(2î − 2) + k) + (2î – k) = (4î – 2ĵ + ok)
Now we have to find a vector which when added to the above vector gives along negative y-axis i.e. -ĵ.
Let the required vector be xî + y + zk
⇒ (4î – 2ƒ + ok) + (xî + yƒ + zk) = −↑
⇒ (4+x)î + (−2 + y) + (0 + z)k
4+ x = 0 ⇒ x = −4 0 + z = 0⇒ z = 0
-2+y=-1⇒ y=1
Therefore the required vector is -4î + ƒ