Q. The unit vector perpendicular to $\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}}$ and coplanar with $\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}}$ and $\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}}$ is
A) $\frac{2\widehat{\dot{\mathbf{i}}} - 5\widehat{\dot{\mathbf{j}}}}{\sqrt{29}}$
B) $2\widehat{\dot{\mathbf{i}}} + 5\widehat{\dot{\mathbf{j}}}$
C) $\frac{1}{\sqrt{2}}(\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}})$
D) $\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}$
Explanation
(c)
$Let\ the\ unit\ vector\ \frac{\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}}{\sqrt{2}}\ is\ perpendicular\ to\ \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}},\ then\ we\ get$
$\frac{\left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right) \bullet \left( \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} \right)}{\sqrt{2}} = \frac{1 - 1}{\sqrt{2}} = 0$
$\therefore\ \frac{\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}}{\sqrt{2}}\ is\ the\ unit\ vector$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.