Vector Algebra Question 193
Question: The vector c directed along the internal bisector of the angle between the vectors $ \mathbf{a}=7\mathbf{i}-4\mathbf{j}-4\mathbf{k} $ and $ \mathbf{b}=-2\mathbf{i}-\mathbf{j}+2\mathbf{k} $ with $ |\mathbf{c}|,=5\sqrt{6}, $ is
Options:
A) $ \frac{5}{3},(\mathbf{i}-7\mathbf{j}+2\mathbf{k}) $
B) $ \frac{5}{3},(5\mathbf{i}+5\mathbf{j}+2\mathbf{k}) $
C) $ \frac{5}{3},(\mathbf{i}+7\mathbf{j}+2\mathbf{k}) $
D) $ \frac{5}{3},(-5\mathbf{i}+5\mathbf{j}+2\mathbf{k}) $
Show Answer
Answer:
Correct Answer: A
Solution:
- The required vector c is given by $ \lambda ( \frac{\mathbf{a}}{|\mathbf{a}|}+\frac{\mathbf{b}}{|\mathbf{b}|} ) $
Now, $ \frac{\mathbf{a}}{|\mathbf{a}|}=\frac{1}{9}(7\mathbf{i}-4\mathbf{j}-4\mathbf{k}) $
and $ \frac{\mathbf{b}}{|\mathbf{b}|}=\frac{1}{3}(-2\mathbf{i}-\mathbf{j}+2\mathbf{k}) $
$ \Rightarrow \mathbf{c}=\lambda ( \frac{1}{9}\mathbf{i}-\frac{7}{9}\mathbf{j}+\frac{2}{9}\mathbf{k} ) $
$ \Rightarrow |\mathbf{c}{{|}^{2}}={{\lambda }^{2}}.\frac{54}{81} $
$ \Rightarrow {{\lambda }^{2}}=225 $ or $ \lambda =\pm 15 $ .
Therefore, $ \mathbf{c}=\pm \frac{5}{3}(\mathbf{i}-7\mathbf{j}+2\mathbf{k}). $
BETA
