A) \[k=\frac{1}{2}\]
B) \[k=-\frac{1}{2}\]
C) \[k=\pm \frac{1}{2}\]
D) K can take any value
Correct Answer: C
Solution :
[c] For \[\left( \frac{1}{\sqrt{2}},\frac{1}{2},k \right)\] to represent direction cosines, we should have \[{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \frac{1}{2} \right)}^{2}}+{{k}^{2}}=1\] or \[\frac{1}{2}+\frac{1}{4}+{{k}^{2}}=1\Rightarrow k=\pm \frac{1}{2}\]You need to login to perform this action.
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