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Learn/ 11th Class Mathematics/ Unit 14 /Numericals

Unit 14: Vectors in Space — Exercise 14 1

11th Class Mathematics · Unit 14: Vectors in Space

1.Let u = 3i + 2j - 5k, v = i - 5j - k, and w = -4i - j + 7k. Find the following: (i) u + 2v + w (ii) v - 3w (iii) |3v + w|
Given
Vectors vec{u} = 3vec{i} + 2vec{j} - 5vec{k}, vec{v} = vec{i} - 5vec{j} - vec{k}, vec{w} = -4vec{i} - vec{j} + 7vec{k}
(i) u + 2v + w = (3vec{i} + 2vec{j} - 5vec{k}) + 2(vec{i} - 5vec{j} - vec{k}) + (-4vec{i} - vec{j} + 7vec{k})
= (3 + 2 - 4)vec{i} + (2 - 10 - 1)vec{j} + (-5 - 2 + 7)vec{k}
Result
Result (i) = vec{i} - 9vec{j}
(ii) v - 3w = (vec{i} - 5vec{j} - vec{k}) - 3(-4vec{i} - vec{j} + 7vec{k})
= (1 + 12)vec{i} + (-5 + 3)vec{j} + (-1 - 21)vec{k}
Result (ii) = 13vec{i} - 2vec{j} - 22vec{k}
(iii) |3v + w| 3vec{v} + vec{w} = 3(vec{i} - 5vec{j} - vec{k}) + (-4vec{i} - vec{j} + 7vec{k}) = -vec{i} - 16vec{j} + 4vec{k}
|3vec{v} + vec{w}| = sqrt{(-1)2 + (-16)2 + 42} = sqrt{1 + 256 + 16}
Result (iii) = sqrt{273}
2.Find the magnitude of the vector v and write the direction cosines of v: (i) v = 3i - 2j + 6k (ii) v = -4i + 4j + 2k (iii) v = -6i + 8j
(i) Magnitude of v = 3i - 2j + 6k |vec{v}| = sqrt{32 + (-2)2 + 62} = sqrt{9 + 4 + 36} = sqrt{49}
Result
(i) Magnitude and direction cosines |vec{v}| = 7; direction cosines 37, -27, 67
(ii) Magnitude of v = -4i + 4j + 2k |vec{v}| = sqrt{(-4)2 + 42 + 22} = sqrt{16 + 16 + 4} = sqrt{36}
(ii) Magnitude and direction cosines |vec{v}| = 6; direction cosines -23, 23, 13
(iii) Magnitude of v = -6i + 8j |vec{v}| = sqrt{(-6)2 + 82} = sqrt{36 + 64} = sqrt{100}
(iii) Magnitude and direction cosines |vec{v}| = 10; direction cosines -35, 45, 0
3.Find t so that |2i + (t - 1)j + tk| = √13
Given
Condition |2vec{i} + (t-1)vec{j} + tvec{k}| = sqrt{13}
Squaring both sides sqrt{4 + (t-1)2 + t2} = sqrt{13}
4 + (t-1)2 + t2 = 13
4 + t2 - 2t + 1 + t2 = 13
2t2 - 2t + 5 = 13
2t2 - 2t - 8 = 0
Simplify t2 - t - 4 = 0
Quadratic formula t = frac{1 ± sqrt{1 + 16}{2} = frac{1 ± sqrt{17}{2}
Result
Solution t = frac{1 + sqrt{17}{2} or t = frac{1 - sqrt{17}{2}
4.Find a unit vector in the direction of v = -i + 4j - 8k
Given
Vector vec{v} = -vec{i} + 4vec{j} - 8vec{k}
Magnitude |vec{v}| = sqrt{(-1)2 + 42 + (-8)2} = sqrt{1 + 16 + 64} = sqrt{81}
Result
Magnitude |vec{v}| = 9
Unit vector hat{vec{v} = frac{-vec{i} + 4vec{j} - 8vec{k}{9} = -19vec{i} + 49vec{j} - 89vec{k}
5.If u = 2i + j - 3k, v = -i + 4j + 2k and w = 3i - 2j + k, find a unit vector parallel to 4u - 3v + 2w
Given
Vectors vec{u} = 2vec{i} + vec{j} - 3vec{k}, vec{v} = -vec{i} + 4vec{j} + 2vec{k}, vec{w} = 3vec{i} - 2vec{j} + vec{k}
Calculate 4u - 3v + 2w 4(2vec{i} + vec{j} - 3vec{k}) - 3(-vec{i} + 4vec{j} + 2vec{k}) + 2(3vec{i} - 2vec{j} + vec{k})
= (8 + 3 + 6)vec{i} + (4 - 12 - 4)vec{j} + (-12 - 6 + 2)vec{k}
Result
Combined vector = 17vec{i} - 12vec{j} - 16vec{k}
