−

−
−→

∣∣
Q.1

(A)
(B)

(C)

(D)

Q.2

→
→ →
If m,

(D)

Q.3

(B)

(C)

(D)

Q.4

→
→

→
n

m + n

(A)

(B)

(C)
2
−2
−1
−4

5
21

10

3
2
→
Quizrr Chapter-Wise Test for JEE Advanced - 2025

If vectors 2i − j + k, i + 2j − 3k and 3i + aj + 5k are coplanar, then the value of a is

→
= 2 + 4 m × n
→

, then the value of

→ →→
[m n r]
2

is equal to
→

The position vector of a point in which a line through the origin perpendicular to the plane 2x − y
the plane r

(A) (1, −1, −1)

(−1, −1, 2)
(4, 2, 2)
( 43 , −2
3
​ , −2
3
)

to the area of ΔAOC is

(A)

(B)
(C)
(D)
3 : 1

3 : 2

2 : 1

4 : 3
⋅ (3^i − 5^j + 2k^) = 6, is

​ ​
​

Let O be an interior point of triangle ABC, such that 2OA + 3OB + 4OC = 0,
→
are non-parallel unit vectors and r is a vector which is perpendicular to m and n such that
→
By: C I P H Ξ R

r = 5 and

− z = 4 meets

then the ratio of the area of ΔABC

@IITJEE_Advanced
Q.5

The resultant of forces P and Q is R. If Q is doubled, then R is doubled. If the direction of Q is reversed, then R
​ ​ ​

2 2 2
is again doubled. Then P :Q :R =
(A) 3:1:1
(B) 2:3:2
(C) 1:2:3
(D) 2:3:1

Q.6

Let a1 , a2 … ..an be the sides of a regular polygon inscribed in a circle of unit radius. If
​ ​ ​

[a1 × a2 + a2 × a3 … .. + an × a1 ] = [a1 ⋅ a2 + a2 ⋅ a3 + … .. + an ⋅ a1 ] smallest possible value of

​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​

n is
(A) 4
(B) 6

(C) 8

(D) 12

Q.7

Let A(−1, 0, 0), B(0, 1, 0), C(0, 0, −1) are the vertices of △ABC. If R and r denote the circumradius and
inradius of △ABC , then the value of (R2 + r2 ) is equal to
(A) 56 ​

(B) 16 ​

(C) 23 ​

(D) 13 ​

Q.8

ABCD is a parallelogram. Point P lies on side BC while point Q lies on CD such that BP = 4PC and
CQ = 4QD. The line PQ intersect the line AC at point E .
​ ​

k1
If AE = ​

k2
​
AC, then value of k1
​ ​ + k2 (where k1 and k2 are coprime numbers), is-
​ ​ ​

(A) 46

(B) 27
(C) 19

(D) 35

@IITJEE_Advanced
Q.9

If (R − S) ⋅ t^ = 0, (R − S) ⋅ n ^ = 0, where t^, n
^ = 0 and (R − S) ⋅ b ^ form a right handed triad of
^, b
orthogonal vectors, then -

(A) R=S
(B) R×S=0
(C) R⋅S=0
(D) R=0

Q.10

Which of the following statement(s) is/are correct :

(A) If a, b, c are non-coplanar and d is any vector, then [dbc]a + [dca]b + [dab]c − [abc]d = 0
(B) If I is incentre of △ABC then ∣BC∣IA + ∣CA∣IB + ∣AB∣IC = 0
(C) Any vector in three dimension can be written as linear combination of three non-coplanar vectors.

(D) In a triangle, if position vector of vertices are a, b, c, then position vector of incentre is a+b+c
3
​

Q.11

i + ^j − 2k^ and ^i − 3^j + k^. The position vector of a

A line passes through the points whose position vectors are ^ ​ ​

point on it at a distance 2 units from the first point is -

(A) ^i − 13 ^
5
j + 16 5
k^
​ ​ ​

(B) ^ 13 ^ 16 ^
i+ 5 j− 5 k
​ ​ ​

(C) ^i + 3^
5
j + 45 k^
​ ​ ​

(D) ^i − 3^ 4^
5j − 5k
​ ​ ​

Q.12

Passage:
^), ^i and 3i^ respectively.
The position vectors of the vertices A, B and C of a tetrahedron ABCD are (^i + ^j + k ​

