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AP EAMCET · Maths · Vector Algebra

Let a=xi^+yj^+zk^ and x=2y. If a=52 and a makes an angle of 135° with the z-axis then a=

  1. A 23i^+3j^-3k^
  2. B 26i^+6j^-6k^
  3. C 25i^+5j^-5k^
  4. D 25i^-5j^-5k^
Verified Solution

Answer & Solution

Correct Answer

(C) 25i^+5j^-5k^

Step-by-step Solution

Detailed explanation

Given a→=xi^+yj^+zk^ a→=x2+y2+z2=5y2+z2 ∵x=2y i.e. 5y2+z2=52 ⇒5y2+z2=50 a→·k^=a→cos135°⇒z=52×-12=-5 Now, 5y2=25⇒y=±5, x=±25 i.e. a→=±25i^±5j^-5k^