Magnitude |4vec{u} - 3vec{v} + 2vec{w}| = sqrt{172 + (-12)2 + (-16)2} = sqrt{289 + 144 + 256} = sqrt{689}
Unit vector frac{17vec{i} - 12vec{j} - 16vec{k}{sqrt{689}
6.Find a vector (i) whose magnitude is 5 and is parallel to 3i + 4j - k, and (ii) whose magnitude is 7 and is parallel to -i + j + k
Given
(i) Reference vector vec{u} = 3vec{i} + 4vec{j} - vec{k}
(i) Magnitude of reference |vec{u}| = sqrt{32 + 42 + (-1)2} = sqrt{9 + 16 + 1} = sqrt{26}
(i) Unit vector hat{vec{u} = frac{3vec{i} + 4vec{j} - vec{k}{sqrt{26}
Result
(i) Vector of magnitude 5 5hat{vec{u} = frac{15vec{i} + 20vec{j} - 5vec{k}{sqrt{26}
Given
(ii) Reference vector vec{u} = -vec{i} + vec{j} + vec{k}
(ii) Magnitude of reference |vec{u}| = sqrt{(-1)2 + 12 + 12} = sqrt{1 + 1 + 1} = sqrt{3}
(ii) Unit vector hat{vec{u} = frac{-vec{i} + vec{j} + vec{k}{sqrt{3}
Result
(ii) Vector of magnitude 7 7hat{vec{u} = frac{-7vec{i} + 7vec{j} + 7vec{k}{sqrt{3}
7.If u = xi + 2j + 3k, v = i + yj - 3k and w = -2i - 3j represent the sides of a triangle, find the values of x and y
Given
Triangle condition For sides of a triangle: vec{u} + vec{v} = -vec{w}
Expand condition (xvec{i} + 2vec{j} + 3vec{k}) + (vec{i} + yvec{j} - 3vec{k}) = -(-2vec{i} - 3vec{j})
(x+1)vec{i} + (2+y)vec{j} + 0vec{k} = 2vec{i} + 3vec{j}
Compare i-coefficients x + 1 = 2 ⇒ x = 1
Compare j-coefficients 2 + y = 3 ⇒ y = 1
Result
Solution x = 1, y = 1
8.The position vectors of points A, B, C and D are u = i + 2j + k, v = 7i + 8j + 4k, w = -i + k, z = i + 2j + 2k respectively. Show that AB is parallel to CD
Given
Position vectors vec{OA} = vec{u} = vec{i} + 2vec{j} + vec{k}, vec{OB} = vec{v} = 7vec{i} + 8vec{j} + 4vec{k}
vec{OC} = vec{w} = -vec{i} + vec{k}, vec{OD} = vec{z} = vec{i} + 2vec{j} + 2vec{k}
Calculate AB vec{AB} = vec{OB} - vec{OA} = (7vec{i} + 8vec{j} + 4vec{k}) - (vec{i} + 2vec{j} + vec{k})
Result
vec{AB} = 6vec{i} + 6vec{j} + 3vec{k}
Calculate CD vec{CD} = vec{OD} - vec{OC} = (vec{i} + 2vec{j} + 2vec{k}) - (-vec{i} + vec{k})
vec{CD} = 2vec{i} + 2vec{j} + vec{k}
Check parallelism vec{AB} = 3vec{CD} = 3(2vec{i} + 2vec{j} + vec{k})
Conclusion Since vec{AB} = 3vec{CD}, AB is parallel to CD
9.We say that two vectors v and w in space are parallel if there is a scalar c such that v = cw the vectors point in the same direction if c > 0, and the vectors point in the opposite direction if c < 0. (a) Find two vectors of length 2 parallel to the vector v = 2i - 4j + 4k (b) Find the constant α so that the vectors v = i - 3j + 4k and w = αi + 9j - 12k are parallel (c) Find a vector of length 5 in the direction opposite that of v = i - 2j + 3k (d) Find a and b so that the vectors 3i - j + 4k and ai + bj - 2k are parallel
(a) Vector v = 2i - 4j + 4k |vec{v}| = sqrt{22 + (-4)2 + 42} = sqrt{4 + 16 + 16} = sqrt{36} = 6
(a) Unit vector hat{vec{v} = frac{2vec{i} - 4vec{j} + 4vec{k}{6} = 13vec{i} - 23vec{j} + 23vec{k}
