Question
Let $$\overrightarrow r \times \overrightarrow a = \overrightarrow b \times \overrightarrow a $$ and $$\overrightarrow r .\overrightarrow c = 0,$$ where $$\overrightarrow a .\overrightarrow b \ne 0.$$ Then $$\overrightarrow r $$ is equal to :
A.
$$\overrightarrow b + t\overrightarrow a $$ where $$t$$ is a scalar
B.
$$\overrightarrow b - \frac{{\overrightarrow b .\overrightarrow c }}{{\overrightarrow a .\overrightarrow c }}\overrightarrow a $$
C.
$$\overrightarrow a - \overrightarrow c $$
D.
none of these
Answer :
$$\overrightarrow b - \frac{{\overrightarrow b .\overrightarrow c }}{{\overrightarrow a .\overrightarrow c }}\overrightarrow a $$
Solution :
$$\eqalign{
& {\text{Here }}\left( {\overrightarrow r - \overrightarrow b } \right) \times \overrightarrow a = 0.\,{\text{So, }}\left( {\overrightarrow r - \overrightarrow b } \right)||\overrightarrow a \cr
& \therefore \,\overrightarrow r - \overrightarrow b = t\overrightarrow a {\text{ or }}\overrightarrow r = \overrightarrow b + t\overrightarrow a \cr
& {\text{But }}\overrightarrow r .\overrightarrow c = 0 \cr
& \therefore \,0 = \overrightarrow b .\overrightarrow c + t\overrightarrow a .\overrightarrow c \cr
& \therefore \,t = - \frac{{\overrightarrow b .\overrightarrow c }}{{\overrightarrow a .\overrightarrow c }} \cr
& \therefore \,\overrightarrow r = \overrightarrow b - \frac{{\overrightarrow b .\overrightarrow c }}{{\overrightarrow a .\overrightarrow c }}\overrightarrow a \cr} $$