Vector Algebra Question 2
Question: If a unit vector lies in yz?plane and makes angles of $ 30^{o} $ and $ 60^{o} $ with the positive y-axis and z-axis respectively, then its components along the co-ordinate axes will be
Options:
A) $ \frac{\sqrt{3}}{2},\frac{1}{2},,0 $
B) $ 0,\frac{\sqrt{3}}{2},\frac{1}{2} $
C) $ \frac{\sqrt{3}}{2},0,\frac{1}{2} $
D) $ 0,\frac{1}{2},,\frac{\sqrt{3}}{2} $
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Answer:
Correct Answer: B
Solution:
- Let unit vector be $ y\mathbf{i}+z\mathbf{k}, $ then $ \sqrt{y^{2}+z^{2}}=1 $ ?..(i) Since given that $ \cos 30{}^\circ =\frac{(y\mathbf{j}+z\mathbf{k}),.,(y\mathbf{j})}{|y\mathbf{j}+z\mathbf{k}||y\mathbf{j}|} $
$ \Rightarrow \frac{y^{2}}{( \sqrt{y^{2}+z^{2}} ),y}=\frac{\sqrt{3}}{2}\Rightarrow y=\frac{\sqrt{3}}{2} $ , $ (\because ,\sqrt{y^{2}+z^{2}}=1 $ by (i)) Similarly, $ \cos 60{}^\circ =\frac{(y\mathbf{j}+z\mathbf{k}),.,z\mathbf{k}}{|y\mathbf{j}+z\mathbf{k}||z\mathbf{k}|}\Rightarrow z=\frac{1}{2} $ Hence the components of unit vector are $ 0,\frac{\sqrt{3}}{2},\frac{1}{2}. $ Trick : Since the vector lies in $ yz- $ plane, so it will be either $ 0\mathbf{i}+\frac{\sqrt{3}}{2}\mathbf{j}+\frac{1}{2}\mathbf{k} $ or $ 0\mathbf{i}+\frac{1}{2}\mathbf{j}+\frac{\sqrt{3}}{2}\mathbf{k}. $ But the vector $ \frac{\sqrt{3}}{2}\mathbf{j}+\frac{1}{2}\mathbf{k} $ makes angle $ 30{}^\circ $ with $ y- $ axis and that of $ 60{}^\circ $ with z-axis.
BETA