Result
(a) Two vectors of length 2 ± 2hat{vec{v} = 23vec{i} - 43vec{j} + 43vec{k} and -23vec{i} + 43vec{j} - 43vec{k}
(b) Vectors v = i - 3j + 4k and w = αi + 9j - 12k For parallelism: vec{w} = cvec{v}
(b) From j-component 9 = c(-3) ⇒ c = -3
(b) From k-component (check) -12 = c(4) ⇒ c = -3 checkmark
(b) Solution α = c(1) = -3
(c) Vector v = i - 2j + 3k |vec{v}| = sqrt{12 + (-2)2 + 32} = sqrt{1 + 4 + 9} = sqrt{14}
(c) Unit vector in opposite direction -hat{vec{v} = -frac{vec{i} - 2vec{j} + 3vec{k}{sqrt{14}
(c) Vector of length 5 in opposite direction -5hat{vec{v} = frac{-5vec{i} + 10vec{j} - 15vec{k}{sqrt{14}
(d) Vectors v = 3i - j + 4k and w = ai + bj - 2k For parallelism: vec{w} = cvec{v}
(d) From k-component -2 = 4c ⇒ c = -12
(d) From i-component a = 3c = 3left(-12right) = -32
(d) From j-component b = -c = -left(-12right) = 12
(d) Solution a = -32, b = 12
10.A spacecraft moves from point (120, 240, -50) to point (130, 210, 80) in kilometers. What is the magnitude of the displacement vector in kilometers?
Given
Initial position A = (120, 240, -50)
Final position B = (130, 210, 80)
Displacement vector vec{AB} = (130 - 120, 210 - 240, 80 - (-50)) = (10, -30, 130)
In vector form vec{AB} = 10vec{i} - 30vec{j} + 130vec{k}
Magnitude |vec{AB}| = sqrt{102 + (-30)2 + 1302} = sqrt{100 + 900 + 16900} = sqrt{17900}
Simplify = sqrt{100 × 179} = 10sqrt{179}
Result
Magnitude |vec{AB}| = 10sqrt{179} km
11.Find the direction cosines for the given vector: (i) u = -6i + 3j + 2k (ii) v = 4i + 2j - 5k (iii) PQ, where P = (9, 3, 13) and Q = (11, 6, 19)
(i) Vector u = -6i + 3j + 2k |vec{u}| = sqrt{(-6)2 + 32 + 22} = sqrt{36 + 9 + 4} = sqrt{49} = 7
Result
(i) Direction cosines -67, 37, 27
(ii) Vector v = 4i + 2j - 5k |vec{v}| = sqrt{42 + 22 + (-5)2} = sqrt{16 + 4 + 25} = sqrt{45} = 3sqrt{5}
(ii) Direction cosines frac{4}{3sqrt{5}, frac{2}{3sqrt{5}, frac{-5}{3sqrt{5}
(iii) Vector from P(9,3,13) to Q(11,6,19) vec{PQ} = (11-9, 6-3, 19-13) = (2, 3, 6)
(iii) Magnitude |vec{PQ}| = sqrt{22 + 32 + 62} = sqrt{4 + 9 + 36} = sqrt{49} = 7
(iii) Direction cosines 27, 37, 67
12.Which of the following triples can be the direction angles of a single vector: (i) 45°, 45°, 60° (ii) 30°, 45°, 60° (iii) 45°, 60°, 60°
Given
Direction angle property cos2α + cos2β + cos2gamma = 1
(i) α = 45°, β = 45°, γ = 60° cos2(45°) + cos2(45°) + cos2(60°) = left(frac{1}{sqrt{2}right)2 + left(frac{1}{sqrt{2}right)2 + left(12right)2
= 12 + 12 + 14 = 54 neq 1
Result
(i) Result Not valid direction angles
(ii) α = 30°, β = 45°, γ = 60° cos2(30°) + cos2(45°) + cos2(60°) = left(frac{sqrt{3}{2}right)2 + left(frac{1}{sqrt{2}right)2 + left(12right)2
= 34 + 12 + 14 = 32 neq 1
(ii) Result Not valid direction angles
(iii) α = 45°, β = 60°, γ = 60° cos2(45°) + cos2(60°) + cos2(60°) = left(frac{1}{sqrt{2}right)2 + left(12right)2 + left(12right)2
= 12 + 14 + 14 = 1 checkmark
(iii) Result Valid direction angles