Unit Vectors

A unit vector along any nonzero vector v is found by dividing v by its magnitude, producing a vector of length one that preserves direction. This normalization step is foundational across JEE Advanced vector problems involving directions and resolved components. First add componentwise: a + b = (2 + 1)i + (-1 + 1)j + (2 - 1)k = 3i + 0j + k = 3i + k. Its magnitude is sqrt(3^2 + 0^2 + 1^2) = sqrt(9 + 1) = sqrt(10). Dividing gives the unit vector (3i + k)/sqrt(10). Option (3i + k)/sqrt(8) uses an incorrect magnitude, forgetting the contribution of one squared component. Option (3i - k)/sqrt(10) flips the sign of the k-component, which would point in the wrong direction. Option (3i + j + k)/sqrt(11) wrongly retains a j-term that actually cancels during addition. The vanishing j-component is the crucial observation: the opposite signs of the j-parts annihilate each other. Plausibility check: the magnitude of the proposed unit vector is sqrt(9/10 + 1/10) = sqrt(10/10) = 1, confirming it is genuinely a unit vector.