Let a → = x i ^ + y j ^ + z k ^ and x = 2 y . If a → = 5 2 and a → makes an angle of 135…

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. 23i^+3j^-3k^
  2. 26i^+6j^-6k^
  3. 25i^+5j^-5k^
  4. 25i^-5j^-5k^

Solution

Given a=xi^+yj^+zk^

a=x2+y2+z2=5y2+z2 x=2y

i.e. 5y2+z2=52

5y2+z2=50

a·k^=acos135°z=52×-12=-5

Now, 5y2=25y=±5, x=±25

i.e. a=±25i^±5j^-5k^

Asked in: AP EAMCET 2022 (04 Jul Shift 1)

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