 # JEE Main Maths Vector Algebra Previous Year Questions With Solutions

## What is Vector Algebra?

An algebra is a set of mathematical rules. And in order to use vector algebra, we have to know the rules. Vectors are used to represent a quantity with a direction. We have already learnt about, basic Concepts of Vectors, its components, vectors type, addition of Vectors, scalar and vector product of two vectors etc. Those studying a calculus-based physics course also have to consider how to multiply vectors and other vector operations. This article covers the position vector of a point, scalar component of a vector, relation between direction ratios, direction cosines and magnitude, parallelogram law of vector, unit vector, multiplication of a vector by a constant, scalar product, cross product and coplanar vectors.

The vector algebra questions from the previous years of JEE Main are present on this page, along with the detailed solution for each question. About 2-3 questions are asked from this topic in JEE Examination.

## JEE Main Maths Vector Algebra Previous Year Questions With Solutions

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

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

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 $[\begin{vmatrix} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3} \end{vmatrix}]^2$ is equal to ________.

Solution:

|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 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) = (a1b1 + a2b2 + a3b3)2 …..(iv)

Now,

$[\begin{vmatrix} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3} \end{vmatrix}]^2$ = $\begin{vmatrix} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3} \end{vmatrix}$ * $\begin{vmatrix} a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3} \end{vmatrix}$

= $\begin{vmatrix} [a_{1}]^2+[a_{2}]^2+[a_{3}]^2 & [a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}] & 0\\ [b_{1}a_{1}+b_{2}a_{2}+b_{3}a_{3}] & [b_{1}]^2+[b_{2}]^2+[b_{3}]^2 & 0\\ 0 & 0 & 1 \end{vmatrix}$

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

= [(Σa21) * (Σb21)] / ,

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 _________.

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

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.

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

Question 5: If the vectors ai + j + k, i + bj + k and i + j + ck (a ≠ b ≠ c ≠ 1) are coplanar, then the value of  / [1 − a] +  / [1 − b] +  / [1 − c] = _________.

Solution:

Since $\begin{vmatrix} a &1 &1 \\ 1&b &1 \\ 1&1 &c\end{vmatrix}$ = 0

Applying R2 → R2 − R1 and R3 → R3 − R1, we get

$\begin{vmatrix} a &1 &1 \\ 1-a&b-1 &0 \\ 1-a&0 &c-1\end{vmatrix}$ = 0

On expanding, we get

a (b − 1) (c − 1) − (1 − a) (c − 1) − (1 − a) (b − 1) = 0

On dividing by (1 − a) (1 − b) (1 − c), we get

[a] / [1 − a] +  / [1 − b] +  / [1 − c] = 0

⇒  / [1 − a] +  / [1 − b] +  / [1 − c]

= { / [1 − a]} − {[a] / [1 − a]}

= 1

Question 6: 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) = ___________.

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

Question 7: 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 ____________.

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

⇒ 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).

Question 8: 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| = _________.

Solution:

a × b = $\begin{vmatrix} i & j & k \\ 2 &1 &-2 \\ 1&1 &0 \end{vmatrix}$ = 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.

Question 9: 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

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

Solution:

$\begin{vmatrix} a & a & c \\ 1 &0 &1 \\ c&c &b \end{vmatrix}$ = 0

$\begin{vmatrix} a &0 &c \\ 1 &-1 &1 \\ c&0 &b \end{vmatrix}$

=0 {Applying C2 → C2 − C1}

⇒ a (−b) + c (c) = 0

⇒ c2 = ab

Hence, c is the geometric mean of a and b.

Question 10: a. [(b + c) × (a + b + c)] is equal to ______.

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

Question 11: 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 ________.

Solution:

Let $\overrightarrow{OA}$ = P1 i,

$\overrightarrow{CB}$ = −P1 i,

$\overrightarrow{OB}$ = −P1i + Pj

$\frac{\overrightarrow{OB}.{j}}{OB}=cos60^o$

⇒ [(−P1i + Pj) . j] / [√P21 + P2] = [1 / 2]

⇒ 2P = [√P21 + P2]

⇒ P1 = P√3

$|\overrightarrow{OB}|$ = [√P21 + P2]

= √[P2 + 3P2]

= 2P

Question 12: 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 ________.

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

Question 13: 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 _________.

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].

Question 14: The magnitudes of mutually perpendicular forces a, b and c are 2, 10 and 11, respectively. Then the magnitude of its resultant is ______.

Solution:

R = √[22 + 102 + 112].

= √[4 + 100 + 121]

= 15