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Unit vector in the direction of $$$\left\langle 8 t, - \frac{6}{t^{2}}, 0\right\rangle$$$
Your Input
Find the unit vector in the direction of $$$\mathbf{\vec{u}} = \left\langle 8 t, - \frac{6}{t^{2}}, 0\right\rangle$$$.
Solution
The magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \frac{2 \sqrt{16 t^{6} + 9}}{t^{2}}$$$ (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 $$$\mathbf{\vec{e}} = \left\langle \frac{4 t^{3}}{\sqrt{16 t^{6} + 9}}, - \frac{3}{\sqrt{16 t^{6} + 9}}, 0\right\rangle$$$ (for steps, see vector scalar multiplication calculator).
Answer
The unit vector in the direction of $$$\left\langle 8 t, - \frac{6}{t^{2}}, 0\right\rangle$$$A is $$$\left\langle \frac{4 t^{3}}{\sqrt{16 t^{6} + 9}}, - \frac{3}{\sqrt{16 t^{6} + 9}}, 0\right\rangle = \left\langle \frac{4 t^{3}}{\left(16 t^{6} + 9\right)^{0.5}}, - \frac{3}{\left(16 t^{6} + 9\right)^{0.5}}, 0\right\rangle.$$$A