Given,
→a=^i+^j+^k,→b=2^j−3^k+^k
^a=1√3(^i+^j+^k)unitvector
Component of →b along →a is →b.^a=(2^j−3^k+^k).[1√3(^i+^j+^k)]=1√3(2−3+1)=0.
Component of →b normal to →a is →b×^a
ijk2−31111 = −4^i−1^j+5^k
(i) Component of →b along →a is zero.
(ii) Component of →b normal to →a is −4^i−1^j+5^k .