JEE Main Vector Questions with Solutions - Download PDF

The top 10 JEE Main Vector questions that are highly expected in JEE Mains 2024 cover Vector Algebra, Vector Components, Cross Product, Position Vectors, Scalar Vectors, Multiplication Of Vectors, Vector Triple Product, Subtraction Of Vector, Direction Cosines And Magnitude, Orthogonal Vector.

JEE Main 2024 Session 2 is scheduled to be conducted from Apr 4 to Apr 15, 2024. The JEE Main 2024 admit cards for the exams on Apr 4, Apr 5, and Apr 6 have been released by the National Testing Agency on Mar 31, 2024. The admit cards for the other exams shall be released shortly.

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• JEE Main Vector Questions with Answers - Download PDF
• Top 10 JEE Main Vector Questions with Answers

JEE Main Vector Questions with Answers - Download PDF

The PDF containing all the top 10 JEE Main Vector questions with answers is listed below for students to download. Students can directly download the PDF by clicking on it.

Top 10 JEE Mains Vector Questions with Answers

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Top 10 JEE Main Vector Questions with Solutions

The top 10 JEE Main Vector questions with answers are listed below for students to practise to prepare the Vector topics.

Question 1: 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

Answer: |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

Also Check: JEE Main Important Formulas

Question 2: If three non-zero vectors are a = a₁ i + a₂ j + a₃ k, b = b₁ i + b₂ j + b₃ k and c = c₁ i + c₂ j + c₃ k. If c is the unit vector perpendicular to the vectors a and b and the angle between a and b is π / 6, then it is equal to ________.

Answer: |c| = 1, we have |c|² = 1 or c²₁ + c²₂ + c²₃ = 1 …..(i)

Again, since c ⊥ a and c ⊥ b, we have c * a = 0

= a₁c₁ + a₂c₂ + a₃c₃ = 0 …..(ii) and

c * b = 0 = b₁c₁ + b₂c₂ + b₃c₃ = 0 ..…(iii)

Also, since the angle between a and b is π / 6, we have a * b = a₁b₁ + a₂b₂ + a₃b₃

|a| * |b| * cos [π / 6] = a₁b₁ + a₂b₂ + a₃b₃

[3 / 4] (a²₁ + a²₂ + a²₃) (b²₁ + b²₂ + b²₃) = (a₁c₁ + a₂c₂ + a₃c₃)2 …..(iv)

= [1 / 4] (a²₁ + a²₂ + a²₃) (b²₁ + b²₂ + b²₃), {Using (iv)}

= [(Σa²₁) * (Σb²₁)] / [4],

where Σa²₁  = a²₁ + a²₂ + a²₃ and Σb²₁ = b²₁ + b²₂ + b²₃.

Question 3: 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 __?

Answer: 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, the 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

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Question 4: 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.

Answer: 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

Question 5: 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) = ____?

Answer: 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

Question 6: 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 ___?

Answer: |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

= c₂ (x − q + x − r + x − p) − (p . x) p − (q . x) q − (r . x) r = 0

= c₂ {3x − (p + q + r)} − [(p . x) p + (q . x) q + (r . x) r] = 0

which is satisfied by x = [1 / 2] (p + q+ r).

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Question 7: a. [(b + c) × (a + b + c)] is equal to ___?

Answer: 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

Question 8: The horizontal force and the force inclined at an angle 60o with the vertical, whose resultant is in the vertical direction of ‘P’ kg, are ___?

Answer: Let F1 be the horizontal force, and P is the vertical force.

Let F2 be the force inclined at 600 with vertical.

The result is P.

So we can write F1 i + (-F2 cos 30 i + F2 cos 60 j) = Pj

Equating coefficients

F₁ – (√3/2) F₂ = 0

F₂/2 = P

F₂ = 2P

So F₁ = √3 P

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Question 9: 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 ____?

Answer: 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

Question 10: The magnitudes of mutually perpendicular forces a, b and c are 2, 10 and 11, respectively. Then, the magnitude of its resultant is ______.

Answer:R = √[22 + 102 + 112].

= √[4 + 100 + 121]

= 15

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