Fill in the Blanks: Vector Algebra and Three Dimensional Geometry - JEE Advanced

Q. 1. Let  be vectors of length 3, 4, 5 respectively. Let  be perpendicular to   Then the length of vector                 (1981 - 2 Marks)

Ans. 5√2

Solution. 

Adding (1), (2) and (3) we get

= 50 

Q. 2. The unit vector perpendicular to the plane determined by P(1, -1, 2), Q (2, 0, -1) and R(0, 2, 1) is .......          (1983 - 1 Mark)

Ans. 

Solution. Required unit vector, 

Q. 3. The area of the triangle whose vertices are A (1, -1, 2), B (2, 1, -1), C( 3, - 1, 2) is .......            (1983 - 1 Mark)

Ans. 

Solution.

Q. 4. A, B, C and D, are four points in a plane with position vectors a, b, c and d respectively such that

The point D,  then, is the ................... of the triangle ABC.                 (1984 - 2 Marks)

Ans. orthocen tre

Solution. Given that are position vectors of points A, B, C and D respectively, such tha

Clearly D is orthocentre of DΔABC

Q. 5.   and the vectors   are non -coplanar, then the product abc = .......              (1985 - 2 Marks)

Ans. -1

Solution. 

Operating   in first determinant

Also given that the vectors   are noncoplanar 

i.e., 

∴ We must have 1 + abc = 0 ⇒ abc = - 1

Q. 6. If   are three non-coplanar vectors, then -  

               (1985 - 2 Marks)

Ans. 0

Solution. As given that   are three noncoplan ar vectors,  therefore, 

Also by the property of scalar triple product we have

Q. 7.  are given vectors, then a vector B satisfying the equations   and           (1985 - 2 Marks)

Ans. 

Solution.  

Using equations (1) and (2) we get

1 + z + z + z = 3

⇒ z = 2/3 ⇒ y = 2/3, x =5/3

Q. 8. If the vectors   (a ≠ b ≠ c ≠ 1) are coplanar, then the value of                 (1987 - 2 Marks)

Ans. 1

Solution. Given that the vectors   and   where a ≠ b ≠ c ≠ 1 are coplanar

Taking (1 - a), (1 - b), (1 - c) common from R₁, R₂ and R₃ respectively.

But a ≠ b ≠ c ≠ 1 (given)

Q. 9.  be two vectors perpendicular to each other in the xy-plane. All vectors in the same plane having projections 1 and 2 along   respectively,, are given by ........                  (1987 - 2 Marks)

Ans. 

Solution. 

   ...(1)

Now, let   be the required vectors.

Then as per question 

Projection of 

⇒  4x + 3y = 5            ..(2)

Also, projection of 

⇒ 3λx - 4λy = 10λ
⇒  3x - 4y = 10              ...(3)

Solving (2) and (3), we get x = 2, y = - 1

∴ The required vector is 

Q. 10. The components of a vector   along and perpendicular to a non-zero vector   ..........and .......respectively..              (1988 - 2 Marks)

Ans. 

Solution.  Component of 

Component of 

Q. 11. Given that  and                 (1991 -  2 Marks)

Ans. 

Solution. 

Using equations (1) and (2) we get

1 + z + z + z = 3

⇒ z = 2/3 ⇒ y = 2/3, x =5/3

Q. 12. A unit vector coplanar with  and perpendicular to                    (1992 -  2 Marks)

Ans. 

Solution. Let   be a unit vector, coplanar with   and   and also perpendicular to 

Solving the above by cross multiplication method, we get 

As  is a unit vector, therefore 

∴ The required vector is 

Q. 13. A unit vector perpendicular to the plane determined by the points P(1, - 1, 2) Q(2, 0, -1) and R(0, 2, 1) is .......                   (1994 -  2 Marks)

Ans. 

Solution. We have position vectors of points  

Now any vector perpendicular to the plane formed by pts

PQR is given by 

∴ Unit vector normal to plane 

Q. 14. A nonzero vector  is parallel to the line of intersection of the plane determined by the vectors  and the plane determined by the vectors   The angle between  and the vector                (1996 - 2 Marks)

Ans. 

Solution. Eqⁿ of plane containing vectors 

 
Similarly, eqⁿ of plane containing vectors

⇒ (x - 1) (-1 - 0) - (y + 1) (1 - 0) + z (0 + 1) = 0
⇒ - x + 1 - y - 1 + z = 0
⇒ x + y - z = 0                    ....(2)

Since  parallel to (1) and (2) 

a₃ = 0 and a₁ + a₂ - a₃ = 0 ⇒ a₁ = - a₂ , a₃ = 0

∴ a vector in direction of 

Now  if θ is the angle between   then 

Q. 15. If   are any two non-collinear unit vectors and   any vector, then                 (1996 - 2 Marks)

Ans. 

Solution. Let us consider 

Q. 16. Let OA = a, OB = 10 a + 2b and OC = b where O, A and C are non-collinear points. Let p denote the area of the quadrilateral OABC, and let q denote the area of the parallelogram with OA and OC as adjacent sides. If p = kq, then k = .......              (1997 - 2 Marks)

Ans. 6

Solution. q = area of parallelogram with 

adjacent sides 

and p = area of quadrilateral OABC