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How do you find the unit vector in the same direction of the given vector $\overrightarrow{a}=\left( -10,6,-7 \right)$?

Answer
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Hint: We start solving the problem by making use of the result that the magnitude of the vector $\left( x,y,z \right)$ is $\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$ to find the magnitude of the given vector $\overrightarrow{a}=\left( -10,6,-7 \right)$. We then make use of the fact that the unit vector in the direction of the vector $\overrightarrow{r}=\left( x,y,z \right)$ is defined as \[\dfrac{\overrightarrow{r}}{\left| \overrightarrow{r} \right|}\]. We use this result to find the required unit vector of the given vector $\overrightarrow{a}=\left( -10,6,-7 \right)$.

Complete step by step answer:
According to the problem, we are asked to find the unit vector in the same direction of the given vector $\overrightarrow{a}=\left( -10,6,-7 \right)$.
Let us find the magnitude of the given vector $\overrightarrow{a}$.
We know that the magnitude of the vector $\left( x,y,z \right)$ is $\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$. Let us use this result to find the magnitude of the given vector $\overrightarrow{a}$.
So, the magnitude of the given vector $\overrightarrow{a}$ is $\left| \overrightarrow{a} \right|=\sqrt{{{\left( -10 \right)}^{2}}+{{6}^{2}}+{{\left( -7 \right)}^{2}}}=\sqrt{100+36+49}=\sqrt{185}$ ---(1).
We know that the unit vector in the direction of the vector $\overrightarrow{r}=\left( x,y,z \right)$ is defined as \[\dfrac{\overrightarrow{r}}{\left| \overrightarrow{r} \right|}\]. Using this result, we get the unit vector in the direction of vector $\overrightarrow{a}=\left( -10,6,-7 \right)$ as \[\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}\].
From equation (1), we get the unit vector as $\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}=\dfrac{\left( -10,6,-7 \right)}{\sqrt{185}}$.

$\therefore $ We have found the unit vector in the direction of given vector $\overrightarrow{a}=\left( -10,6,-7 \right)$ as $\dfrac{\left( -10,6,-7 \right)}{\sqrt{185}}$.

Note: We can verify the obtained result by finding the magnitude of the obtained vector as the magnitude of the unit vectors is 1 and the angle of the obtained unit vector with given vector should be 0 as both are in the same direction. We can also solve the problem by making use of the fact that the vector in the direction of the vector $\overrightarrow{x}$ is $\lambda \overrightarrow{x}$ and then equating its magnitude to 1. Similarly, we can expect problems to find the unit vector opposite to the given vector $\overrightarrow{a}=\left( -10,6,-7 \right)$.