Let →p,→q,→r be three unit vectors such that →p×→q=→r. If →a is any vector such that [→a→q→r]=1,[→a→r→p]=2 and [→a→p→q]=3 then →a is
A
→p+3→q+→r
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B
3→p+2→q+→r
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C
→p−2→q−3→r
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D
→p+2→q+3→r
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Solution
The correct option is D→p+2→q+3→r (→p×→q).→r=|→r|2⇒[→p→q→r]=1 Let →a=x→p+y→q+z→r Take dot product with →p×→q,→q×→r and →r×→p [→a→p→q]=1=x Similarly y=2,z=3 ⇒→a=→p+2→q+3→r