Vector Algebra Question 361
Question: If the non-zero vectors a and b are perpendicular to each other, then the solution of the equation $ \mathbf{r}\times \mathbf{a}=\mathbf{b} $ is given by
Options:
A) $ \mathbf{r}=x\mathbf{a}+\frac{1}{\mathbf{a},.\mathbf{a}}(\mathbf{a}\times \mathbf{b}) $
B) $ \mathbf{r}=x\mathbf{b}-\frac{1}{\mathbf{b},.\mathbf{b}}(\mathbf{a}\times \mathbf{b}) $
C) $ \mathbf{r}=x\mathbf{a}\times \mathbf{b} $
D) $ \mathbf{r}=x\mathbf{b}\times \mathbf{a} $
Show Answer
Answer:
Correct Answer: A
Solution:
- Since $ \mathbf{a},\mathbf{b} $ and $ \mathbf{a}\times \mathbf{b} $ are non-coplanar, hence $ \mathbf{r}=x\mathbf{a}+y\mathbf{b}+z(\mathbf{a}\times \mathbf{b}) $ for some scalars $ x,y $ and $ z. $ Now, $ \mathbf{b}=\mathbf{r}\times \mathbf{a}={ x\mathbf{a}+y\mathbf{b}+z(\mathbf{a}\times \mathbf{b}) }\times \mathbf{a} $ $ =y(\mathbf{b}\times \mathbf{a})+z[(\mathbf{a}\times \mathbf{b})\times \mathbf{a}] $ $ =-y(\mathbf{a}\times \mathbf{b})-z,[\mathbf{a}\times (\mathbf{a}\times \mathbf{b})] $ $ =-y(\mathbf{a}\times \mathbf{b})-z,[(\mathbf{a},.,\mathbf{b})\mathbf{a}-(\mathbf{a},.,\mathbf{a}),\mathbf{b}] $ $ =-y(\mathbf{a}\times \mathbf{b})+z(\mathbf{a},.,\mathbf{a})\mathbf{b} $ , $ { \because ,\mathbf{a},.,\mathbf{b}=0 } $
$ \Rightarrow y=0 $ and $ z=\frac{1}{(\mathbf{a},.,\mathbf{a})}\Rightarrow \mathbf{r}=x\mathbf{a}+\frac{1}{\mathbf{a},.,\mathbf{a}}(\mathbf{a}\times \mathbf{b}) $ .
BETA
