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# Fill in the Blanks: Vector Algebra and Three Dimensional Geometry | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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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

and p = area of quadrilateral OABC

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