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Step-by-Step Solution
Step 1: Understand the Problem
We have a unit vector
$ \hat{u} = x \hat{i} + y \hat{j} + z \hat{k} $
that makes angles
$ \frac{\pi}{2}, \frac{\pi}{3}, \text{ and } \frac{2\pi}{3} $
respectively with the vectors:
$ \displaystyle \frac{1}{\sqrt{2}}\,\hat{i} + \frac{1}{\sqrt{2}}\,\hat{k} $
$ \displaystyle \frac{1}{\sqrt{2}}\,\hat{j} + \frac{1}{\sqrt{2}}\,\hat{k} $
$ \displaystyle \frac{1}{\sqrt{2}}\,\hat{i} + \frac{1}{\sqrt{2}}\,\hat{j} $
We also have
$ \vec{v} = \frac{1}{\sqrt{2}}\,\hat{i} + \frac{1}{\sqrt{2}}\,\hat{j} + \frac{1}{\sqrt{2}}\,\hat{k} $
and need to find
$ \bigl|\hat{u} - \vec{v}\bigr|^2. $
Step 2: Relate Angles to Dot Products
The dot product of two vectors
$ \vec{a} $
and
$ \vec{b} $
making angle
$ \theta $
is given by
$ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta.$
Since
$ \hat{u} $
is a unit vector ($|\hat{u}|=1$), the dot product with any unit vector
$ \vec{w} $
becomes
$ \hat{u} \cdot \vec{w} = |\vec{w}|\cos\theta = \cos\theta $
(since $|\vec{w}|=1$ in each case).
Angle $ \frac{\pi}{2} $ with $ \displaystyle \frac{1}{\sqrt{2}}\,\hat{i} + \frac{1}{\sqrt{2}}\,\hat{k} $
⇒
$ \cos\frac{\pi}{2}=0. $
Denote
$ \vec{A} = \bigl(\tfrac{1}{\sqrt{2}},0,\tfrac{1}{\sqrt{2}}\bigr). $
Then
$ \hat{u} \cdot \vec{A} = 0. $
Angle $ \frac{\pi}{3} $ with $ \displaystyle \frac{1}{\sqrt{2}}\,\hat{j} + \frac{1}{\sqrt{2}}\,\hat{k} $
⇒
$ \cos\frac{\pi}{3}=\frac{1}{2}. $
Denote
$ \vec{B} = \bigl(0,\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}}\bigr). $
Then
$ \hat{u} \cdot \vec{B} = \frac{1}{2}. $
Angle $ \frac{2\pi}{3} $ with $ \displaystyle \frac{1}{\sqrt{2}}\,\hat{i} + \frac{1}{\sqrt{2}}\,\hat{j} $
⇒
$ \cos\frac{2\pi}{3}=-\frac{1}{2}. $
Denote
$ \vec{C} = \bigl(\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}},0\bigr). $
Then
$ \hat{u} \cdot \vec{C} = -\frac{1}{2}. $
Step 3: Write Down the Equations
Let $ \hat{u} = (x, y, z). $ Then:
$ \hat{u} \cdot \vec{A} = x\,\tfrac{1}{\sqrt{2}} + z\,\tfrac{1}{\sqrt{2}} = 0
\;\Longrightarrow\; x + z = 0 \;\Longrightarrow\; z=-x. $
$ \hat{u} \cdot \vec{B} = y\,\tfrac{1}{\sqrt{2}} + z\,\tfrac{1}{\sqrt{2}} = \tfrac{1}{2}
\;\Longrightarrow\; y + z = \tfrac{\sqrt{2}}{2}. $
$ \hat{u} \cdot \vec{C} = x\,\tfrac{1}{\sqrt{2}} + y\,\tfrac{1}{\sqrt{2}} = -\tfrac{1}{2}
\;\Longrightarrow\; x + y = -\tfrac{\sqrt{2}}{2}. $
Also, since $ \hat{u} $ is a unit vector,
$ x^2 + y^2 + z^2 = 1. $
Step 4: Solve for x, y, and z
From
$ z=-x, $
substitute into the remaining two linear equations:
$ x + y = -\tfrac{\sqrt{2}}{2} \;\Rightarrow\; y = -\tfrac{\sqrt{2}}{2} - x.$
$ y + z = \tfrac{\sqrt{2}}{2}. $
Substitute
$ y = -\tfrac{\sqrt{2}}{2} - x $
and
$ z = -x: $
$$
(-\tfrac{\sqrt{2}}{2} - x) + (-x) = \tfrac{\sqrt{2}}{2}.
$$
$$
-\tfrac{\sqrt{2}}{2} - 2x = \tfrac{\sqrt{2}}{2}.
$$
$$
-2x = \tfrac{\sqrt{2}}{2} + \tfrac{\sqrt{2}}{2} = \sqrt{2}.
$$
$$
x = -\tfrac{\sqrt{2}}{2}.
$$
Hence,
$ z = -x = \tfrac{\sqrt{2}}{2}, $
and therefore
$$
y = -\tfrac{\sqrt{2}}{2} - \Bigl(-\tfrac{\sqrt{2}}{2}\Bigr) = 0.
$$
So,
$ \hat{u} = \Bigl(-\tfrac{1}{\sqrt{2}},\,0,\,\tfrac{1}{\sqrt{2}}\Bigr). $
Step 5: Compute $ \hat{u} - \vec{v} $
Recall
$ \vec{v} = \Bigl(\tfrac{1}{\sqrt{2}},\,\tfrac{1}{\sqrt{2}},\,\tfrac{1}{\sqrt{2}}\Bigr). $
Then:
$$
\hat{u} - \vec{v}
= \Bigl(-\tfrac{1}{\sqrt{2}} - \tfrac{1}{\sqrt{2}},\, 0 - \tfrac{1}{\sqrt{2}},\, \tfrac{1}{\sqrt{2}} - \tfrac{1}{\sqrt{2}}\Bigr)
= \Bigl(-\tfrac{2}{\sqrt{2}},\, -\tfrac{1}{\sqrt{2}},\, 0\Bigr)
= \Bigl(-\sqrt{2},\, -\tfrac{1}{\sqrt{2}},\, 0\Bigr).
$$
Step 6: Calculate $ \bigl|\hat{u} - \vec{v}\bigr|^2 $
$$
\bigl|\hat{u} - \vec{v}\bigr|^2
= \bigl(-\sqrt{2}\bigr)^2
+ \Bigl(-\tfrac{1}{\sqrt{2}}\Bigr)^2
+ 0^2
= 2 + \tfrac{1}{2}
= \tfrac{5}{2}.
$$
Final Answer
The value of
$ \bigl|\hat{u} - \vec{v}\bigr|^2 $
is
$ \displaystyle \frac{5}{2} .
