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Learn/ 11th Class Mathematics/ Unit 14 /Numericals

Unit 14: Vectors in Space — Exercise 14 4

11th Class Mathematics · Unit 14: Vectors in Space

1.(i) Find the volume of the parallelepiped for which the given vectors are three edges. u = 3i + 2k, v = i + 2j + k, w = -j + 4k (ii) Find the volume of the parallelepiped for which the given vectors are three edges. u = i - 4j - k, v = i - j - 2k, w = 2i - 3j + k (iii) Find the volume of the parallelepiped for which the given vectors are three edges. u = i - 2j + 3k, v = 2i - j - k, w = j + k
Given
(i) Vectors u = 3i + 2k, v = i + 2j + k, w = -j + 4k
Formula
Volume of parallelepiped V = |u · (v × w)|
u·(v×w) determinant = begin{vmatrix} 3 & 0 & 2 1 & 2 & 1 0 & -1 & 4 end{vmatrix}
= 3(8+1) - 0 + 2(-1-0) = 27 - 2
Result
Volume V = 25 cubic units
Given
(ii) Vectors u = i - 4j - k, v = i - j - 2k, w = 2i - 3j + k
Formula
Volume of parallelepiped V = |u · (v × w)|
u·(v×w) determinant = begin{vmatrix} 1 & -4 & -1 1 & -1 & -2 2 & -3 & 1 end{vmatrix}
= 1(-1-6) + 4(1+4) - 1(-3+2) = -7 + 20 + 1
Result
Volume V = 14 cubic units
Given
(iii) Vectors u = i - 2j + 3k, v = 2i - j - k, w = j + k
Formula
Volume of parallelepiped V = |u · (v × w)|
u·(v×w) determinant = begin{vmatrix} 1 & -2 & 3 2 & -1 & -1 0 & 1 & 1 end{vmatrix}
= 1(-1+1) + 2(2-0) + 3(2-0) = 0 + 4 + 6
Result
Volume V = 10 cubic units
2.Verify that a·b × c = b·c × a = c·a × b if a = 3i - j + 5k, b = 4i + 3j - 2k and c = 2i + 5j + k
Given
Vectors a = 3i - j + 5k, b = 4i + 3j - 2k, c = 2i + 5j + k
Formula
a·b × c a · (b × c)
a·b × c = begin{vmatrix} 3 & -1 & 5 4 & 3 & -2 2 & 5 & 1 end{vmatrix} = 3(3+10) + 1(4+4) + 5(20-6) = 39 + 8 + 70 = 117
b·c × a = begin{vmatrix} 4 & 3 & -2 2 & 5 & 1 3 & -1 & 5 end{vmatrix} = 4(25+1) - 3(10-3) - 2(-2-15) = 104 - 21 + 34 = 117
c·a × b = begin{vmatrix} 2 & 5 & 1 3 & -1 & 5 4 & 3 & -2 end{vmatrix} = 2(2-15) - 5(-6-20) + 1(9+4) = -26 + 130 + 13 = 117
Result
Verification a · b × c = b · c × a = c · a × b = 117
3.Prove that the vectors i - 2j + 3k, -2i + 3j - 4k and i - 3j + 5k are coplanar.
Given
Vectors a = i - 2j + 3k, b = -2i + 3j - 4k, c = i - 3j + 5k
Formula
Condition for coplanarity a · (b × c) = 0
a·b × c = begin{vmatrix} 1 & -2 & 3 -2 & 3 & -4 1 & -3 & 5 end{vmatrix}
= 1(15-12) - (-2)(-10+4) + 3(6-3) = 3 - 12 + 9 = 0
Result
Conclusion a · (b × c) = 0 hence vectors are coplanar
4.(i) Find the constant α such that the vectors i - j + k, i - 2j - 3k and 3i - αj + 5k are coplanar. (ii) Find the constant α such that the vectors i - 2αj - k, i - 2j + 2k and αi - 2j + k are coplanar.
Given
(i) Vectors u = i - j + k, v = i - 2j - 3k, w = 3i - αj + 5k
Formula
Condition for coplanarity u · (v × w) = 0
u·(v×w) = begin{vmatrix} 1 & -1 & 1 1 & -2 & -3 3 & -α & 5 end{vmatrix}
= 1(-10-3α) + 1(5+9) + 1(-α+6) = -10-3α+14-α+6 = 10-4α
Setting = 0 10 - 4α = 0 ⇒ α = 52
Result
Constant α = 52
Given
(ii) Vectors u = i - 2αj - k, v = i - 2j + 2k, w = αi - 2j + k
Formula
Condition for coplanarity u · (v × w) = 0
Determinant = begin{vmatrix} 1 & -2α & -1 1 & -2 & 2 α & -2 & 1 end{vmatrix}
= 1(-2+4) + 2α(1-2α) - 1(-2+2α) = 2 + 2α - 4α2 + 2 - 2α = 4 - 4α2
Setting = 0 4 - 4α2 = 0 ⇒ α2 = 1 ⇒ α = ± 1
Result
Constant α = ± 1
5.Prove that the points whose position vectors are A(-6i + 3j + 2k), B(3i - 2j + 4k), C(5i + 7j + 3k) and D(-13i + 17j - k) are coplanar.
