Tutorial 1 — All Problems Solved
Tutorial 1 · Problem 1 — Three points A(4,3,2), B(−2,0,5), C(7,−2,1)
(a) Position vector from origin to A. (b) Unit vector to midpoint of AB. (c) Perimeter of triangle ABC.
(a) Position vector OA:
Ā = A − O = 4âx + 3ây + 2âz
(b) Unit vector to midpoint of AB:
MAB = (A+B)/2 = ((4+(−2))/2, (3+0)/2, (2+5)/2) = (1, 1.5, 3.5)
MAB = 1âx + 1.5ây + 3.5âz
|MAB| = √(1² + 1.5² + 3.5²) = √(1 + 2.25 + 12.25) = √15.5 = 3.937
âM = MAB/|MAB| = (1âx + 1.5ây + 3.5âz) / 3.937
(c) Perimeter of triangle ABC:
BA = A−B = (6,3,−3) → |BA| = √(36+9+9) = √54 = 7.348
CB = B−C = (−9,2,4) → |CB| = √(81+4+16) = √101 = 10.050
AC = C−A = (3,−5,−1) → |AC| = √(9+25+1) = √35 = 5.916
Perimeter = 7.348 + 10.050 + 5.916
(a) Ā = 4âx + 3ây + 2âz
(b) âM = (1âx + 1.5ây + 3.5âz) / √15.5
(c) Perimeter = 23.314 length units
Tutorial 1 · Problem 2 — Points A(8,−5,4) and B(−2,3,2)
(a) Distance A→B. (b) Unit vector A→B. (c) Unit vector origin→midpoint. (d) Intersection with plane z=3.
(a) Distance A to B:
AB = B−A = (−10, 8, −2)
|AB| = √(100+64+4) = √168 ≈ 12.96 units
(b) Unit vector A→B:
âAB = AB/|AB| = (−10âx + 8ây − 2âz) / √168
(c) Unit vector to midpoint:
MAB = (A+B)/2 = (3, −1, 3)
|MAB| = √(9+1+9) = √19
âM = (3âx − ây + 3âz) / √19
(d) Intersection with z=3 plane:
Line equation: (x−8)/(−10) = (y+5)/8 = (z−4)/(−2) = t
Set z=3: (3−4)/(−2) = t = 1/2
x = 8 + (−10)(1/2) = 3
y = −5 + 8(1/2) = −1
(a) |AB| = √168 ≈ 12.96
(b) âAB = (−10âx+8ây−2âz)/√168
(c) âM = (3âx−ây+3âz)/√19
(d) Intersection point P = (3, −1, 3)
Tutorial 2 — Vector Problems (P1–P4)
Tutorial 2 · Problem 1 — Interior angles of triangle A(1,3,−2), B(−2,4,5), C(0,−2,1)
Find all interior angles using the dot product.
Angle at A (between AB and AC):
AB = B−A = −3âx+1ây+7âz, |AB| = √(9+1+49) = √59
AC = C−A = −1âx−5ây+3âz, |AC| = √(1+25+9) = √35
AB·AC = (−3)(−1)+(1)(−5)+(7)(3) = 3−5+21 = 19
cos θA = 19/(√59×√35) = 19/√2065 = 19/45.44 → θA = 65.28°
Angle at B (between BA and BC):
BA = A−B = 3âx−1ây−7âz, |BA| = √59
BC = C−B = 2âx−6ây−4âz, |BC| = √(4+36+16) = √56
BA·BC = (3)(2)+(−1)(−6)+(−7)(−4) = 6+6+28 = 40
cos θB = 40/(√59×√56) = 40/57.48 → θB = 45.90°
Angle at C:
θC = 180° − 65.28° − 45.90° = 68.82°
✔ Verify: θA+θB+θC = 65.28+45.90+68.82 = 180° ✓
θA = 65.28° | θB = 45.90° | θC = 68.82°
Tutorial 2 · Problem 2 — Vector projections & angle (A,B,C,D given)
Given: A(10,12,−6), B(16,8,−2), C(8,1,4), D(−2,−5,8).
(a) Projection of (AB+BC) onto AD. (b) Projection onto DC. (c) Angle between DA and DC.