The altitude from the vertex D to the opposite face ABC meets the median AF of △ABC at E.AD = 4 and
AE = 2 3. ​

Question:
Area of ΔABC

(A) 2 sq. units

​

(B) 2 2 sq. units

​

1
(C) sq. units
2
​

(D) 2sq. units

@IITJEE_Advanced
Q.13

Passage:
^), ^i and 3i^ respectively.
The position vectors of the vertices A, B and C of a tetrahedron ABCD are (^i + ^j + k ​

The altitude from the vertex D to the opposite face ABC meets the median AF of △ABC at E.AD = 4 and
AE = 2 3. ​

Question:
Volume of the tetrahedron DABC =
(A) 2
3
cubic units
​

(B) 2 2 cubic units

​

3
​

(C) 2 2 cubic units ​

2
(D) 3 cubic units
​

Q.14

Passage:
^), ^i and 3i^ respectively.
The position vectors of the vertices A, B and C of a tetrahedron ABCD are (^i + ^j + k ​

The altitude from the vertex D to the opposite face ABC meets the median AF of △ABC at E.AD = 4 and
AE = 2 3. ​

Question:
If F and E are on the same side of A, then the positive vector of E is -

(A) ^
3^i − ^j − k ​

(B) 3^i + ^j + k
^ ​

(C) −^i + 3^j + 3k^ ​

i − 3^j − 3k^
(D) ^ ​

Q.15

Passage:
b× c c×a a×b
The three vectors a ′, b ′, c ′ defined by a′ = ​, b′ = ​
, c′ = ​, where a , b , c are
[a b c] [a b c] [a b c]
non-coplanar are called the reciprocal system to the vectors a, b, c .
Question:
a × a′ + b × b′ + c × c′ equals −
(A) 0
(B) a + b + c ​

[a b c]

(C) [abc](a + b + c)
(D) None of these

@IITJEE_Advanced
Q.16

Passage:
b× c c×a a×b
The three vectors a ′, b ′, c ′ defined by a′ = ​ , b′ = ​
, c′ = ​ , where a , b , c are
[a b c] [a b c] [a b c]

non-coplanar are called the reciprocal system to the vectors a, b, c .

Question:
a ′ × b ′ + b ′ × c ′ + c ′ × a ′ equals −
(A) a +b+ c
(B) a × (b × c)
(C) a + b + c ​

[a b c]

(D) (a × b) × c

Q.17

Passage:
b× c c×a a×b
The three vectors a ′, b ′, c ′ defined by a′ = , ​ b′ = , ​ c′ = , where
​ a , b , c are
[a b c] [a b c] [a b c]
non-coplanar are called the reciprocal system to the vectors a, b, c .
Question:
(a + b + c) ⋅ (a′ + b′ + c′ )

(A) 3

(B) 2

(C) 1

(D) 0

Q.18

@IITJEE_Advanced
 a and b are two unit vectors inclined at an angle θ to each other. Then −

(A) (A) - s, (B) - q, (C) - r, (D) - t

(B) (A) - t, (B) - t, (C) - q, (D) - q

(C) (A) - r, (B) - p, (C) - p, (D) - r

(D) (A) - r, (B) - t, (C) - p, (D) - q

Q.19

Let O be an interior point of ΔABC such that OA + 2OB + 3OC = 0 , then find the ratio of area of
△ABC to area of △AOC.

Q.20

A cube in the first octant has sides OP , OQ and OR of length 1, along the x-axis, y-axis and z-axis, respectively,
where O(0, 0, 0) is the origin. Let the centre of the cube be at S( , , ) and T be the vertex of the cube opposite
1

2
1

2
1

−
−
−
− → → → → →
→ → →
O S OT to the origin
p = SP , q = SQ r = SR such that
t = ST lies on the diagonal . If , and , then the value of
→ → → →
(p × q) × (r × t) is :

Answers & Solutions

Q.1 Answer:
−4
Solution:

If given vectors are coplanar, then there exists two scalar quantities x and y such that 2i − j + k =
x(i + 2j − 3k) + y(3i + aj + 5k) ....(1)

