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Learn/ 11th Class Mathematics/ Unit 14 /Numericals

Unit 14: Vectors in Space — Exercise 14 3

11th Class Mathematics · Unit 14: Vectors in Space

1.(i) Compute the cross product a×b and b×a. Check your answer by showing that each a and b are perpendicular to a×b and b×a. a = 2i + j - k, b = i - j + k (ii) Compute the cross product a×b and b×a. Check your answer by showing that each a and b are perpendicular to a×b and b×a. a = i + 3j + 2k, b = 2i - j + 1k (iii) Compute the cross product a×b and b×a. Check your answer by showing that each a and b are perpendicular to a×b and b×a. a = 2i - 2j + k, b = -i + j + 3k (iv) Compute the cross product a×b and b×a. Check your answer by showing that each a and b are perpendicular to a×b and b×a. a = -4i + j - 2k, b = 2i + j + k
Given
(i) Vectors a = 2i + j - k, quad b = i - j + k
Formula
Cross product a×b a × b = begin{vmatrix} i & j & k 2 & 1 & -1 1 & -1 & 1 end{vmatrix}
= i(1-1) - j(2+1) + k(-2-1)
Result
a×b = -3j - 3k
Perpendicularity check a · (a × b) = (2i + j - k) · (-3j - 3k) = 0 - 3 + 3 = 0 ⇒ a perp a × b
b · (a × b) = (i - j + k) · (-3j - 3k) = 0 + 3 - 3 = 0 ⇒ b perp a × b
Formula
Cross product b×a b × a = begin{vmatrix} i & j & k 1 & -1 & 1 2 & 1 & -1 end{vmatrix}
= i(1-1) - j(-1-2) + k(1+2)
Result
b×a = 3j + 3k
Given
(ii) Vectors a = i + 3j + 2k, quad b = 2i - j + k
Formula
Cross product a×b a × b = begin{vmatrix} i & j & k 1 & 3 & 2 2 & -1 & 1 end{vmatrix}
= i(3+2) - j(1-4) + k(-1-6)
Result
a×b = 5i + 3j - 7k
Verification a · (a × b) = (i + 3j + 2k) · (5i + 3j - 7k) = 5 + 9 - 14 = 0 ⇒ a perp a × b
b · (a × b) = (2i - j + k) · (5i + 3j - 7k) = 10 - 3 - 7 = 0 ⇒ b perp a × b
Formula
Cross product b×a b × a = begin{vmatrix} i & j & k 2 & -1 & 1 1 & 3 & 2 end{vmatrix}
= i(-2-3) - j(4-1) + k(6+1)
Result
b×a = -5i - 3j + 7k
Given
(iii) Vectors a = 2i - 2j + k, quad b = -i + j + 3k
Formula
Cross product a×b a × b = begin{vmatrix} i & j & k 2 & -2 & 1 -1 & 1 & 3 end{vmatrix}
= i(-6-1) - j(6+1) + k(2-2)
Result
a×b = -7i - 7j
Verification a · (a × b) = (2i - 2j + k) · (-7i - 7j) = -14 + 14 + 0 = 0 ⇒ a perp a × b
b · (a × b) = (-i + j + 3k) · (-7i - 7j) = 7 - 7 + 0 = 0 ⇒ b perp a × b
Formula
Cross product b×a b × a = begin{vmatrix} i & j & k -1 & 1 & 3 2 & -2 & 1 end{vmatrix}
= i(1+6) - j(-1-6) + k(2-2)
Result
b×a = 7i + 7j
Given
(iv) Vectors a = -4i + j - 2k, quad b = 2i + j + k
Formula
Cross product a×b a × b = begin{vmatrix} i & j & k -4 & 1 & -2 2 & 1 & 1 end{vmatrix}
= i(1+2) - j(-4+4) + k(-4-2)
Result
a×b = 3i - 6k
Verification a · (a × b) = (-4i + j - 2k) · (3i - 6k) = -12 + 0 + 12 = 0 ⇒ a perp a × b
b · (a × b) = (2i + j + k) · (3i - 6k) = 6 + 0 - 6 = 0 ⇒ b perp a × b
Formula
Cross product b×a b × a = begin{vmatrix} i & j & k 2 & 1 & 1 -4 & 1 & -2 end{vmatrix}
= i(-2-1) - j(-4+4) + k(2+4)
Result
b×a = -3i + 6k
2.(i) Find a unit vector perpendicular to the plane containing a and b. Also find sine of the angle between them. a = i + 6j - 3k, b = 2i + j + 3k (ii) Find a unit vector perpendicular to the plane containing a and b. Also find sine of the angle between them. a = -i - j - k, b = 2i - 3j + 4k
Given
(i) Vectors a = i + 6j - 3k, quad b = 2i + j + 3k
Formula
Cross product a × b = begin{vmatrix} i & j & k 1 & 6 & -3 2 & 1 & 3 end{vmatrix}
= i(18+3) - j(3+6) + k(1-12)
Result
a×b = 21i - 9j - 11k
Magnitude |a × b| = sqrt{212 + (-9)2 + (-11)2} = sqrt{441 + 81 + 121} = sqrt{643}
Formula
Unit vector hat{n} = a × b|a × b| = frac{21i - 9j - 11k}{sqrt{643}
= frac{21}{sqrt{643}i - frac{9}{sqrt{643}j - frac{11}{sqrt{643}k
