Let x ,   y are real numbers. If a → = ( sin x ) i ^ + ( sin y ) j ^ and b → = ( cos x ) i…

Let x, y are real numbers. If a→=(sinx)i^+(siny)j^ and b→=(cosx)i^+(cosy)j^, then |a→×b→| is
  1. 0
  2. Greater than one
  3. Less than or equal to 1
  4. Less than 1

Solution

Given, a→=(sinx)i^+(siny)j^ and b→=(cosx)i^+(cosy)j^

Then, a→×b→=sinxcosy-cosxsinyk^=sinx-yk^

Now, a→×b→=sin2x-y=sinx-y

We know that value of sine lies between -1, 1

⇒a→×b→≤1

Asked in: AP EAMCET 2021 (19 Aug Shift 2)

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