Comparing coefficient of i, j and k on both sides of (1) we get x + 3y = 2, 2x + ay = −1, −3x +

5y = 1 .....(2)
@IITJEE_Advanced
5

∣
Solving first and third equations, we get x

Q.2 Answer:

Solution:
→

=
→→ →
m + n

1 + 1 + 2 cos

⇒ tan θ =

[m n r]

→
r

= 25 ×

Q.3 Answer:
( 34 , −2
​

3
Solution:
, −2
3
​ )
2(1/2) + a(1/2) = −1 ⇒ a = −4

is parallel to m
Let θ be angle between m and n
→ →

2
2

→
m

5
2

​
2

= 5
1

→
n sin

Vector perpendicular to 2x − y
Also line is collinear with n.
So equation of line is r

So, λ(6 + 5 − 2)
⇒λ= 2
3
→

→ →

2
→

θ
→
× n

→
= 2 + 4 m × n
→

θ = 2 + 4 × sin θ

→
r. (m × n)

2
2

= λ(2^i − ^j − k^) ...(1)

Now line (1) meet the plane r ⋅ (3^

​
→

i − 5^j + 2k^) = 6

=6

Hence p.v. of the point 'A' is 32 (2^i − ^j − k

Q.4 Answer:
^) ​
− z = 4 is

​
= 1/2, y = 1/2

Since the vectors are coplanar, therefore these values of x and y will satisfy the equation 2x + ay

​
n = 2^i − ^j − k^
​
= −1

@IITJEE_Advanced
−
−
−

−
−

−
−
→ → →

ΔAOC =

ΔABC =
1

1
→

→
→
OA × OC =

→
BA × BC =
3 : 1

Solution:

1 →

=
→ →
Area of

1
2

∣
Let position vectors of O, A, B, C are O, a, b, c respectively.

2OA + 3OB + 4OC = 0

→ →

→ →

Area of ΔAOC

Q.5 Answer:
2:3:2
Solution:

​
→
⇒ 2a + 3b + 4c = 0

→ →
Area of
a × c

→ → →
a × b + b × c + c × a

a × (

4→ →
c × a +

c × a

Area of ΔABC

​
=
→ →
−4c−2a

2→

P + Q = R ⇒ P 2 + Q2 + 2P ⋅ Q = R2 ...(i)
2

) + (

→ → →
→ →
−4c−2a

c × a + c × a

∣P + 2Q∣ = 2∣R∣ ⇒ P 2 + 4Q2 + 4P ⋅ Q = 4R2 ...(ii)

​

∣P − Q∣ = 2∣R∣ ⇒ P 2 + Q2 − 2P ⋅ Q = 4R2 ...(iii)

(ii) +2( iii )

or P 2
​

(i) + (iii) ⇒ 2P 2 + 2Q2 − 5R2 = 0 ...(iv)

⇒ 3P 2 + 6Q2 − 12R2 = 0 ...(v)
+ 2Q 2 − 4R 2 = 0
​

By cross multiplication from (iv), (v), p2

Q.6 Answer:
8
Solution:
2
​
​
​

​
2

→ → →
) × c + c × a

​
→→→→

​
​

=
​

Q2
3
​

​
​

= R2
2
.
​
​

@IITJEE_Advanced
 2π
∣a1 × a2 ∣ = ∣a1 ∣ ∣a2 ∣ sin
​ ​ ​ ​ ​

n
2π
∣a1 ⋅ a2 ∣ = ∣a1 ∣ ∣a2 ∣ cos
​ ​ ​ ​ ​

n
2π 2π
n sin = n cos ​ ​

n n ​

2π 2π
tan ( ) = 1 ⇒
π
= mπ +
4
​ ​ ​

n n
8
n=
1 + 4m
​

Smallest possible value of n corresponds to m = 0 ⇒ n = 8

Q.7 Answer:
5
6
​

Solution:

Clearly △ABC is equilateral.

r 1 1 1 1
⇒ =4⋅ ⋅ ⋅ =
2 2 2 2
​ ​ ​ ​ ​

​
R ​

⇒ 2r = R
abc 2 2 2 1
Also, R = 4Δ
= ​

= 3
⇒r=
4⋅ 43 ⋅2 6
​ ​ ​ ​ ​

1 2 5
∴ R2 + r 2 = 6
​
+ 3
​
= 6
​

Q.8 Answer:
46
Solution:

