Write a vector in the direction of the $\widehat i - 2\widehat j + 2\widehat k$ that has magnitude $9$ units.

Hint: In the question we have to find the vector in the direction by using the given vector and magnitude. For that, first we are going to find the unit vector in the direction of the given vector and then we find the required vector to multiply the magnitude to the unit vector in the direction of a given vector.
A vector is a quantity which has both magnitudes, as well as direction. A vector which has a magnitude of 1 is a unit vector.
\[{\text{Unit vector = }}\dfrac{{{\text{Vector}}}}{{{\text{Vector's magnitude}}}}\]
$ \Rightarrow \widehat a = \dfrac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}$ where, $\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$ and $\left| {\overrightarrow a } \right| = \sqrt {{x^2} + {y^2} + {z^2}} $

Complete step-by-step answer:
From the question, the given vector is $\widehat i - 2\widehat j + 2\widehat k$ that has magnitude $9$ units
Let us consider $\overrightarrow a = \widehat i - 2\widehat j + 2\widehat k$
Here $x = 1,{\text{ }}y = - 2,{\text{ }}z = 2$
Unit vector in the direction of a vector $\overrightarrow a $
$ \Rightarrow \widehat a = \dfrac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}$
Substitute the vector in above equation we get,
\[ \Rightarrow \widehat a = \dfrac{{\widehat i - 2\widehat j + 2\widehat k}}{{\sqrt {{{(1)}^2} + {{( - 2)}^2} + {{(2)}^2}} }}\]
Simplifying the term by taking square we get,
\[ \Rightarrow \widehat a = \dfrac{{\widehat i - 2\widehat j + 2\widehat k}}{{\sqrt {(1) + (4) + (4)} }}\]
Adding the values in the root we get,
\[ \Rightarrow \widehat a = \dfrac{{\widehat i - 2\widehat j + 2\widehat k}}{{\sqrt 9 }}\]
Taking square root to solve the above equation we get,
\[ \Rightarrow \widehat a = \dfrac{{\widehat i - 2\widehat j + 2\widehat k}}{3}\]
Hence, a vector in the direction of the $\overrightarrow a $ with magnitude $9$ units
\[ \Rightarrow 9\left( {\dfrac{{\widehat i - 2\widehat j + 2\widehat k}}{3}} \right)\]
\[ \Rightarrow 3\left( {\widehat i - 2\widehat j + 2\widehat k} \right)\]
Or
\[ \Rightarrow 3\widehat i - 6\widehat j + 6\widehat k\]
Hence, we get the required result.
$\therefore $A vector in the direction of the $\widehat i - 2\widehat j + 2\widehat k$ that has magnitude $9$ units is \[3\left( {\widehat i - 2\widehat j + 2\widehat k} \right)\] or \[3\widehat i - 6\widehat j + 6\widehat k\]

Note: We can solve this problem by using another method.
That is,
Let the given vector as $\overrightarrow a = \widehat i - 2\widehat j + 2\widehat k$
Here $x = 1,{\text{ }}y = - 2,{\text{ }}z = 2$
$ \Rightarrow \widehat a = \dfrac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}}$ where, $\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$ and $\left| {\overrightarrow a } \right| = \sqrt {{x^2} + {y^2} + {z^2}} $
The vector in the direction of $\overrightarrow a $ with magnitude of $9$ is $9 \times \widehat a$
Hence the required vector,
\[ \Rightarrow 9 \times \dfrac{{\widehat i - 2\widehat j + 2\widehat k}}{{\sqrt {{{(1)}^2} + {{( - 2)}^2} + {{(2)}^2}} }}\]
Simplifying the term by taking square we get,
\[ \Rightarrow 9 \times \dfrac{{\widehat i - 2\widehat j + 2\widehat k}}{{\sqrt {(1) + (4) + (4)} }}\]
Adding the values in the root we get,
\[ \Rightarrow 9 \times \dfrac{{\widehat i - 2\widehat j + 2\widehat k}}{{\sqrt 9 }}\]
Taking square root to solve the above equation we get,
\[ \Rightarrow 9 \times \dfrac{{\widehat i - 2\widehat j + 2\widehat k}}{3}\]
Simplifying we get,
\[ \Rightarrow 3\left( {\widehat i - 2\widehat j + 2\widehat k} \right)\]
Or
\[ \Rightarrow 3\widehat i - 6\widehat j + 6\widehat k\]
$\therefore $A vector in the direction of the $\widehat i - 2\widehat j + 2\widehat k$ that has magnitude $9$ units is \[3\left( {\widehat i - 2\widehat j + 2\widehat k} \right)\] or \[3\widehat i - 6\widehat j + 6\widehat k\].