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The unit vectors orthogonal to $-i+2j+2k$ and making equal angles with x and y axis are
$\begin{align}
  & A.\ \text{ }\dfrac{\pm 1}{3}(\hat{i}+\hat{j}-\hat{k}) \\
 & B.\ \text{ }\dfrac{\pm 1}{3}(2\hat{i}+2\hat{j}-\hat{k}) \\
 & C.\text{ }\dfrac{\pm 1}{3}(2\hat{i}-2\hat{j}-\hat{k}) \\
 & D.\text{ }\dfrac{\pm 1}{4}(\hat{i}+\hat{j}-\hat{k}) \\
\end{align}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: To find a unit vector with the same direction as a given vector, we divide the vector by its magnitude.
For example, consider a vector v = (1, 4) which has the magnitude of |v|. If we divide each component of vector v by |v| we will get the unit vector uv which is in the same direction as v.

Complete step by step answer:
Let the required vector be $a\widehat{i}+b\widehat{j}+c\widehat{k}$
∵ It makes an equal angle with the x and y-axis.
⟹ \[\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\]

By the property -Equal terms from the denominator cancels each other
 ⟹ a=b
∵It is a unit vector.
⟹ \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1-(i)\]
and it is orthogonal to \[-\widehat{i}+\mathbf{2}\widehat{j}+\mathbf{2}\widehat{k~}\Rightarrow -a+2b+2c=0\]
∵a=b
⟹ −a+2a+2c=0
⟹ 2c=−a
⟹ c=2−a​
Put the value b=a, c=\[\dfrac{-a}{2}\] In the equation (i):
⟹ \[{{a}^{2}}+{{a}^{2}}+\dfrac{{{a}^{2}}}{4}=1\]
Simplify the above equation –

Take LCM (Do cross-multiplication)
$\begin{align}
  & 9{{a}^{2}}=4 \\
 & {{a}^{2}}=\dfrac{4}{9} \\
 & \therefore a=\dfrac{\pm 2}{3} \\
\end{align}$ (Take square-root on both the sides of the equations)

Place the value of a in the general form –
$\begin{align} & a\widehat{i}+b\widehat{j}+c\overset\frown{k}=a\widehat{i}+a\widehat{j}-\dfrac{a}{2}\overset\frown{k} \\
 & a\widehat{i}+b\widehat{j}+c\overset\frown{k}=\pm \dfrac{2}{3}\widehat{i}\pm \dfrac{2}{3}\widehat{j}-\dfrac{1}{3}\widehat{k} \\
\end{align}$
(Take $\pm \dfrac{1}{3}$ common from all the three terms)
$a\widehat{i}+b\widehat{j}+c\overset\frown{k}=\pm \dfrac{1}{3}(2\widehat{i}+2\widehat{j}-\widehat{k})$
Therefore, the required solution is - The unit vectors orthogonal to $-i+2j+2k$ and making equal angles with x and y axes are $a\widehat{i}+b\widehat{j}+c\overset\frown{k}=\pm \dfrac{1}{3}(2\widehat{i}+2\widehat{j}-\widehat{k})$

Hence, from the given multiple choices, option A is the correct answer.

Note: A unit vector is a vector that has a magnitude of the 1 unit. A unit vector is also known as the direction vector. It is represented by using the lowercase letter with a cap (‘^’) symbol along with it. A vector can be represented in the space using unit vectors. Any vector can become a unit vector by dividing it by the vector’s magnitude as follows:
\[unit\;vector=\dfrac{vector}{magnitude\;of\;the\;vector}\]
A unit vector p ̂having the same direction as vector p⃗ is given as:
\[\widehat{p}\]= \[\overrightarrow{p}\]| \[\overrightarrow{p}\]|
Where,
\[\widehat{p}\]represents a unit vector
\[\overrightarrow{p}\] represents the vector
$\left| p \right|$ represents the magnitude of the vector

It must be kept in mind that any two-unit vectors \[\widehat{p}\] and \[\widehat{q}\] must not be considered as the equal unit vectors just because they have the same magnitude. Since the direction in which the vectors are taken might be different therefore these unit vectors are different from each other. Therefore, to define a vector both the magnitude and the direction should be specified.