@IITJEE_Advanced
 μb
+ 4d
E = bλ+1
+d
= 5μ+15
​ ​

​ ​

⇒ μ = 4 and λ = 214 ​

1 4 1
∴ 5(μ+1) = λ+1 and 5(μ+1)
μ
=
​

λ+1
​ ​ ​ ​

AE = ( λ+1
λ
AC) = ​
21
25
AC
​

Q.9 Answer:
R=S
Solution:

(R − S) ⋅ t^ = 0
⇒ either R − S = 0 or R − S is perpendicular to t^.
Similarly, the other two conditions. But R − S cannot be perpendicular to all three t^, n ^ since they
^ and b
are themselves mutually perpendicular.
∴ R − S = 0 ⇒ R = S and hence R × S = 0

Q.10 Answer:

If a, b, c are non-coplanar and d is any vector, then [dbc]a + [dca]b + [dab]c − [abc]d = 0
Solution:

(A) [db c]a + [dca]b + [dab]c − [abc]d

= ([bcd]a − [bca]d) + ([dab]c − [dac]b) = (b × c) × (a × d) + (d × a) × (c × b) ​

= ( d × a ) × ( b × c ) − (d × a ) × (b × c ) = 0
(B)

aa+bb+cc
I≡ a+b+c
​

Q.11 Answer:
^i + j 16 k^
13 ^
5 − 5
​ ​ ​

@IITJEE_Advanced
Solution:

AB
OP = r = OA ± 2 ⋅ ​

∣AB∣

^ ^ ^ OB − OA ^ ^ ^ 2(−4^j + 3k ^)
=i j + − 2 k ± 2 ⋅ = i j+ − 2 k ±
​

​ ​

∣ − 4^j + 3k ^∣
​ ​ ​ ​

∣ OB − OA ∣ ​

=^i + ^j − 2k^ ± 1 (−8^j + 6k ^) = ^i + (1 − 8 ) ^j + k ^ (−2 + 6 )

5 5 5
​ ​ ​ ​ ​ ​

^ i.e., ^i − 3^j − 4 k
or ^i + (1 + 85 ) ^j + (−2 − 65 ) k ^ or ^i + 13^j − 16 k
^
5 5 5 5
​ ​ ​ ​ ​ ​ ​ ​ ​

Q.12 Answer:
2 sq. units
​

Solution:

1
(i) Area of ΔABC
= 2
∣AB ​ × AC∣
∣ ^i ^j k^ ​
∣
1
= 2 0 −1 −1
​ ​ ​ ​ ​ ​ = 21 ∣ − 2^j + 2k^∣ =
​ ​
1
2 ​ ×2 2= ​ 2 sq. units
​

∣ 2 −1 −1 ∣
Volume of the tetrahedron = 13 (area of ΔABC)× height = 31 ( ​ ​ 2) × 2 =
​
2 2
3
​

​cubic units
(ii) AD = 4 and AE = 2 3 ​

Hence, DE = 16 − 12 = 2
​

@IITJEE_Advanced
 = 2i^
(iii) P.V. of F
^i−^j−k^
AF = ^i − ^j − k
^, AF
^ = ​

3
​

^ ^ ^
​

​
^)
Since, A = 2 3, AE = 2 3 (i−j−3 k) = 2(^i − ^j − k ​
​

​ ​
​

P.V. of E = P.V. of A + AE = ^i + ^j + k^ + 2(^i − ^j − k

^) = 3^i − ^j − k
^
​

​ ​ ​

Q.13 Answer:
2 2
3
cubic units
​

Solution:

(i) Area of ΔABC = 21 ∣AB × AC∣ ​

∣ ^i ^j k^ ∣​

= 12 0 −1 −1 = 21 ∣ − 2^j + 2k^∣ = 12 × 2 2 = 2 sq. units

​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​

∣ 2 −1 −1 ∣
(ii) AD = 4 and AE = 2 3 ​

Hence, DE = 16 − 12 = 2
​

Volume of the tetrahedron = 31 ( area of ΔABC)× height = 31 ( 2) × 2 =

​ ​ ​
2 2
3 cubic units
​

Q.14 Answer:
^
3^i − ^j − k​

Solution:

