MereTeacher

JEE • NEET • FOUNDATION

JEE · NEET Previous Year Question

$The angle between vector Q and the resultant of (2 Q + 2 P) and (2 Q - 2 P) is :$

Topic: JEE PYQ

Answer: B. Full step-by-step solution below.

A $tan^{- 1} (2Q-2P)2Q+2P$
B $0^{\circ}$
C $tan^{- 1} (P / Q)$
D $tan^{- 1} (2 Q / P)$

Correct answer: B

Solution

(b) R = (2 Q + 2 P) + (2 Q - 2 P) R = 4 Q Angle between Q and R is zero.

More Vectors & Basic Mathematics PYQs

$A force is represented by F = a x^{2} + b t^{1/2} Where x = distance and t = time. The dimensions of b^{2} / a are :$
$A = 3 i^+ 4 j^Find a vector perpendicular to A in the plane of A$
$Given below are two statements : Statement (I) : Planck's constant and angular momentum have same dimensions. Statement (II) : Lin\dots$
$If A = 3 i + j + 2 k and B = 2 i - 2 j + 4 k then value of | A \times B | will be-$
$The position of a particle is given by r = (i + 2 j - k) and momentum p = (3 i + 4 j - 2 k) . The angular momentu\dots$