Sine of angle sin θ = |a × b||a||b|
= frac{sqrt{643}{sqrt{1 + 36 + 9} · sqrt{4 + 1 + 9} = frac{sqrt{643}{sqrt{46} · sqrt{14}
Result
sin θ = frac{sqrt{643}{sqrt{644} = frac{sqrt{643}{sqrt{644}
Given
(ii) Vectors a = -i - j - k, quad b = 2i - 3j + 4k
Formula
Cross product a × b = begin{vmatrix} i & j & k -1 & -1 & -1 2 & -3 & 4 end{vmatrix}
= i(-4-3) - j(-4+2) + k(3+2)
Result
a×b = -7i + 2j + 5k
Magnitude |a × b| = sqrt{(-7)2 + 22 + 52} = sqrt{49 + 4 + 25} = sqrt{78}
Formula
Unit vector hat{n} = frac{-7i + 2j + 5k}{sqrt{78}
Sine of angle sin θ = |a × b||a||b|
|a| = sqrt{1 + 1 + 1} = sqrt{3}, quad |b| = sqrt{4 + 9 + 16} = sqrt{29}
sin θ = frac{sqrt{78}{sqrt{3} · sqrt{29} = frac{sqrt{78}{sqrt{87}
Result
sin θ = frac{sqrt{78}{sqrt{87}
3.(i) Find the area of the triangle formed by the points P, Q and R. P(2,3,5), Q(1,2,0), R(4,1,2) (ii) Find the area of the triangle formed by the points P, Q and R. P(0,0,1), Q(2,-1,2), R(-1,3,2)
Given
(i) Points P(2,3,5), quad Q(1,2,0), quad R(4,1,2)
Formula
Vector PQ overrightarrow{PQ} = Q - P = (1-2)i + (2-3)j + (0-5)k = -i - j - 5k
Vector PR overrightarrow{PR} = R - P = (4-2)i + (1-3)j + (2-5)k = 2i - 2j - 3k
Cross product overrightarrow{PQ} × overrightarrow{PR} = begin{vmatrix} i & j & k -1 & -1 & -5 2 & -2 & -3 end{vmatrix}
= i(3-10) - j(3+10) + k(2+2)
Result
Cross product = -7i - 13j + 4k
Magnitude |overrightarrow{PQ} × overrightarrow{PR}| = sqrt{49 + 169 + 16} = sqrt{234}
Formula
Area of triangle Δ = 12|overrightarrow{PQ} × overrightarrow{PR}| = 12sqrt{234}
Result
Area = frac{3sqrt{26}{2} sq.units
Given
(ii) Points P(0,0,1), quad Q(2,-1,2), quad R(-1,3,2)
Formula
Vector PQ overrightarrow{PQ} = (2-0)i + (-1-0)j + (2-1)k = 2i - j + k
Vector PR overrightarrow{PR} = (-1-0)i + (3-0)j + (2-1)k = -i + 3j + k
Cross product overrightarrow{PQ} × overrightarrow{PR} = begin{vmatrix} i & j & k 2 & -1 & 1 -1 & 3 & 1 end{vmatrix}
= i(-1-3) - j(2+1) + k(6-1)
Result
Cross product = -4i - 3j + 5k
Magnitude |overrightarrow{PQ} × overrightarrow{PR}| = sqrt{16 + 9 + 25} = sqrt{50}
Formula
Area of triangle Δ = 12|overrightarrow{PQ} × overrightarrow{PR}| = 12sqrt{50}
Result
Area = 12(5sqrt{2}) = frac{5sqrt{2}{2} sq.units
4.(i) Find the area of parallelogram whose vertices are: A(1,1,1), B(4,2,3), C(5,6,7), D(2,5,5) (ii) Find the area of parallelogram ABCD whose vertices are: A(4,5,6), B(1,3,2), C(-2,0,1), D(1,2,5)
Given
(i) Vertices A(1,1,1), B(4,2,3), C(5,6,7), D(2,5,5)
Formula
Vector AB overrightarrow{AB} = (4-1)i + (2-1)j + (3-1)k = 3i + j + 2k
Vector AC overrightarrow{AC} = (5-1)i + (6-1)j + (7-1)k = 4i + 5j + 6k
Cross product overrightarrow{AB} × overrightarrow{AC} = begin{vmatrix} i & j & k 3 & 1 & 2 4 & 5 & 6 end{vmatrix}
= i(6-10) - j(18-8) + k(15-4)
Result
Cross product = -4i - 10j + 11k
Magnitude |overrightarrow{AB} × overrightarrow{AC}| = sqrt{16 + 100 + 121} = sqrt{237}
Area of parallelogram Δ = |overrightarrow{AB} × overrightarrow{AC}| = sqrt{237} sq.units
Given
(ii) Vertices A(4,5,6), B(1,3,2), C(-2,0,1), D(1,2,5)
Formula
Vector AB overrightarrow{AB} = (1-4)i + (3-5)j + (2-6)k = -3i - 2j - 4k
Vector AC overrightarrow{AC} = (-2-4)i + (0-5)j + (1-6)k = -6i - 5j - 5k
Cross product overrightarrow{AB} × overrightarrow{AC} = begin{vmatrix} i & j & k -3 & -2 & -4 -6 & -5 & -5 end{vmatrix}
= i(10-20) - j(15-24) + k(15-12)
Result
Cross product = -10i + 9j + 3k
Magnitude |overrightarrow{AB} × overrightarrow{AC}| = sqrt{100 + 81 + 9} = sqrt{190}
Area of parallelogram Δ = sqrt{190} sq.units
5.If the cross product of the vectors u = 7i - 4j + 5k and v = ai - b j + 3k is zero, then find the values of a and b.