1
(i) Area of ΔABC
= 2
∣AB ​ × AC∣
∣ ^i ^j k^ ​
∣
= 12 0 −1 −1
​ ​ ​ ​ ​ ​ = 21 ∣ − 2^j + 2k^∣ =
​ ​
1
2
​ ×2 2= ​ 2 sq. units
​

∣ 2 −1 −1 ∣
(ii)
@IITJEE_Advanced
 AD = 4 and AE = 2 3 ​

Hence, DE = 16 − 12 = 2
​

2 2
Volume of the tetrahedron = 31 ( area of ΔABC)× height = 31 ( 2) × 2 =
​ ​ ​

3
cubic units
​

(iii) P.V. of F = 2i^

AF = ^i − ^j − k ^ = ^i−^j−k^
^, AF ​

3
​ ​

^ ^ ^
​

Since, A = 2 3, AE = 2 3 (i−j−3 k) = 2(^i − ^j − k

​
^) ​
​

​ ​
​

P.V. of E = P.V. of A + AE = ^i + ^j + k^ + 2(^i − ^j − k

^) = 3^i − ^j − k
^
​

​ ​ ​

Q.15 Answer:

0
Solution:

(b× c ) ( a ⋅ c ) b −( a ⋅ b ) c
a × a′ = a × ​ = ​

[a b c] [a b c]

Similarly, b× b′ = ( b ⋅ a ) c −( b ⋅ c ) a and ​ c × c′ = ( c ⋅ b ) a −( c ⋅ a ) b
​

[a b c] [a b c]
∴ Σa × a ′ = 0 (∵ a ⋅ b = b ⋅ a etc. )
( b × c )×( c × a ) ( b × c ⋅ a ) c −( b × c ⋅ c ) a [b c a] c c
(ii) a′ × b′ = ​ = ​ = ​ = ​

[a b c ]2 [a b c ]2 [a b c ]2 [a b c]

(∵ [ b c a ] = [ a b c ] and b × c ⋅ c = 0)
Similarly, we can find b′ × c′ and c′ × a′ as a and b
respectively.
[abc] [abc]
​ ​

a+b+ c
∴ ∑ a′ × b′ = ​

[a b c]
a⋅(b×c) a⋅(c×a)
(iii) a ⋅ a′ =
[abc]
= [[abc
​
abc]
]
​
= 1 and a ⋅ b′ = [abc]
​
=0
a ⋅ c ′ = 0 in a similar way.
Hence, (a + b + c) ⋅ (a′ + b′ + c′ ) = 1 + 1 + 1 = 3.

Q.16 Answer:
a+b+ c
​

[a b c]
Solution:

(b× c ) ( a ⋅ c ) b −( a ⋅ b ) c
a × a′ = a × ​ = ​

[a b c] [a b c]
b a c b c a ( c ⋅ b ) a −( c ⋅ a ) b
Similarly, b× b′ = ( ⋅ ) −( ⋅ ) and ​ c × c′ = ​

[a b c] [a b c]
∴ Σa × a ′ = 0 (∵ a ⋅ b = b ⋅ a etc. )
( b × c )×( c × a ) ( b × c ⋅ a ) c −( b × c ⋅ c ) a [b c a] c
(ii) a′ × b′ = ​
= ​
= ​
= c
​

[a b c ]2 [a b c ]2 [a b c ]2 [a b c]
(∵ [ b c a ] = [ a b c ] and b × c ⋅ c = 0)
Similarly, we can find b′ × c′ and c′ × a′ as
a b
and respectively.
[ ] [abc]
​ ​

abc
a+b+ c
∴ ∑ a′ × b′ = ​

[a b c]
a⋅(b×c)
(iii) a ⋅ a′ = = [[abc]] = 1 and a ⋅ b′ = a⋅(c×a)
=0
[abc] [abc] @IITJEE_Advanced
​ ​ ​

abc
 a ⋅ c ′ = 0 in a similar way.
Hence, (a + b + c) ⋅ (a′ + b′ + c′ ) = 1 + 1 + 1 = 3.

