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Q. 1. If are vectors in space given by and then find the value of (2010)
Ans. 5
Solution.
Q. 2. If the distance between the plane Ax – 2y + z = d and the plane containing the lines and , then find d. (2010)
Ans. 6
Solution. The equation of plane containing the lines
∴ Distance between x – 2 y +z=0 and Ax –2 y + z = d
= Perpendicular distance between parallel planes (∴ A= 1)
Q. 3. Let be three given vectors. is a vector such that then the value of (2011)
Ans. 9
Solution.
Q. 4. If are unit vectors satisfying (2012)
Ans. 3
Solution. are units vectors such that
Q. 5. Consider the set of eight vectors Three non coplanar vectors can be chosen from V in 2^{p} ways. Then p is (JEE Adv. 2013)
Ans. 5
Solution. Given 8 vectors are
(1, 1, 1), (–1, –1, –1); (–1, 1, 1), (1, –1, –1); (1, –1, 1),
(–1, 1, –1); (1, 1, –1), (–1, –1, 1)
These are 4 diagonals of a cube and their opposites.
For 3 non coplanar vectors first we select 3 groups of diagonals and its opposite in ^{4}C_{3} ways. Then one vector from each group can be selected in 2 × 2 × 2 ways.
∴ Total ways = ^{4}C_{3} × 2 × 2× 2 = 32 = 25
∴ p = 5
Q. 6. A pack contains n cards numbered from 1 to n . Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is k, then k – 20 = (JEE Adv. 2013)
Ans. 5
Solution. Let k, k + 1 be removed from pack.
∴ (1 + 2 + 3 + ... + n) – (k + k + 1) = 1224
Q. 7. Let be three noncoplanar unit vectors suchthat the angle between every pair of them is π/3. If where p, q and r are scalars, then the value of (JEE Adv. 2014)
Ans. 4
Solution.
Taking its dot product with we get
From (1) and (3), p = r Using (2) q = – p
Q. 8. Suppose that are three noncoplanar vectors in Let the components of a vector be 4, 3 and 5, respectively. If the components of this vector are x, y and z, respectively, then the value of 2x + y + z is (JEE Adv. 2015)
Ans. 9
Solution.
⇒ – x + y – z = 4
x – y – z = 3
x + y + z = 5
Solving above equation
∴ 2x + y + z = 9
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