Given
Position vectors A = -6i + 3j + 2k, B = 3i - 2j + 4k, C = 5i + 7j + 3k, D = -13i + 17j - k
AB vector overrightarrow{AB} = OB - OA = 9i - 5j + 2k
AC vector overrightarrow{AC} = OC - OA = 11i + 4j + k
AD vector overrightarrow{AD} = OD - OA = -7i + 14j - 3k
Coplanarity condition overrightarrow{AB} · (overrightarrow{AC} × overrightarrow{AD}) = begin{vmatrix} 9 & -5 & 2 11 & 4 & 1 -7 & 14 & -3 end{vmatrix}
= 9(-12-14) - (-5)(-33+7) + 2(154+28) = 9(-26) + 5(-26) + 2(182) = -234 - 130 + 364 = 0
Result
Conclusion overrightarrow{AB} · (overrightarrow{AC} × overrightarrow{AD}) = 0 hence vectors are coplanar
6.(i) Find the value of 2i × 2j · k (ii) Find the value of 3j · k × i (iii) Find the value of [k i j] (iv) Find the value of [i i k] (v) Prove that u·(v×w) + v·(w×u) + w·(u×v) = 3u·(v×w)
Given
(i) Expression 2i × 2j · k
Simplify = 4(i × j) · k = 4k · k
= 4(1) = 4
Result
Value 4
Given
(ii) Expression 3j · (k × i)
Simplify = 3j · j = 3(1) = 3
Result
Value 3
Given
(iii) Scalar triple product [k i j]
Evaluate = k · (i × j) = k · k = 1
Result
Value 1
Given
(iv) Scalar triple product [i i k]
Evaluate = i · (i × k) = i · (-j) = 0
Result
Value
Given
(v) Identity to prove u · (v × w) + v · (w × u) + w · (u × v) = 3u · (v × w)
Formula
Property u · (v × w) = v · (w × u) = w · (u × v)
L.H.S. = u · (v × w) + u · (v × w) + u · (v × w) = 3u · (v × w)
Result
Conclusion L.H.S = R.H.S
7.Find volume of the Tetrahedron with the vertices: (i) (0,1,2), (3,2,1), (1,2,1) and (5,5,6) (ii) (2,1,8), (3,2,9), (2,1,4) and (3,3,10)
Given
(i) Vertices A(0,1,2), B(3,2,1), C(1,2,1), D(5,5,6)
AB vector overrightarrow{AB} = 3i + j - k
AC vector overrightarrow{AC} = i + j - k
AD vector overrightarrow{AD} = 5i + 4j + 4k
Formula
Volume formula V = 16|overrightarrow{AB} · (overrightarrow{AC} × overrightarrow{AD})|
Determinant = 16begin{vmatrix} 3 & 1 & -1 1 & 1 & -1 5 & 4 & 4 end{vmatrix} = 16[3(4+4) - 1(4+5) - 1(4-5)]
= 16[24 - 9 + 1] = 166 = 83
Result
Volume V = 83 cubic units
Given
(ii) Vertices A(2,1,8), B(3,2,9), C(2,1,4), D(3,3,10)
AB vector overrightarrow{AB} = i + j + k
AC vector overrightarrow{AC} = 0i + 0j - 4k
AD vector overrightarrow{AD} = i + 2j + 2k
Formula
Volume formula V = 16|overrightarrow{AB} · (overrightarrow{AC} × overrightarrow{AD})|
Determinant = 16begin{vmatrix} 1 & 1 & 1 0 & 0 & -4 1 & 2 & 2 end{vmatrix} = 16[(0+8) - (0+4) + (0-0)]
= 16(8-4) = 46 = 23
Result
Volume V = 23 cubic unit
8.Prove that the points whose position vectors are A(3i + 2j - k), B(i - 2j + k), C(6i + 4j - 2k) and D(9i + 6j - 3k) are coplanar.
Given
Position vectors A = 3i + 2j - k, B = i - 2j + k, C = 6i + 4j - 2k, D = 9i + 6j - 3k
AB vector overrightarrow{AB} = -2i - 4j + 2k
AC vector overrightarrow{AC} = 3i + 2j - k
AD vector overrightarrow{AD} = 6i + 4j - 2k
Formula
Coplanarity test overrightarrow{AB} · (overrightarrow{AC} × overrightarrow{AD}) = begin{vmatrix} -2 & -4 & 2 3 & 2 & -1 6 & 4 & -2 end{vmatrix}
= -2(-4+4) - (-4)(-6+6) + 2(12-12) = 0
Result
Conclusion overrightarrow{AB} · (overrightarrow{AC} × overrightarrow{AD}) = 0 hence vectors are coplanar
9.Prove that for any three non-zero vectors u, v and w, (u + v)·[(v + w)×(w + u)] = 2[u v w]
Given
Identity (u + v) · [(v + w) × (w + u)] = 2[u v w]
Expand L.H.S. = (u + v) · [v × w + v × u + w × w + w × u]
Using w × w = 0 = (u + v) · [v × w + v × u + w × u]
Distribute = u · (v × w) + u · (v × u) + u · (w × u) + v · (v × w) + v · (v × u) + v · (w × u)
Using a·(a×b) = 0 = u · (v × w) + 0 + 0 + 0 + 0 + v · (w × u)
Rewrite v·(w×u) = u · (v × w) + u · (v × w) = 2[u v w]
Result
Conclusion L.H.S = R.H.S
10.Consider a parallelepiped determined by the vectors u = 2i + 4j - 3k, v = 5i - 3j + 6k and w = 4i - 7j - 2k. If the base of the parallelepiped is defined by the vectors u and v then find the height of the parallelepiped.
