To find the vector that is perpendicular to the \(z\) axis and makes equal angles with the \(x\) and \(y\) axes, let us break down the problem step by step:
- The vector is perpendicular to the \(z\) axis, implying it has no \(\hat{k}\) component. Therefore, the vector can be expressed as \(\vec{v} = a \hat{i} + b \hat{j}\).
- It is given that the vector makes equal angles with the x and y axes. For this, the components in the \(x\) and \(y\) directions should be equal. Thus, \(a = b\).
- Given that the magnitude of the vector is 3, we can express this as: \(|\vec{v}| = \sqrt{a^2 + b^2} = 3\). Substitute \(a = b\), we have: \(\sqrt{2a^2} = 3\), which simplifies to \(a = \frac{3}{\sqrt{2}}\).
- Therefore, the components of the vector are \(a = \frac{3}{\sqrt{2}}\) and \(b = \frac{3}{\sqrt{2}}\). Hence, the vector is: \(\vec{v} = \frac{3}{\sqrt{2}}\hat{i} + \frac{3}{\sqrt{2}}\hat{j}\).
- To simplify, multiply numerator and denominator by \(\sqrt{2}\): \(\vec{v} = \frac{3\sqrt{2}}{2}\hat{i} + \frac{3\sqrt{2}}{2}\hat{j}\).
Thus, the correct answer is \(\frac{3\sqrt{2}}{2}\hat{i} + \frac{3\sqrt{2}}{2}\hat{j}\), which matches the option:
\(\dfrac{3\sqrt{2}}{2}\hat{i}+\dfrac{3\sqrt{2}}{2}\hat{j}\)
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