Q.17 Answer:
3
Solution:

(b× c ) ( a ⋅ c ) b −( a ⋅ b ) c
a × a′ = a × ​ = ​

[a b c] [a b c]
( ⋅ ) −( ⋅ c)a ( c ⋅ b ) a −( c ⋅ a ) b
Similarly, b× b′ = b a c b
and ​ c × c′ = ​

[a b c] [a b c]
∴ Σa × a ′ = 0 (∵ a ⋅ b = b ⋅ a etc. )
( b × c )×( c × a ) ( b × c ⋅ a ) c −( b × c ⋅ c ) a [b c a] c c
(ii) a′ × b′ = ​ = ​ = ​ = ​

[ a b c ]2 [ a b c ]2 [ a b c ]2 [a b c]

(∵ [ b c a ] = [ a b c ] and b × c ⋅ c = 0)
Similarly, we can find b′ × c′ and c′ × a′ as a and b
respectively.
[abc] [ ]
​ ​

abc
a+b+ c
∴ ∑ a′ × b′ = ​

[a b c]
a⋅(b×c) [abc] a⋅(c×a)
(iii) a ⋅ a′ = = = 1 and a ⋅ b′ = =0
[abc] [abc] [abc]
​ ​ ​

a ⋅ c ′ = 0 in a similar way.
Hence, (a + b + c) ⋅ (a′ + b′ + c′ ) = 1 + 1 + 1 = 3.

Q.18 Answer:
(A) - r, (B) - t, (C) - p, (D) - q
Solution:

(A) − r, (B) − t, (C) − p, (D) − q

(A) ∣ a+ b ∣2 < 2 ⇒ 1 + 1 + 2 a ⋅ b < 2 ⇒ a ⋅ b < 0
⇒ cos θ < 0 ⇒ π2 < θ < π ​

(B) −2 ⋅ a ⋅ b < 0 ⇒ a ⋅ b > 0 ⇒ 0 < θ < π2 ​

(C)
∣ a + b ∣ < 1 ⇒ 2 a ⋅ b < −1 ⇒ a ⋅ b < − 21 ⇒ cos θ < − 12 ​ ​

⇒ −1 < cos θ < − 21 ⇒ 2π 3 <θ <π

​ ​

(D) 1 < ∣a − b∣ < 2 ⇒ ∣a − b∣ > 1 and ∣a − b∣ < 2

​ ​

⇒ −2a ⋅ b > −1 and −2a ⋅ b < 0

1
⇒ a ⋅b < 2
​and a ⋅ b >0
π π
⇒ 3
​ <θ< 2
​

Q.19 Answer:
—
Solution:

@IITJEE_Advanced
 1
area of (ΔABC) 2∣
OA×OB∣+∣OB×OC∣+∣OC×OA∣ ∣OA×OB∣+∣OB×OC∣
= = +1
​

area of (ΔOAC) 1
​ ​ ​

2 ∣OA×OC∣ ∣OA×OC∣
​

Given : OA + 2OB + 3OC = 0

2OB × OA + 3OC × OA = 0
∣OB × OA∣ ∣OC × OA∣
⇒ = = λ.....(1)
3 2
​ ​

OB × OA + 3OB × OC = 0
​ ​

∣OB × OA∣ ∣OB × OC∣

⇒ = = λ.......(2)
3 1
​ ​

ΔABC 3λ + λ
⇒ Ratio of area of Δ s i.e., = +1=3
ΔAOC 2λ
​ ​

Q.20 Answer:
—
Solution:

− →
→ 1 1 1 1
p = SP = ( ,− ,− ) = (î − ĵ − k̂)
2 2 2 2

− →
→ 1 1 1 1
q = SQ = (− , ,− ) = (−î + ĵ − k̂)
2 2 2 2

− →
→ 1 1 1 1
r = SR = (− ,− , ) = (−î − ĵ + k̂)
2 2 2 2

−
→ →
1 1 1 1
t = ST = ( , , ) = (î + ĵ + k̂)
2 2 2 2

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∣
=
→ →

16
→ →
(p × q) × (r × t) =
1
4
î

−1

(2î + 2ĵ) × (−2î + 2ĵ)

ĵ

−1

=
−1

−1

k̂

2
k̂

=
×

2
1
4
î

−1

1
ĵ

−1

1
k̂

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