Mathematics MCQ on Vector Algebra for JEE and Engineering Exam 2022
MCQ on Vector Algebra: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject.
In this post we are providing you MCQ on Vector Algebra, which will be beneficial for you in upcoming JEE and Engineering Exams.
MCQ on Vector Algebra
Q1. 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 ________.
a) 20
b) 25
c) -25
d) none
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
Q2. 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.
a) p = 1 or −1 / 3
b) p = 1 or −1 / 4
c) p = -1 or −1 / 3
d) p = 1 or 1 / 3
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 4p^{2 }+ 1 = (p + 1)^{2 }+ 1
⇒ p + 1 = ± 2p
⇒ p = 1 or −1 / 3
Q3. 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 _________.
a) a = [−i / 2] + [j / 2] + [k / √2].
b) a = [−i / 2] − [j / 2] – [k / √2].
c) a = [−i / 2] − [j / 2] + [k / √6].
d) a = [−i / 2] − [j / 2] + [k / √2].
Solution:
Let a = li + mj + nk, where l^{2 }+ m^{2} + n^{2} = 1. a makes an angle π / 4 with z−axis.
Hence, n = 1 / √2, l^{2 }+ m^{2} = 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].
Q4. Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies equation p × {(x − q) × p} + q × {(x − r) × q} + r × {(x − p) × r} = 0, then x is given by ____________.
a) x = [-1 / 2] (p + q+ r).
b) x = [1 / 2] (p + q+ r).
c) x = [1 / 2] (p – q+ r).
d) none
Solution:
|p| = |q| = |r| = c, (say) and
p . q = 0 = p . r = q . r
p × |( x − q) × p |+ q × |(x − r) × q| + r × |( x − p) × r| = 0
⇒ (p . p) (x − q) − {p . (x − q)} p + . . . . . . . . . = 0
⇒ c^{2 }(x − q + x − r + x − p) − (p . x) p − (q . x) q − (r . x) r = 0
⇒ c^{2 }{3x − (p + q + r)} − [(p . x) p + (q . x) q + (r . x) r] = 0
which is satisfied by x = [1 / 2] (p + q+ r).
Q5. If a, b and c are unit vectors, then |a − b|^{2} + |b − c|^{2} + |c − a|^{2} does not exceed
a) 4
b) 9
c) 8
d) 6
Solution:
|a − b|^{2} + |b − c|^{2} + |c − a|^{2 }= 2 (a^{2 }+ b^{2} + c^{2}) − 2 (a * b + b * c + c * a)
= 2 * 3 − 2 (a * b + b * c + c * a)
= 6 − {(a + b + c)^{2 }− a^{2}− b^{2 }− c^{2}}
= 9 − |a + b + c| 2 ≤ 9
Q6. a. [(b + c) × (a + b + c)] is equal to ______.
a) 1
b) -1
c) 0
d) none
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
Q7. 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 _________.
a) [−2 / 3] i + [11 / 5] j
b) [−2 / 5] i + [18 / 5] j
c) [2 / 5] i + [11 / 4] j
d) [−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
Q8. The magnitudes of mutually perpendicular forces a, b and c are 2, 10 and 11, respectively. Then the magnitude of its resultant is ______.
a) 14
b) 15
c) -15
d) none
Solution:
R = √[2^{2} + 10^{2} + 11^{2}].
= √[4 + 100 + 121]
= 15
Q9. 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) = ___________.
a) 0
b)-a
c) a
d) none
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 = a_{1}i + a_{2}j + a_{3}k
Now a . b = a . i = a_{1}, a . c = a . j = a_{2} and {[a] . [b × c] / [|b × c|]} = a . k = a_{3}
Thus, (a . b) b + (a . c) c + {[a] . [(b × c) / |b × c|] * [(b × c)|]}
= a_{1}b + a_{2}c + a_{3} [b × c] / [|b × c|]
= a_{1}i + a_{2}j + a_{3}k
= a
Q10. 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 30^{o}, then |(a × b) × c| = _________.
a) 4/5
b) 2/3
c) 3/2
d) none
Solution:
a × b = $∣∣∣∣∣∣∣ i j k ∣∣∣∣∣∣∣ $ = 2i − 2j + k
Therefore, |a × b| = √[4 + 4 + 1] = 3
|c − a| = 2√2
⇒ |c − a|^{2} = (c − a)^{2 }= 8
⇒ |c|^{2} − 2 a . c + |a|^{2} = 8
⇒|c|^{2} − 2|c| + 9 = 8
⇒|c|^{2 }− 2|c| + 1 = 0
⇒|c| = 1
Hence, |(a × b) × c| = |a × b| |c|sin 30^{∘} = 3 / 2.
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