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Question: Let \(\overset{\rightarrow}{A} = \frac{1}{\sqrt{2}}\)cosq\(\widehat{i} + \frac{1}{\sqrt{2}}\sin\thet...

Let A→=12\overset{\rightarrow}{A} = \frac{1}{\sqrt{2}}cosqi^+12sin⁡θ 6muj^\widehat{i} + \frac{1}{\sqrt{2}}\sin\theta\mspace{6mu}\widehat{j} be any vector. What will be the unit vector n^\widehat{n} in the direction of A→\overset{\rightarrow}{A}?

A

cosqi^\widehat{i} + sin qj^\widehat{j}

B

– cosqi^\widehat{i} - sin qj^\widehat{j}

C

1/21/\sqrt{2} (cosqi^\widehat{i} + sin qj^\widehat{j})

D

1/21/\sqrt{2} (cosqi^\widehat{i} - sin qj^\widehat{j})

Answer

cosqi^\widehat{i} + sin qj^\widehat{j}

Explanation

Solution

Q A→\overset{\rightarrow}{A} = 12cos⁡θi^+12sin⁡θj^\frac{1}{\sqrt{2}}\cos\theta\widehat{i} + \frac{1}{\sqrt{2}}\sin\theta\widehat{j}

\ A→\overset{\rightarrow}{A} = 12cos⁡2θ+sin⁡2θ=12\frac{1}{\sqrt{2}}\sqrt{\cos^{2}\theta + \sin^{2}\theta} = \frac{1}{\sqrt{2}}

Ž n^=A→∣A→∣\widehat{n} = \frac{\overset{\rightarrow}{A}}{|\overset{\rightarrow}{A}|} = cos⁡θi^+sin⁡θj^\cos\theta\widehat{i} + \sin\theta\widehat{j}