AB = B−A = 6âx−4ây+4âz
BC = C−B = −8âx−7ây+6âz
M = AB+BC = −2âx−11ây+10âz
AD = D−A = −12âx−17ây+14âz, |AD| = √(144+289+196) = √629
DC = C−D = 10âx+6ây−4âz, |DC| = √(100+36+16) = √152
(a) Projection of M onto AD:
Scalar proj = M·âAD = M·AD/|AD| = [(−2)(−12)+(−11)(−17)+(10)(14)]/√629
= [24+187+140]/√629 = 351/√629 = 351/25.08 = 13.996
Vector proj P = (351/629)×AD = (351/629)(−12âx−17ây+14âz)
P = −6.696âx − 9.486ây + 7.812âz
(b) Projection onto DC:
Scalar = M·DC/|DC| = [(−2)(10)+(−11)(6)+(10)(−4)]/√152 = [−20−66−40]/√152 = −126/12.33
Vector P = (−126/152)(10âx+6ây−4âz)
P = −8.289âx − 4.974ây + 3.316âz
(c) Angle between DA and DC:
DA = −AD = 12âx+17ây−14âz
DA·DC = (12)(10)+(17)(6)+(−14)(−4) = 120+102+56 = 278
cos θ = 278/(√629×√152) = 278/309.3 → θD = 25.96°
(a) P_AD = −6.696âx−9.486ây+7.812âz
(b) P_DC = −8.289âx−4.974ây+3.316âz
(c) θD = 25.96°
Tutorial 2 · Problem 3 — Cross product, perpendicular unit vectors, triangle area
Given: r₁=(7,3,−2), r₂=(−2,7,−3), r₃=(0,2,3).
(a) Unit vector ⊥ to r₁ and r₂. (b) Unit vector ⊥ to (r₁−r₂) and (r₂−r₃). (c) Area of triangle defined by r₁ and r₂.
(a) N = r₁×r₂:
N = |âx ây âz / 7 3 −2 / −2 7 −3|
= [(3)(−3)−(−2)(7)]âx − [(7)(−3)−(−2)(−2)]ây + [(7)(7)−(3)(−2)]âz
= [−9+14]âx − [−21−4]ây + [49+6]âz
= 5âx + 25ây + 55âz
|N| = √(25+625+3025) = √3675 = 60.62
âN = (5âx+25ây+55âz) / 60.62
(b) M₁ = r₁−r₂ = 9âx−4ây+1âz, M₂ = r₂−r₃ = −2âx+5ây−6âz:
N₂ = M₁×M₂ = |(9)(−6)−(1)(5)|âx − |(9)(−6)−(1)(−2)|ây + |(9)(5)−(−4)(−2)|âz
= [−54−5]âx − [−54+2]ây + [45−8]âz = −59âx... let me redo:
N₂x = (−4)(−6)−(1)(5) = 24−5 = 19
N₂y = −[(9)(−6)−(1)(−2)] = −[−54+2] = 52
N₂z = (9)(5)−(−4)(−2) = 45−8 = 37
N₂ = 19âx+52ây+37âz, |N₂| = √(361+2704+1369) = √4434 = 66.59
âN₂ = (19âx+52ây+37âz) / 66.59
(c) Area of triangle (r₁ and r₂ as sides):
Area = ½|N| = ½×60.62 = 30.31 area units
(a) âN = (5âx+25ây+55âz)/√3675
(b) âN₂ = (19âx+52ây+37âz)/√4434
(c) Area = 30.31 sq. units
Tutorial 2 · Problem 4 — Triangle with vertex A(−4,2,5), AM=(20,18,−10), AN=(−10,8,15)
(a) Unit vector ⊥ to triangle plane. (b) Unit vector in plane ⊥ to AN. (c) Unit vector through A bisecting MN.
