Given r×b=c×b ⇒‌‌(r−c)×b=0 ⇒r−c is parallel to b. ⇒(r−c)=λb or r=c+λb....(i) Also, ‌‌r⋅a=0‌‌⇒(c+λb)⋅a=0 ⇒‌‌c⋅a+λb⋅a=0 or ‌‌λ=−(‌
câ‹…a
bâ‹…a
) ....(ii) So, from Eq. (i) we substitute λλ ' from Eq. (ii), we get r=c−(‌