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