Matrix-Match Type Questions: Vector Algebra and Three Dimensional Geometry - 2 | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

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DIRECTIONS (Q. 7-9) : Each question has matching lists. The codes for the lists have choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Q. 7. Match List I with List II and select the correct answer using the code given below the lists :

List I                                                                                                                             List II

P. Volume of parallelepiped determined by vectors                                               1. 100
 Then the volume of the parallelepiped determined by vectors

Q. Volume of parallelepiped determined by vectors                       2. 30
 Then the volume of the parallelepiped determined by vectors

R. Area of a triangle with adjacent sides determined by vectors                           3. 24
 Then the area of the triangle with adjacent sides
 determined by vectors 

S. Area of a parallelogram with adjacent sides determined by vectors                 4. 60
 Then the area of the parallelogram with adjacent sides
 determined by vectors

Ans. (c)

Solution. 

Q. 8. Consider the lines and the planes P₁ : 7x + y + 2z = 3, P₂ = 3x + 5y – 6z = 4. Let ax + by + cz = d be the equation of the plane passing through the point of intersection of lines L₁ and L₂, and perpendicular to planes P₁ and P₂.

Match List I with List II and select the correct answer using the code given below the lists :

Ans. (a)

Solution. Any point on L₁ is (2λ + 1, – λ, λ – 3) and that on L₂ is (μ + 4, μ – 3, 2μ –3)

For point of intersection of L₁ and L₂ 2λ + 1 = μ + 4, – λ = μ – 3, λ – 3 = 2μ – 3 ⇒ λ = 2, μ = 1

∴ Intersection point of L₁ and L₂ is (5, –2, – 1)

∵ ax + by + cz = d is perpendicular to p₁ & p₂

∴ 7a + b + 2c = 0 and 3a + 5b – 6c = 0

∴ Equation of plane is x – 3y – 2z = d

As it passes through (5, –2, –1)

∴ 5 + 6 + 2 = d = 13

∴ a = 1, b = –3, c = –2, d = 13

or (P)  → (3) (Q)  → (2)  (R)  → (4) (S)  → (1)

Q. 9. Match List I with List II and select the correct answer using the code given below the lists :

List - I                                                                                                               List - II

                               1. 1

Q. Let A₁, A₂, ..., A_n (n > 2) be the vertices of a regular                                  2. 2
 polygon of n sides with its  centre at the origin. Let

be the position vector of the point A_k, k = 1, 2, ....., n.

 then the minimum value of n is

R. If the normal from the point P(h, 1) on the ellipse                                        3. 8
 is perpendicular  to the line x + y = 8, then the
 value of h is

S. Number of positive solutions satisfying the equation                                      4. 9
 

Ans. (a)

Solution. 

Q(3) ... A₁, A₂, A₃, ... A_n are the vertices of a regular 

 are their position vectors.

Hence given equation reduces to

It is perpendicular to x + y = 8

∴ Only one +ve solution is there

Hence (a) is the correct option.

DIRECTIONS (Q. 10-11) :  Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :

If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s then the correct darkening of bubbles will look like the given.

Q. 10. Match the following :

Column I                                                                                                                   Column II

(A) In R², if the magnitude of the projection vector of the vector                         (p) 1
  then possible value
 of |α| is/are

(B) Let a and b be real numbers such that the function                                           (q) 2

Then possible value of a is (are)

(C) Let  ω ≠ 1 be a complex cube root of unity.                                                        (r) 3

If 
 then possible value (s) of n is (are) 

(D) Let the harmonic mean of two positive real numbers a and b be 4.                 (s) 4
 If q is a positive real number such that a, 5, q, b is an arithmetic
 progression, then the value(s) of | q – a | is (are)

Ans. (A) → q; (B) → p, q; (C) → p, q, s, t; (D) → q, t

Solution. 

∴–3a – 2 = b + a²
⇒ a² – 3a + 2 = 0 (using (i))
⇒ a = 1, 2

⇒ 4n + 3 should be an integer other than multiple of 3.

∴ n = 1, 2, 4, 5

Putting values of a and b in eqn(i) 

Q. 11. Match the following :

Column I                                                                                                                Column II

(A) In a triangle DXYZ, let a, b, and c be the lengths                                             (p) 1
 of the sides opposite to the angles X, Y and Z,
 respectively. If 2(a² – b²) = c² and 
 , then possible values of n for which  cos(nπλ) = 0 is (are) 

(B) In a triangle DXYZ, let a, b and c be the lengths of the                                      (q) 2
 sides opposite to the angles X, Y, and Z respectively. If
 1 + cos 2X – 2cos 2Y = 2 sin X sin Y, then possible value
 (s) of a/b is (are)

(C) In R², let   be the position                                      (r) 3
 vectors of X , Y and Z with respect to the origin O,
 respectively. If the distance of Z from the bisector of the
 acute angle of  then possible
 value(s) of |β| is (are)

(D) Suppose that F(a) denotes the area of the region bounded                              (s) 5
 by x = 0, x = 2, y² = 4x and y =| αx- 1 | + | αx- 2 | +αx ,
 where a∈ {0, 1} . Then the value(s) of   , when
 a = 0 and a = 1, is (are)

                                                                                                                                       (t) 6

Ans. (A)→p, r, s;  (B) → p; (C) → p, q; (D)→s, t

Solution. (A) 2(a² – b²) = c² 
⇒ 2(sin²x – sin²y) = sin²z
⇒ 2sin(x + y) sin(x – y) = sin²z
⇒ 2sin(x – y) = sin z     (∵sin(x + y) = sin z)

(B) 1 + cos2X – 2cos2Y = 2sinXsinY
⇒ 2cos²X – 2cos2Y = 2sinXsinY
⇒ 1 – sin²X – 1 + 2sin²Y = sinXsinY

⇒ sin²X + sinXsinY – 2sin²Y = 0
⇒ (sinX – sinY) (sinX + 2sinY) = 0

By symmetry, acute angle bisector of ÐXOY is y = x.

∴ Distance of Z from bisector

∴ |β| = 1, 2 (D)

For α = 0, y = 3

For α = 1, y = |x – 1| + |x – 2| + x

Case I 

F(aα) is the area bounded by x = 0, x = 2, y² = 4x and y = 3

Case II

F(α) is the area bounded by x = 0, x = 2, y² = 4x and y = |x – 1| + |x – 2| + x