Let→a=^i−^j,→b=^j−^k,→c=^k−^i. If →d is a unit vector such that →a⋅→d=0=[→b→c→d], then →d
A
(^i+^j−2^k)√6
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B
(−^i+^j−2^k)√6
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C
(−^i−^j+2^k)√6
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D
(^i+^j+2^k)√6
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Solution
The correct options are A(^i+^j−2^k)√6 C(−^i−^j+2^k)√6 Let →d=xi+yj+zk Given that →a.→d=x−y=0 ...(1) and [→b→c→d]=∣∣
∣∣01−1−101xyz∣∣
∣∣=−(−z−x)−(−y)=x+y+z=0 Using (1): z=−2x Since, x2+y2+z2=1 Therefore, x2+y2+4x2=1 ⇒x=±1√6 Therefore, →d=±(^i+^j−2^k)√6