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Question
Let →a=2→i+^j+^k, →b=^i+2^j−^k and a unit vector →c be coplanar. If →c is perpendicular to →a, then →c is equal to
Let →a=2→i+^j+^k, →b=^i+2^j−^k and a unit vector →c be coplanar. If →c is perpendicular to →a, then →c is equal to
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Solution
The correct option is A 1√2(−^j+^k)
It is given that →c is coplanar with →a and →b, we take →c=p→a+q→b (i)
where, p and q are scalars.
Since, →c⊥→a⇒→c.→a=0
Taking dot product of →a in Eq. (i), we get
→c.→a=p→a.→a+q→b.→a⇒0=p|→a|2+q|→b.→a|⎡⎢
⎢
⎢
⎢
⎢⎣∵→a=2^i+^j+^k|→a|=√22+1+1=√6.→a.→b=(2^i+^j+^k).(^i+2^j−^k)=2+2−1=3⎤⎥
⎥
⎥
⎥
⎥⎦
⇒0=p.6+q.3⇒q=−2p
On putting in Eq. (i), we get
→c=p→a+→b(−2p)⇒→c=p→a−2p→b⇒→c=p(→a−2→b)⇒→c=p[(2^i+^j+^k)−2(^i+2^j−^k)]⇒→c=p(−3^j+3^k)⇒|→c|=p√(−3)2+32⇒|→c|2=p2(√18)2⇒|→c|2=p2.18⇒1=p2.18⇒1=p2.18 [∵|→c|=1]⇒p2=118⇒p=±13√2∴→c=±1(−^j+^k√2
It is given that →c is coplanar with →a and →b, we take →c=p→a+q→b (i)
where, p and q are scalars.
Since, →c⊥→a⇒→c.→a=0
Taking dot product of →a in Eq. (i), we get
→c.→a=p→a.→a+q→b.→a⇒0=p|→a|2+q|→b.→a|⎡⎢ ⎢ ⎢ ⎢ ⎢⎣∵→a=2^i+^j+^k|→a|=√22+1+1=√6.→a.→b=(2^i+^j+^k).(^i+2^j−^k)=2+2−1=3⎤⎥ ⎥ ⎥ ⎥ ⎥⎦
⇒0=p.6+q.3⇒q=−2p
On putting in Eq. (i), we get
→c=p→a+→b(−2p)⇒→c=p→a−2p→b⇒→c=p(→a−2→b)⇒→c=p[(2^i+^j+^k)−2(^i+2^j−^k)]⇒→c=p(−3^j+3^k)⇒|→c|=p√(−3)2+32⇒|→c|2=p2(√18)2⇒|→c|2=p2.18⇒1=p2.18⇒1=p2.18 [∵|→c|=1]⇒p2=118⇒p=±13√2∴→c=±1(−^j+^k√2
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