1
Question
Let →a=^j−^k and →c=^i−^j−^k, then the vector →b satisfying →a×→b+→c=0 and →a.→b=3 is
Let →a=^j−^k and →c=^i−^j−^k, then the vector →b satisfying →a×→b+→c=0 and →a.→b=3 is
Open in App
Solution
The correct option is B −^i+^j−2^k
→a×→b=−→c
→a×(→a×→b)=−→a×→c
⇒(→a.→b)→a−(→a.→a)→b=→c×→a
∴→b=1(→a.→a)[3→a+→a×→c]
→a.→a=(^j−^k)(^j−^k)=1+1=2
→a×→c=∣∣
∣
∣∣^i^j^k01−11−1−1∣∣
∣
∣∣
=^i(−1−1)−^j(0+1)+^k(0−1)
=−2^i−^j−^k
∴→b=12[3(^j−^k)+(−2^i−^j−^k)]
=12[3^j−3^k−2^i−^j−^k]
=12[−2^i+2^j−4^k]
=−^i+^j−2^k
→a×→b=−→c
→a×(→a×→b)=−→a×→c
⇒(→a.→b)→a−(→a.→a)→b=→c×→a
∴→b=1(→a.→a)[3→a+→a×→c]
→a.→a=(^j−^k)(^j−^k)=1+1=2
→a×→c=∣∣ ∣ ∣∣^i^j^k01−11−1−1∣∣ ∣ ∣∣
=^i(−1−1)−^j(0+1)+^k(0−1)
=−2^i−^j−^k
∴→b=12[3(^j−^k)+(−2^i−^j−^k)]
=12[3^j−3^k−2^i−^j−^k]
=12[−2^i+2^j−4^k]
=−^i+^j−2^k
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
