If →x+→c×→y=→a and →y+→c×→x=→b, where →c is a non - zero vector, then which of the following is correct
A
→x=(→b×→c)+→a+(→c⋅→a)→c1+→c⋅→c
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
→x=(→c×→b)+→b+(→c⋅→a)→c1+→c⋅→c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
→x=(→a×→c)+→b+(→c⋅→a)→c1+→c⋅→c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
→x=(→a×→c)+→b+(→c⋅→a)→c1−→c⋅→c
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A→x=(→b×→c)+→a+(→c⋅→a)→c1+→c⋅→c Given: →x+→c×→y=→a⋯(i)→y+→c×→x=→b⋯(ii)
Taking cross with →c, of equation (ii) We have →c×→y+→c×(→c×→x)=→c×→b
Now, using (i), we have ⇒(→a−→x)+(→c⋅→x)→c−(→c⋅→c)→x=→c×→b⋯(iii)
Also taking dot product with →c in equation (i), we have ⇒→c⋅→x+→c⋅(→c×→y)=→c⋅→a or →c⋅→x+0=→c⋅→a or →c⋅→x=→c⋅→a⋯(iv)
Using (iv) in (iii), we have ⇒(→a−→x)+(→c⋅→a)→c−(→c⋅→c)→x=→c×→b ⇒→x(1+(→c⋅→c))=→b×→c+→a+(→c⋅→a)⋅→c ⇒→x=(→b×→c)+→a+(→c⋅→a)⋅→c1+→c⋅→c