Vector Algebra Question 275
Question: The equation of the plane containing the lines $ \mathbf{r}={{\mathbf{a}}_1}+\lambda {{\mathbf{a}}_2} $ and $ \mathbf{r}={{\mathbf{a}}_2}+\lambda {{\mathbf{a}}_1} $ is
Options:
A) $ [\mathbf{r},\ {{\mathbf{a}}_1}\ ,{{\mathbf{a}}_2}]=0 $
B) $ [\mathbf{r}\ ,{{\mathbf{a}}_1}\ ,{{\mathbf{a}}_2}]={{\mathbf{a}}_1}.\ {{\mathbf{a}}_2} $
C) $ [\mathbf{r}\ ,{{\mathbf{a}}_2}\ ,{{\mathbf{a}}_1}]={{\mathbf{a}}_1}.\ {{\mathbf{a}}_2} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- The required plane passes through a point having position vector $ {{\mathbf{a}}_1} $ and is parallel to the vectors $ {{\mathbf{a}}_1} $ and $ {{\mathbf{a}}_2} $ . If $ \mathbf{r} $ is the position vector of any point on the plane, then $ \mathbf{r}-{{\mathbf{a}}_1},{{\mathbf{a}}_1},{{\mathbf{a}}_2} $ are coplanar. Therefore, $ (\mathbf{r}-{{\mathbf{a}}_1}).({{\mathbf{a}}_1}\times {{\mathbf{a}}_2})=0 $
Þ $ [\mathbf{r}{{\mathbf{a}}_1},{{\mathbf{a}}_2}] $ = $ [{{\mathbf{a}}_1},{{\mathbf{a}}_1},{{\mathbf{a}}_2}]\Rightarrow [\mathbf{r},{{\mathbf{a}}_1},{{\mathbf{a}}_2}]=0 $ Hence, the required plane is $ [\mathbf{r}{{\mathbf{a}}_1},{{\mathbf{a}}_2}]=0 $ .
BETA
