Vector Algebra Question 187
Question: Unit vectors a, b and c are coplanar. A unit vector d is perpendicular to them. If $ (\mathbf{a}\times \mathbf{b})\times (\mathbf{c}\times \mathbf{d})=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k} $ and the angle between a and b is $ 30^{o} $ , then c is
[Roorkee Qualifying 1998]
Options:
A) $ \frac{(\mathbf{i}-2\mathbf{j}+2\mathbf{k})}{3} $
B) $ \frac{(2\mathbf{i}+\mathbf{j}-\mathbf{k})}{3} $
C) $ \frac{(-\mathbf{i}+2\mathbf{j}-2\mathbf{k})}{3} $
D) $ \frac{(-\mathbf{i}+2\mathbf{j}+\mathbf{k})}{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
- Since $ \mathbf{a},,\mathbf{b},,\mathbf{c} $ are coplanar, hence $ [\mathbf{a}\mathbf{b}\mathbf{c}]=0 $
Given $ (\mathbf{a}\times \mathbf{b})\times (\mathbf{c}\times \mathbf{d})=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k} $
$ \Rightarrow [(\mathbf{a}\times \mathbf{b}),.,\mathbf{d}],\mathbf{c}-[(\mathbf{a}\times \mathbf{b}),.,\mathbf{c}]\mathbf{d}=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k} $
$ \Rightarrow [(|\mathbf{a}||\mathbf{b}|\sin 30{}^\circ ),\mathbf{\hat{n}},.,\mathbf{d}],\mathbf{c}-0=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k} $
$ \Rightarrow [ (1)(1)( \frac{1}{2} ) ][|\mathbf{\hat{n}}||\mathbf{d}|\cos \theta ],\mathbf{c}=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k} $
$ \Rightarrow [(\mathbf{a}\times \mathbf{b}),.,\mathbf{d}],\mathbf{c}-[(\mathbf{a}\times \mathbf{b}),.,\mathbf{c}]\mathbf{d}=\frac{1}{6}\mathbf{i}-\frac{1}{3}\mathbf{j}+\frac{1}{3}\mathbf{k} $ ,
Where $ \mathbf{\hat{n}} $ and $ \mathbf{d} $ are unit perpendicular vector and angle between $ \mathbf{\hat{n}} $ and $ \mathbf{d} $ may be 0 or $ \pi $ .
When $ \theta =0{}^\circ , $ $ \mathbf{c}=\frac{1}{3}[\mathbf{i}-2\mathbf{j}+2\mathbf{k}] $
When $ \theta =\pi , $ $ \mathbf{c}=\frac{1}{3}[-\mathbf{i}+2\mathbf{j}-2\mathbf{k}] $ .
BETA