Given
Vectors u = 2i + 4j - 3k, v = 5i - 3j + 6k, w = 4i - 7j - 2k
Formula
Volume (u × v) · w = |u × v||w|cosθ
(u × v)·w = begin{vmatrix} 2 & 4 & -3 5 & -3 & 6 4 & -7 & -2 end{vmatrix}
= 2(6+42) - 4(-10-24) - 3(-35+12) = 96 + 136 + 69 = 301
u × v magnitude |u × v| = sqrt{152 + (-27)2 + (-26)2} = sqrt{255 + 729 + 676} = sqrt{1630}
Height formula h = |ucosθ| = (u × v) · w|u × v|
Result
Height h = frac{301}{sqrt{1630} units
11.A mechanic applies a force of 50 pounds along the positive x-axis on a wrench connected to a bolt. The pivot point of the wrench is at the origin (0, 0, 0) and the force is applied at the point (0ft, 2ft, 3ft). Determine the torque produced by this force about the pivot point.
Given
Force vector F = 50i + 0j + 0k
Position vector r = 0i + 2j + 3k
Formula
Torque formula tau = r × F
Cross product tau = begin{vmatrix} i & j & k 0 & 2 & 3 50 & 0 & 0 end{vmatrix}
= i(0) - j(-150) + k(-100) = 150j - 100k
Result
Torque tau = 150j - 100k pound feet
12.A drone flies from point (1, 2, 5) to point (4, 6, 9), with each unit representing a meter. What is the magnitude of the displacement the drone experienced during this flight?
Given
Start point A(1, 2, 5)
End point B(4, 6, 9)
Displacement vector overrightarrow{AB} = (4-1)i + (6-2)j + (9-5)k = 3i + 4j + 4k
Formula
Magnitude |overrightarrow{AB}| = sqrt{32 + 42 + 42}
= sqrt{9 + 16 + 16} = sqrt{41}
Result
Displacement magnitude |overrightarrow{AB}| = sqrt{41} m
13.The vector u = 50i + 75j + 65k shows how many belts, pants and shirts were sold at a store. The vector w = 1500i + 3500j + 3000k shows the price (in rupees) of each item. Find u·w and explain what the result tells us in real life.
Given
Sales quantities u = 50i + 75j + 65k
Prices in rupees w = 1500i + 3500j + 3000k
Formula
Dot product u · w = (50)(1500) + (75)(3500) + (65)(3000)
= 75000 + 262500 + 195000 = 532500
Result
Total revenue u · w = 532500 rupees
Interpretation This result shows the total revenue from the sales of all items in rupees
14.A force F = (20, -10, 30) N is applied at a point P(2, -1, 4) in 3D space. The pivot point is at M(1, 2, -3). Calculate the torque produced by this force about the pivot point M.
Given
Force vector F = 20i - 10j + 30k
Applied point P(2, -1, 4)
Pivot point M(1, 2, -3)
Position vector r = overrightarrow{MP} = (2-1)i + (-1-2)j + (4-(-3))k = i - 3j + 7k
Formula
Torque formula tau = r × F
Cross product tau = begin{vmatrix} i & j & k 1 & -3 & 7 20 & -10 & 30 end{vmatrix}
= i(-90+70) - j(30-140) + k(-10+60) = -20i + 110j + 50k
Result
Torque tau = -20i + 110j + 50k Nm
15.(a) An electric shop sells three types of appliances: Fans, Heaters and Ovens. The monthly sales quantities are 500 units of Fans, 300 units of Heaters and 200 units of Ovens. The profit per unit for each appliance is Rs.500 for Fans, Rs.400 for Heaters and Rs.2,000 for Ovens. Represent the monthly sales quantities and the profit per unit as vectors. (b) Calculate the total monthly profit using vector operations.
Given
Sales quantities Fans: 500, Heaters: 300, Ovens: 200
Profit per unit (Rs.) Fans: 500, Heaters: 400, Ovens: 2,000
(a) Sales vector u = 500i + 300j + 200k or [500, 300, 200]
(a) Profit vector v = 500i + 400j + 2000k or [500, 400, 2000]
Formula
(b) Total profit calculation u · v = (500i + 300j + 200k) · (500i + 400j + 2000k)
= (500)(500) + (300)(400) + (200)(2000)
= 250000 + 120000 + 400000 = 770000
Result
Total monthly profit Rs. 770,000