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JEE Maths Quiz on Vector Algebra
JEE Maths Quiz on Vector Algebra : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you quiz on Vector Algebra.
Quiz on Vector Algebra
Q1. The magnitudes of mutually perpendicular forces a, b and c are 2, 10 and 11, respectively. Then the magnitude of its resultant is ______. c) 15 Solution: R = √[22 + 102 + 112]. = √[4 + 100 + 121] = 15
a) 10
b) 12
c) 15
d) 17
Q2. A unit vector a makes an angle π / 4 with z-axis. If a + i + j is a unit vector, then a is equal to _________. d) [−i / 2] − [j / 2] + [k / √2] Solution: Let a = li + mj + nk, where l2 + m2 + n2 = 1. a makes an angle π / 4 with z−axis. Hence, n = 1 / √2, l2 + m2 = 1 / 2 …..(i) Therefore, a = li + mj + k / √2 a + i + j = (l + 1) i + (m + 1) j + k / √2 Its magnitude is 1, hence (l + 1)2 + (m + 1)2 = 1 / 2 …..(ii) From (i) and (ii), 2lm = 1 / 2 ⇒ l = m = −1 / 2 Hence, a = [−i / 2] − [j / 2] + [k / √2].
a) [−i / 2] − [j / 2] – [k / √2]
b) [−i / 2] + [j / 2] + [k / √2]
c) [i / 2] − [j / 2] + [k / √2]
d) [−i / 2] − [j / 2] + [k / √2]
Q3. Let a, b and c be vectors with magnitudes 3, 4 and 5 respectively and a + b + c = 0, then the values of a . b + b . c + c . a is ________. b) -25 Solution: Since a + b + c = 0 On squaring both sides, we get |a|2 + |b|2 + |c|2 + 2 (a . b + b . c + c . a) = 0 ⇒ 2 (a . b + b . c + c . a) = − (9 + 16 + 25) ⇒ a . b + b . c + c . a = −25
a) 25
b) -25
c) -15
d) 15
Q4. a. [(b + c) × (a + b + c)] is equal to ______. a) 0 Solution: a. [(b + c) × (a + b + c)] = a . (b × a + b × b + b × c) + a . (c × a + c × b + c × c) = [aba] + [abb] + [a b c] + [aca] + [a c b] + [a c c] = 0 + 0 + [abc] + 0 − [abc] + 0 = 0
a) 0
b) 1
c) 2
d) 4
Q5. Let a, b, c be distinct non-negative numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk lie in a plane, then c is b) The geometric mean of a and b
a) The arithmetic mean of a and b
b) The geometric mean of a and b
c) The harmonic mean of a and b
d) Equal to zero
Q6. Let a = 2i + j − 2k and b = i + j. If c is a vector such that a . c = |c|, |c − a| = 2√2 and the angle between (a × b) and c is 30o, then |(a × b) × c| = _________. a) 3/2
a) 3/2
b) 2/3
c) 3/4
d) 4/5
Q7. If b and c are any two non-collinear unit vectors and a is any vector, then (a . b) b + (a . c) c + [a . (b × c) / |b × c|] (b × c) = ___________. d) a Solution: Let i be a unit vector in the direction of b, j in the direction of c. Note that b = i and c = j We have b × c = |b| |c| sinαk = sinαk, where k is a unit vector perpendicular to b and c. ⇒ |b × c| = sinα ⇒ k = [b × c] / [|b × c|] Any vector a can be written as a linear combination of i, j and k. Let a = a1i + a2j + a3k Now a . b = a . i = a1, a . c = a . j = a2 and {[a] . [b × c] / [|b × c|]} = a . k = a3 Thus, (a . b) b + (a . c) c + {[a] . [(b × c) / |b × c|] * [(b × c)|]} = a1b + a2c + a3 [b × c] / [|b × c|] = a1i + a2j + a3k = a
a) b
b) a2
c) b2
d) a
Q8. Let b = 4i + 3j and c be two vectors perpendicular to each other in the xy-plane. All vectors in the same plane having projections 1 and 2 along b and c respectively are given by _________. c) [−2 / 5] i + [11 / 5] j Solution: Let r = λb + μc and c = ± (xi + yj). Since c and b are perpendicular, we have 4x + 3y = 0 ⇒ c = ±x (i − 43j), {Because, y = [−4 / 3]x} Now, projection of r on b = [r . b] / [|b|] = 1 ⇒ [(λb + μc) . b] / [|b|] = [λb . B] / [|b|] = 1 ⇒ λ = 1 / 5 Again, projection of r on c = [r . c] / [|c|] = 2 This gives μx = [6 / 5] ⇒ r = [1 / 5] (4i + 3j) + [6 / 5] (i − [4 / 3]j) = 2i−j or r = [1 / 5] (4i + 3j) − [6 / 5] (i − [4 / 3]j) = [−2 / 5] i + [11 / 5] j
a) [2 / 5] i + [7 / 5] j
b) [−2 / 5] i – [11 / 5] j
c) [−2 / 5] i + [11 / 5] j
d) [2 / 5] i + [11 / 5] j
Q9. If a, b and c are unit vectors, then |a − b|2 + |b − c|2 + |c − a|2 does not exceed ______. d) 9 Solution: |a − b|2 + |b − c|2 + |c − a|2 = 2 (a2 + b2 + c2) − 2 (a * b + b * c + c * a) = 2 * 3 − 2 (a * b + b * c + c * a) = 6 − {(a + b + c)2 − a2− b2 − c2} = 9 − |a + b + c| 2 ≤ 9
a) 4
b) 6
c) 8
d) 9
Q10. A vector has components 2p and 1 with respect to a rectangular cartesian system. The system is rotated through a certain angle about the origin in the anti-clockwise sense. If a has components p + 1 and 1 with respect to the new system, then find p. Solution: If x, y are the original components; X, Y the new components and α is the angle of rotation, then x = X cosα − Y sinα and y = X sinα + Y cosα Therefore, 2p = (p + 1) cosα − sinα and 1 = (p + 1) sinα + cosα Squaring and adding, we get 4p2 + 1 = (p + 1)2 + 1 ⇒ p + 1 = ± 2p ⇒ p = 1 or −1 / 3
a) 1
b) -1/3
c) option (a) or (b)
d)None of these
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