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Unit vector in the direction of $$$\left\langle 7, 2 t, 3 t^{2}\right\rangle$$$
Your Input
Find the unit vector in the direction of $$$\mathbf{\vec{u}} = \left\langle 7, 2 t, 3 t^{2}\right\rangle$$$.
Solution
The magnitude of the vector is $$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{9 t^{4} + 4 t^{2} + 49}$$$ (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 $$$\mathbf{\vec{e}} = \left\langle \frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}, \frac{2 t}{\sqrt{9 t^{4} + 4 t^{2} + 49}}, \frac{3 t^{2}}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right\rangle$$$ (for steps, see vector scalar multiplication calculator).
Answer
The unit vector in the direction of $$$\left\langle 7, 2 t, 3 t^{2}\right\rangle$$$A is $$$\left\langle \frac{7}{\sqrt{9 t^{4} + 4 t^{2} + 49}}, \frac{2 t}{\sqrt{9 t^{4} + 4 t^{2} + 49}}, \frac{3 t^{2}}{\sqrt{9 t^{4} + 4 t^{2} + 49}}\right\rangle = \left\langle \frac{7}{\left(9 t^{4} + 4 t^{2} + 49\right)^{0.5}}, \frac{2 t}{\left(9 t^{4} + 4 t^{2} + 49\right)^{0.5}}, \frac{3 t^{2}}{\left(9 t^{4} + 4 t^{2} + 49\right)^{0.5}}\right\rangle.$$$A