(a) V₁ = AM×AN:
V₁x = (18)(15)−(−10)(8) = 270+80 = 350
V₁y = −[(20)(15)−(−10)(−10)] = −[300−100] = −200
V₁z = (20)(8)−(18)(−10) = 160+180 = 340
V₁ = 350âx−200ây+340âz, |V₁| = √(122500+40000+115600) = √278100 = 527.3
âV₁ = (350âx−200ây+340âz) / 527.3
(b) V₂ = V₁×AN (in plane, ⊥ to AN):
V₂ = |âx ây âz / 350 −200 340 / −10 8 15|
V₂x = (−200)(15)−(340)(8) = −3000−2720 = −5720
V₂y = −[(350)(15)−(340)(−10)] = −[5250+3400] = −8650
V₂z = (350)(8)−(−200)(−10) = 2800−2000 = 800
|V₂| = √(32718400+74822500+640000) = √108180900 = 10400.9
âV₂ = (−5720âx−8650ây+800âz) / 10400.9
(c) Bisector through A toward midpoint K of MN:
M = AM+A = (20−4, 18+2, −10+5) = (16, 20, −5)
N = AN+A = (−10−4, 8+2, 15+5) = (−14, 10, 20)
K = (M+N)/2 = (1, 15, 7.5)
AK = K−A = (1−(−4), 15−2, 7.5−5) = 5âx+13ây+2.5âz
|AK| = √(25+169+6.25) = √200.25 = 14.15
âAK = (5âx+13ây+2.5âz) / 14.15
(a) âV₁ = (350âx−200ây+340âz)/527.3
(b) âV₂ = (−5720âx−8650ây+800âz)/10400.9
(c) âAK = (5âx+13ây+2.5âz)/14.15
Del Operator (∇) — Tutorial 4 All Problems
∇ Operations Summary
Del operator (Cartesian)
∇ = (∂/∂x)âx + (∂/∂y)ây + (∂/∂z)âz
Gradient ∇V (scalar→vector)
∇V = (∂V/∂x)âx+(∂V/∂y)ây+(∂V/∂z)âz
Direction of steepest increase. Used: E = −∇V
Divergence ∇·A (vector→scalar)
∇·A = ∂Ax/∂x + ∂Ay/∂y + ∂Az/∂z
Measures source/sink strength at a point
Curl ∇×A (vector→vector)
∇×A = determinant with ∂/∂x,∂/∂y,∂/∂z
Measures circulation/rotation
Field Classification
| Condition | Name | Example |
|---|
| ∇·A = 0 | Solenoidal | B field (always!) |
| ∇×A = 0 | Irrotational / Conservative | Static E field |
| Both = 0 | Laplacian (harmonic) | V in free space |
📌 Key Identities
- ∇×(∇V) = 0 always (curl of any gradient = 0)
- ∇·(∇×A) = 0 always (divergence of any curl = 0)
Tutorial 4 · Problem 1a — Gradient of V = e^(2x+3y)cos(5z) at P(0.1,−0.2,0.4)
∂V/∂x = 2e^(2x+3y)cos(5z)
∂V/∂y = 3e^(2x+3y)cos(5z)
∂V/∂z = −5e^(2x+3y)sin(5z)
At P(0.1,−0.2,0.4):
e^(2×0.1+3×(−0.2)) = e^(0.2−0.6) = e^(−0.4) = 0.6703
cos(5×0.4) = cos(2 rad) = −0.4161
sin(5×0.4) = sin(2 rad) = 0.9093
∂V/∂x|P = 2×0.6703×(−0.4161) = −0.5579
∂V/∂y|P = 3×0.6703×(−0.4161) = −0.8369
∂V/∂z|P = −5×0.6703×0.9093 = −3.0476
∇V|P = −0.5579âx − 0.8369ây − 3.0476âz
Tutorial 4 · Problem 1b — Gradient of V = 6xz·sin(3y) at P(0.3,0.2,0.1)
∂V/∂x = 6z·sin(3y)
∂V/∂y = 18xz·cos(3y)
∂V/∂z = 6x·sin(3y)
At P(0.3,0.2,0.1):
sin(3×0.2) = sin(0.6 rad) = 0.5646
cos(3×0.2) = cos(0.6 rad) = 0.8253
∂V/∂x|P = 6×0.1×0.5646 = 0.3388
∂V/∂y|P = 18×0.3×0.1×0.8253 = 0.4457
∂V/∂z|P = 6×0.3×0.5646 = 1.0163
∇V|P = 0.339âx + 0.446ây + 1.016âz
Tutorial 4 · Problem 2 — Divergence & Curl of Ā = e^(xy)âx + sin(xy)ây + cos²(xy)âz
Divergence ∇·A:
∂Ax/∂x = y·e^(xy)
∂Ay/∂y = x·cos(xy)
∂Az/∂z = 0
∇·A = ye^(xy) + x·cos(xy)
Curl ∇×A:
x-component: ∂Az/∂y − ∂Ay/∂z = −2x·sin(xy)cos(xy) − 0 = −x·sin(2xy)
y-component: ∂Ax/∂z − ∂Az/∂x = 0 − (−2y·cos(xy)sin(xy)) = y·sin(2xy)
z-component: ∂Ay/∂x − ∂Ax/∂y = y·cos(xy) − x·e^(xy)
∇·A = ye^(xy) + x·cos(xy)
∇×A = [−x·sin(2xy)]âx + [y·sin(2xy)]ây + [y·cos(xy)−x·e^(xy)]âz