Given
Condition u × v = 0
Vectors u = 7i - 4j + 5k, quad v = ai - bj + 3k
Formula
Cross product u × v = begin{vmatrix} i & j & k 7 & -4 & 5 a & -b & 3 end{vmatrix} = 0
i(-12 + 5b) - j(21 - 5a) + k(-7b + 4a) = 0
For i component -12 + 5b = 0 ⇒ b = 125
For j component 21 - 5a = 0 ⇒ a = 215
Result
Solution a = 215, quad b = 125
6.Which vectors, if any, are perpendicular or parallel? (i) u = 5i - j + k, v = j - 5k, w = -15i + 3j - 3k (ii) u = i + 2j - k, v = -i + j + k, w = -π/2 i - πj + π/2 k
Given
(i) Vectors u = 5i - j + k, quad v = j - 5k, quad w = -15i + 3j - 3k
Check w and u w = -15i + 3j - 3k = -3(5i - j + k) = -3u
Result
Relationship u and w are parallel
Check u·v u · v = (5i - j + k) · (j - 5k) = 0 - 1 - 5 = -6 neq 0
Check v·w v · w = (j - 5k) · (-15i + 3j - 3k) = 0 + 3 + 15 = 18 neq 0
Given
(ii) Vectors u = i + 2j - k, quad v = -i + j + k, quad w = -pi2i - pi j + pi2k
Check u·v u · v = (i + 2j - k) · (-i + j + k) = -1 + 2 - 1 = 0
Result
u and v relationship u and v are perpendicular
Check v·w v · w = (-i + j + k) · left(-pi2i - pi j + pi2kright)
= pi2 - pi + pi2 = 0
v and w relationship v and w are perpendicular
13.Find the moment about the point M(1,-3,3) of the force represented by AB where the coordinates of points A(4,3,-1) and B(-1,3,7) are given.
Given
Force vector overrightarrow{AB} = (-1-4)i + (3-3)j + (7-(-1))k = -5i + 8k
Position vector from M r = overrightarrow{OA} - overrightarrow{OM} = (4i + 3j - k) - (i + 3j - 3k) = 3i + 6j - 4k
Formula
Moment about M M = r × F = begin{vmatrix} i & j & k 3 & 6 & -4 -5 & 0 & 8 end{vmatrix}
= i(48 - 0) - j(24 - 20) + k(0 + 30)
Result
Moment M = 48i - 4j + 30k
14.A force F = 6i + 4j - 4k is applied at the point A(1,-1,2). Find the moment of the force about the point B(3,-2,3)
Given
Force F = 6i + 4j - 4k
Position vector r r = overrightarrow{BA} = (1-3)i + (-1-(-2))j + (2-3)k = -2i + j - k
Formula
Moment about B M = r × F = begin{vmatrix} i & j & k -2 & 1 & -1 6 & 4 & -4 end{vmatrix}
= i(-4 + 4) - j(8 + 6) + k(-8 - 6)
Result
Moment M = -14j - 14k
15.Given a force F = 2i + j - 3k acting at a point A(1,-2,1). Find the moment of F about the point B(2,0,-2)
Given
Force F = 2i + j - 3k
Position vector r r = overrightarrow{BA} = (1-2)i + (-2-0)j + (1-(-2))k = -i - 2j + 3k
Formula
Moment about B M = r × F = begin{vmatrix} i & j & k -1 & -2 & 3 2 & 1 & -3 end{vmatrix}
= i(6 - 3) - j(3 - 6) + k(-1 + 4)
Result
Moment M = 3i + 3j + 3k
16.A force F = -2i + j - 3k is applied at P(-1,-3,2). Find its moment about the point Q(4,2,2).
Given
Force F = -2i + j - 3k
Position vector r r = overrightarrow{QP} = (-1-4)i + (-3-2)j + (2-2)k = -5i - 5j
Formula
Moment about Q M = r × F = begin{vmatrix} i & j & k -5 & -5 & 0 -2 & 1 & -3 end{vmatrix}
= i(15 - 0) - j(15 - 0) + k(-5 + 10)
Result
Moment M = 15i - 15j - 15k