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Step-by-Step Solution
Step 1: Express the vector u as a linear combination of a and b
Since the vector $ \overrightarrow{u} $ is coplanar with $ \overrightarrow{a} $ and $ \overrightarrow{b} $, we can write:
$ \overrightarrow{u} = x \overrightarrow{a} + y \overrightarrow{b}, $
for some scalars $x$ and $y$.
Step 2: Use the perpendicularity condition with a
Given that $ \overrightarrow{u} $ is perpendicular to $ \overrightarrow{a} $, we have:
$ \overrightarrow{u} \cdot \overrightarrow{a} = 0. $
Substitute $ \overrightarrow{u} = x \overrightarrow{a} + y \overrightarrow{b} $:
$ (x \overrightarrow{a} + y \overrightarrow{b}) \cdot \overrightarrow{a} = 0. $
This simplifies to:
$ x (\overrightarrow{a} \cdot \overrightarrow{a}) + y (\overrightarrow{b} \cdot \overrightarrow{a}) = 0. $
Next, compute each dot product:
$ \overrightarrow{a} \cdot \overrightarrow{a} = | \overrightarrow{a} |^2 $. Here,
$ |\overrightarrow{a}| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{14}, $
so
$ \overrightarrow{a} \cdot \overrightarrow{a} = 14. $
$ \overrightarrow{a} \cdot \overrightarrow{b} = (2\hat{i} + 3\hat{j} - \hat{k}) \cdot (\hat{j} + \hat{k}) = 2 \cdot 0 + 3 \cdot 1 + (-1) \cdot 1 = 2. $
Hence the perpendicularity condition becomes:
$ 14x + 2y = 0 \quad \Longrightarrow \quad 7x + y = 0. \quad (1) $
Step 3: Apply the condition u ⋅ b = 24
We are also given $ \overrightarrow{u} \cdot \overrightarrow{b} = 24. $ Substitute $ \overrightarrow{u} = x \overrightarrow{a} + y \overrightarrow{b} $:
$ (x \overrightarrow{a} + y \overrightarrow{b}) \cdot \overrightarrow{b} = 24. $
This is:
$ x (\overrightarrow{a} \cdot \overrightarrow{b}) + y (\overrightarrow{b} \cdot \overrightarrow{b}) = 24. $
We already have $ \overrightarrow{a} \cdot \overrightarrow{b} = 2. $ Also,
$ \overrightarrow{b} = \hat{j} + \hat{k}, $
so
$ |\overrightarrow{b}|^2 = (\sqrt{1^2 + 1^2})^2 = 2. $
Thus,
$ 2x + 2y = 24 \quad \Longrightarrow \quad x + y = 12. \quad (2) $
Now solve the system of equations (1) and (2):
From (1): $ y = -7x. $
Substitute into (2): $ x + (-7x) = 12 \,\Longrightarrow\, -6x = 12 \,\Longrightarrow\, x = -2. $
Then $ y = -7(-2) = 14. $
Step 4: Compute $ | \overrightarrow{u} |^2 $
The quantity $ | \overrightarrow{u} |^2 $ is given by $ \overrightarrow{u} \cdot \overrightarrow{u} $. Using
$ \overrightarrow{u} = x \overrightarrow{a} + y \overrightarrow{b} $,
we have:
$ \overrightarrow{u} \cdot \overrightarrow{u}
= (x \overrightarrow{a} + y \overrightarrow{b}) \cdot \overrightarrow{u}. $
But $ \overrightarrow{u} \cdot \overrightarrow{a} = 0 $ and $ \overrightarrow{u} \cdot \overrightarrow{b} = 24 $. Hence:
$ \overrightarrow{u} \cdot \overrightarrow{u}
= x (\overrightarrow{u} \cdot \overrightarrow{a}) + y (\overrightarrow{u} \cdot \overrightarrow{b})
= 0 + y \cdot 24
= 14 \times 24
= 336. $
Therefore,
$ | \overrightarrow{u} |^2 = 336, $
which matches the correct answer.
