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Step-by-Step Solution
Step 1: Understand the Given Conditions
We are given two fixed vectors
$ \overrightarrow{a} = \hat{i} + 2\hat{j} - 3\hat{k} $ and
$ \overrightarrow{b} = 2\hat{i} - 3\hat{j} + 5\hat{k} $.
A vector $ \overrightarrow{r} $ satisfies
$ \overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{r} $
and the dot product conditions:
$ \overrightarrow{r} \cdot (\alpha \hat{i} + 2\hat{j} + \hat{k}) = 3 $
$ \overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha \hat{k}) = -1 $
We are asked to find the numerical value of $ \alpha + | \overrightarrow{r} |^2 $.
Step 2: Convert the Cross Product Condition into a Parallel Relation
The equation
$ \overrightarrow{r} \times \overrightarrow{a} = \overrightarrow{b} \times \overrightarrow{r} $
can be rewritten as
$ \overrightarrow{r} \times \overrightarrow{a}
- \overrightarrow{b} \times \overrightarrow{r} = \mathbf{0}. $
Note that
$ \overrightarrow{b} \times \overrightarrow{r} = - ( \overrightarrow{r} \times \overrightarrow{b} ). $
Hence the condition becomes:
$ \overrightarrow{r} \times \overrightarrow{a}
= - \, \overrightarrow{r} \times \overrightarrow{b}
\quad \Longrightarrow \quad
\overrightarrow{r} \times ( \overrightarrow{a} + \overrightarrow{b} ) = \mathbf{0}.
$
The cross product of $ \overrightarrow{r} $ with $ ( \overrightarrow{a} + \overrightarrow{b} ) $ is zero
precisely when $ \overrightarrow{r} $ is parallel to $ ( \overrightarrow{a} + \overrightarrow{b} ) $.
Therefore, we can write:
$ \overrightarrow{r} = \lambda \, ( \overrightarrow{a} + \overrightarrow{b} ) \quad \text{for some real } \lambda.
$
Step 3: Compute $ \overrightarrow{a} + \overrightarrow{b} $
Let us first sum the given vectors $ \overrightarrow{a} $ and $ \overrightarrow{b} $:
$ \overrightarrow{a}
= \hat{i} + 2\hat{j} - 3\hat{k}, \quad
\overrightarrow{b}
= 2\hat{i} - 3\hat{j} + 5\hat{k}.
$
So,
$ \overrightarrow{a} + \overrightarrow{b}
= (\hat{i} + 2\hat{i}) + (2\hat{j} - 3\hat{j}) + (-3\hat{k} + 5\hat{k})
= 3\hat{i} - \hat{j} + 2\hat{k}.
$
Thus,
$ \overrightarrow{r} = \lambda \, (3\hat{i} - \hat{j} + 2\hat{k}).
$
Step 4: Apply the Dot Product Conditions
(a) Using $ \overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha \hat{k}) = -1 $
Substitute $ \overrightarrow{r} = \lambda (3\hat{i} - \hat{j} + 2\hat{k}) $ into
$ \overrightarrow{r} \cdot (2\hat{i} + 5\hat{j} - \alpha \hat{k}) = -1 $:
$
\lambda \, [ (3\hat{i} - \hat{j} + 2\hat{k})
\cdot (2\hat{i} + 5\hat{j} - \alpha \hat{k})
]
= -1.
$
Compute the dot product inside the brackets:
$
(3\hat{i} - \hat{j} + 2\hat{k})
\cdot (2\hat{i} + 5\hat{j} - \alpha \hat{k})
= 3 \cdot 2 + (-1) \cdot 5 + 2 \cdot (-\alpha).
$
So we get
$
6 - 5 - 2\alpha = 1 - 2\alpha.
$
Therefore,
$
\lambda (1 - 2\alpha) = -1.
\quad (1)
$
(b) Using $ \overrightarrow{r} \cdot (\alpha \hat{i} + 2\hat{j} + \hat{k}) = 3 $
Next, substitute $ \overrightarrow{r} = \lambda(3\hat{i} - \hat{j} + 2\hat{k}) $ into
$ \overrightarrow{r} \cdot (\alpha \hat{i} + 2\hat{j} + \hat{k}) = 3 $:
$
\lambda \, [ (3\hat{i} - \hat{j} + 2\hat{k})
\cdot (\alpha \hat{i} + 2\hat{j} + \hat{k})
]
= 3.
$
Compute the dot product:
$
(3\hat{i} - \hat{j} + 2\hat{k})
\cdot (\alpha \hat{i} + 2\hat{j} + \hat{k})
= 3\alpha + (-1) \cdot 2 + 2 \cdot 1
= 3\alpha - 2 + 2
= 3\alpha.
$
Hence,
$
\lambda (3\alpha) = 3
\quad \Longrightarrow \quad
3\alpha \lambda = 3
\quad \Longrightarrow \quad
\alpha \lambda = 1.
\quad (2)
$
Step 5: Solve the System of Equations for $ \lambda $ and $ \alpha $
We have two equations:
$ \lambda(1 - 2\alpha) = -1 \quad (1)$
$ \alpha \lambda = 1 \quad (2)$
From (2), we get $ \alpha = \frac{1}{\lambda}. $ Substitute this into (1):
$
\lambda
\left( 1 - 2 \cdot \frac{1}{\lambda} \right)
= -1
\quad \Longrightarrow \quad
\lambda - 2 = -1
\quad \Longrightarrow \quad
\lambda = 1.
$
Using $ \lambda = 1 $ in $ \alpha \lambda = 1 $, we get
$
\alpha = 1.
$
Step 6: Find $ \overrightarrow{r} $ and Calculate $ \alpha + | \overrightarrow{r} |^2 $
With $ \lambda = 1 $, we have
$
\overrightarrow{r}
= 3\hat{i} - \hat{j} + 2\hat{k}.
$
Next, compute the magnitude squared:
$
| \overrightarrow{r} |^2
= (3)^2 + (-1)^2 + (2)^2
= 9 + 1 + 4
= 14.
$
Since $ \alpha = 1 $, we find
$
\alpha + | \overrightarrow{r} |^2
= 1 + 14
= 15.
$
Final Answer
$ \boxed{15} $
