If →a=^i−^j+^k and →b=^j−^k, then find a vector →c such that →a×→c=→b and →a.→c=3.....
Let →c=x^i+y^j+z^kAlso, →a=^i+^j+^k and ^b=^j−^kFor →a×→c=→b,∣∣
∣
∣∣^i^j^k111xyz∣∣
∣
∣∣=^j−^k⇒ ^i(z−y)−^j(z−x)+^k(y−x)=^j−^k∴ z−y=0 ...(i) x−z=1 ....(ii) x−y=1 ....(iii)Also, →a.→c=3(^i+^j+^k).(x^i+y^j+z^k)=3⇒ x+y+z=3 ....(iv)
On adding Eqs.(ii) and (iii), we get
2x−y−z ...(v)
On solving Eqs. (iv) and (v), we get
x=53∴ y=53−1=23 and z=23Now, →c=53^i+23^j+23^k =13(5^i+2^j+2^k).....