quad |\vec{b}|^2 = 1+16+4 = 21 a⃗⋅b⃗=(2)(1)+(−1)(4)+(3)(−2)=2−4−6=−8 \vec{a}\cdot\vec{b} = (2)(1)+(-1)(4)+(3)(-2) = 2-4-6 = -8 ∣a⃗∣2+2 a⃗⋅b⃗+∣b⃗∣2=14+2(−8)+21=14−16+21=19✓ |\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 F⃗\vec{F} over a displacement d⃗\vec{d} is defined as the scalar product W=F⃗⋅d⃗W = \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 F⃗=⟨5,3⟩\vec{F} = \langle 5, 3 \rangle N acts on an object that undergoes a displacement d⃗=⟨4,−2⟩\vec{d} = \langle 4, -2 \rangle m.

  1. Calculate the work done W=F⃗⋅d⃗W = \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=F⃗⋅d⃗=(5)(4)+(3)(−2)=20−6=14 J 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=∣F⃗∣∣d⃗∣cos⁡90°=0W = |\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)P(2, 1, 2) to point Q(3,2,5)Q(3, 2, 5) (in metres) under a constant force F⃗=7i+5j+2k\vec{F} = 7\mathbf{i} + 5\mathbf{j} + 2\mathbf{k} N.

  1. Find the displacement vector PQ→\overrightarrow{PQ}.
  2. Calculate the work done by the force during this displacement.
Hint
The displacement vector is PQ→=Q−P\overrightarrow{PQ} = Q - P. Work is then W=F⃗⋅PQ→W = \vec{F}\cdot\overrightarrow{PQ}.
View Solution
Solution to question 1: PQ→=⟨3−2,  2−1,  5−2⟩=⟨1,1,3⟩ m \overrightarrow{PQ} = \langle 3-2,\; 2-1,\; 5-2\rangle = \langle 1, 1, 3\rangle \text{ m} Solution to question 2: W=F⃗⋅PQ→=(7)(1)+(5)(1)+(2)(3)=7+5+6=18 J 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 ∣F⃗∣=20|\vec{F}| = 20 N acts on a crate, making an angle of 40°40° with the horizontal floor. The crate is dragged horizontally a distance of 88 m.

  1. Model the displacement as d⃗=⟨8,0⟩\vec{d} = \langle 8, 0 \rangle m and the force as F⃗=⟨20cos⁡40°,20sin⁡40°⟩\vec{F} = \langle 20\cos 40°, 20\sin 40° \rangle N. Compute the work done in component form.
  2. Apply the geometric definition W=∣F⃗∣∣d⃗∣cos⁡θW = |\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 F⃗\vec{F} and d⃗\vec{d} (horizontal) is exactly 40°40°. The fraction of force doing useful work is cos⁡40°\cos 40°.
View Solution
Solution to question 1: F⃗≈⟨15.32,  12.86⟩ N \vec{F} \approx \langle 15.32,\; 12.86\rangle \text{ N} W=F⃗⋅d⃗=(15.32)(8)+(12.86)(0)≈122.6 J W = \vec{F}\cdot\vec{d} = (15.32)(8) + (12.86)(0) \approx 122.6 \text{ J} Solution to question 2: W=20×8×cos⁡40°=160×0.7660≈122.6 J✓ W = 20 \times 8 \times \cos 40° = 160 \times 0.7660 \approx 122.6 \text{ J} \checkmark Both methods agree. Solution to question 3: ∣F⃗∣cos⁡40°∣F⃗∣=cos⁡40°≈0.766=76.6% \frac{|\vec{F}|\cos 40°}{|\vec{F}|} = \cos 40° \approx 0.766 = 76.6\% Only 76.6%76.6\% of the applied force contributes to horizontal motion; the remaining 23.4%23.4\% acts vertically against the floor’s normal reaction.

Frequently asked questions

Is the dot product the same as the scalar product?

Yes. “Dot product” refers to the notation a⃗⋅b⃗\vec a\cdot\vec b; “scalar product” refers to the result, which is a scalar (a number), not a vector. In Rn\mathbb{R}^n both names mean the same operation, which is also the standard inner product.

What does a negative dot product mean?

The angle between the vectors is obtuse (between 90∘90^\circ and 180∘180^\circ), because cos⁡θ<0\cos\theta<0. A positive dot product means an acute angle, and zero means the vectors are perpendicular.

What is the difference between the dot product and the cross product?

The dot product gives a number and measures how much two vectors point the same way. The cross product a⃗×b⃗\vec a\times\vec b (in 3D only) gives a vector perpendicular to both, with length ∣a⃗∣ ∣b⃗∣sin⁡θ|\vec a|\,|\vec b|\sin\theta.