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Unit vector in the direction of $$$\left\langle 4 t^{3}, - \sin{\left(t \right)}, 3\right\rangle$$$
Your Input
Find the unit vector in the direction of $$$\mathbf{\vec{u}} = \left\langle 4 t^{3}, - \sin{\left(t \right)}, 3\right\rangle$$$.
Solution
The magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}$$$ (for steps, see magnitude calculator).
The unit vector is obtained by dividing each coordinate of the given vector by the magnitude.
Thus, the unit vector is $$$\mathbf{\vec{e}} = \left\langle \frac{4 t^{3}}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}, - \frac{\sin{\left(t \right)}}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}, \frac{3}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}\right\rangle$$$ (for steps, see vector scalar multiplication calculator).
Answer
The unit vector in the direction of $$$\left\langle 4 t^{3}, - \sin{\left(t \right)}, 3\right\rangle$$$A is $$$\left\langle \frac{4 t^{3}}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}, - \frac{\sin{\left(t \right)}}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}, \frac{3}{\sqrt{16 t^{6} + \sin^{2}{\left(t \right)} + 9}}\right\rangle = \left\langle \frac{t^{3}}{\left(t^{6} + 0.0625 \sin^{2}{\left(t \right)} + 0.5625\right)^{0.5}}, - \frac{0.25 \sin{\left(t \right)}}{\left(t^{6} + 0.0625 \sin^{2}{\left(t \right)} + 0.5625\right)^{0.5}}, \frac{0.75}{\left(t^{6} + 0.0625 \sin^{2}{\left(t \right)} + 0.5625\right)^{0.5}}\right\rangle.$$$A