The Dot Product — Solutions
-
Q1 — Calculate Dot Products
Given a = 2i + 3j and b = 4i − j.
(a) a · b = 2(4) + 3(−1) = 8 − 3 = 5
(b) a · a = 22 + 32 = 4 + 9 = 13
This equals |a|2 = (√13)2 = 13. The self-dot product always gives the square of the magnitude. ✓
(c) b · b = 42 + (−1)2 = 16 + 1 = 17
This equals |b|2 = (√17)2 = 17. ✓
-
Q2 — Test Perpendicularity
(a) a = 3i − 2j, b = 2i + 3j
a · b = 3(2) + (−2)(3) = 6 − 6 = 0 ✓ Perpendicular.
Notice: the components of b are the components of a swapped and with one sign changed — a quick way to construct a perpendicular vector.
(b) a = 4i + j, b = 2i − 8j
a · b = 4(2) + 1(−8) = 8 − 8 = 0 ✓ Perpendicular.
(c) a = 5i, b = 3j
a · b = 5(0) + 0(3) = 0 ✓ Perpendicular.
The standard basis vectors i and j are always perpendicular by definition — any scalar multiples of them are also perpendicular to each other.
-
Q3 — Find the Angle Between Vectors
a = i, b = i + j.
a · b = 1(1) + 0(1) = 1
|a| = 1. |b| = √(1 + 1) = √2.
cos θ = 1 / (1 × √2) = 1/√2 = √2/2
θ = arccos(1/√2) = 45° exactly.
Geometrically: a lies along the positive x-axis; b = i + j bisects the first quadrant at 45° to both axes. The angle between them is naturally 45°.
-
Q4 — Find k for Perpendicularity
For perpendicularity, a · b = 0:
(ki + 3j) · (2i − 4j) = k(2) + 3(−4) = 2k − 12 = 0
2k = 12 ⇒ k = 6
Check: a = 6i + 3j. a · b = 6(2) + 3(−4) = 12 − 12 = 0 ✓
-
Q5 — Scalar Projection
Find compb(a) where a = 7i + 2j and b = 3i + 4j.
a · b = 7(3) + 2(4) = 21 + 8 = 29
|b| = √(32 + 42) = √25 = 5
compb(a) = 29/5 = 5.8
Interpretation: the component of a in the direction of b has signed length 5.8 units (positive, so it points in the same direction as b).
-
Q6 — Vector Projection
Find projb(a) where a = 7i + 2j and b = 3i + 4j.
a · b = 29 (from Q5). |b|2 = 25.
projb(a) = (29/25)(3i + 4j) = (87/25)i + (116/25)j
This vector is a scalar multiple of b, so it points in the same direction as b. Its magnitude equals the scalar projection: |(87/25, 116/25)| = (29/25)√25 = 29/5 = 5.8 ✓
-
Q7 — Angle in a Triangle
A = (1, 0), B = (4, 0), C = (1, 3).
At vertex A, the two sides are AB and AC:
AB = (4 − 1)i + (0 − 0)j = 3i
AC = (1 − 1)i + (3 − 0)j = 3j
AB · AC = 3(0) + 0(3) = 0
The dot product is zero, so the angle at A is exactly 90°. ✓
Note: A lies directly below C (same x-coordinate) and directly left of B (same y-coordinate), so the right angle at A is evident geometrically too.
-
Q8 — Dot Product Equation
|a| = 5, |b| = 4, a · b = 10.
cos θ = (a · b) / (|a||b|) = 10 / (5 × 4) = 10/20 = 1/2
θ = arccos(1/2) = 60°
-
Q9 — Classifying a Triangle
P = (0, 0), Q = (4, 0), R = (1, 3).
At vertex P: sides PQ = 4i and PR = i + 3j
PQ · PR = 4(1) + 0(3) = 4 > 0 ⇒ angle at P is acute.
At vertex Q: sides QP = −4i and QR = −3i + 3j
QP · QR = (−4)(−3) + (0)(3) = 12 > 0 ⇒ angle at Q is acute.
At vertex R: sides RP = −i − 3j and RQ = 3i − 3j
RP · RQ = (−1)(3) + (−3)(−3) = −3 + 9 = 6 > 0 ⇒ angle at R is acute.
All three angles are acute, so triangle PQR is an acute-angled triangle.
-
Q10 — Rhombus Diagonals are Perpendicular
A = (0, 0), B = (3, 1), C = (4, 4), D = (1, 3).
Part (a): Verify it is a rhombus.
AB = 3i + j. |AB| = √(9 + 1) = √10
BC = i + 3j. |BC| = √(1 + 9) = √10
CD = −3i − j. |CD| = √(9 + 1) = √10
DA = −i − 3j. |DA| = √(1 + 9) = √10
All four sides equal √10, so ABCD is a rhombus. ✓
Part (b): Prove diagonals are perpendicular.
AC = C − A = 4i + 4j
BD = D − B = (1 − 3)i + (3 − 1)j = −2i + 2j
AC · BD = 4(−2) + 4(2) = −8 + 8 = 0
Since AC · BD = 0, the diagonals AC and BD are perpendicular. ✓
This result holds for any rhombus — equal side lengths always force the diagonals to be perpendicular. This was proved generally in the Vector Proofs lesson using |b| = |d| ⇒ AC · BD = |d|2 − |b|2 = 0.