If `vecm, vecn` are non - parallel unit vectors and `vecr` is a vector which is perpendicular to `vecm and vecn` such that `|vecr|=5 and |vecm+vecn|^(2)=2+4|vecmxxvecn|`, then the value of `|[(vecm ,vecn,vecr)]|^(2)` is equal to

Options

A. 7
B. `(21)/(5)`
C. 5
D. `(10)/(3)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have two non-parallel unit vectors \( \vec{m} \) and \( \vec{n} \), and a vector \( \vec{r} \) which is perpendicular to both \( \vec{m} \) and \( \vec{n} \). The magnitude of \( \vec{r} \) is given as \( |\vec{r}| = 5 \). We also have the equation:

\[ |\vec{m} + \vec{n}|^2 = 2 + 4 |\vec{m} \times \vec{n}| \]

### Step 2: Expand the left-hand side
Using the property of magnitudes, we can expand the left-hand side:

\[ |\vec{m} + \vec{n}|^2 = |\vec{m}|^2 + |\vec{n}|^2 + 2 \vec{m} \cdot \vec{n} \]

Since \( \vec{m} \) and \( \vec{n} \) are unit vectors, \( |\vec{m}|^2 = 1 \) and \( |\vec{n}|^2 = 1 \). Thus, we have:

\[ |\vec{m} + \vec{n}|^2 = 1 + 1 + 2 \vec{m} \cdot \vec{n} = 2 + 2 \vec{m} \cdot \vec{n} \]

### Step 3: Set the two equations equal
Now we can set the two expressions equal to each other:

\[ 2 + 2 \vec{m} \cdot \vec{n} = 2 + 4 |\vec{m} \times \vec{n}| \]

Subtracting 2 from both sides gives us:

\[ 2 \vec{m} \cdot \vec{n} = 4 |\vec{m} \times \vec{n}| \]

### Step 4: Relate the dot product and cross product
We know the relationship between the dot product and the cross product:

\[ |\vec{m} \times \vec{n}| = |\vec{m}| |\vec{n}| \sin \theta = \sin \theta \]

where \( \theta \) is the angle between \( \vec{m} \) and \( \vec{n} \). Therefore, we can rewrite the equation as:

\[ 2 \vec{m} \cdot \vec{n} = 4 \sin \theta \]

Using the dot product, we have:

\[ \vec{m} \cdot \vec{n} = \cos \theta \]

Thus, we can substitute:

\[ 2 \cos \theta = 4 \sin \theta \]

### Step 5: Solve for \( \theta \)
Rearranging gives:

\[ \frac{\cos \theta}{\sin \theta} = 2 \quad \Rightarrow \quad \cot \theta = 2 \]

This implies:

\[ \tan \theta = \frac{1}{2} \]

### Step 6: Find \( |\vec{m}, \vec{n}, \vec{r}|^2 \)
The value we need to find is \( |[\vec{m}, \vec{n}, \vec{r}]|^2 \), which can be expressed as:

\[ |[\vec{m}, \vec{n}, \vec{r}]| = |\vec{r}| |\vec{m} \times \vec{n}| \]

Thus:

\[ |[\vec{m}, \vec{n}, \vec{r}]|^2 = |\vec{r}|^2 |\vec{m} \times \vec{n}|^2 \]

### Step 7: Calculate the values
We know \( |\vec{r}| = 5 \), so \( |\vec{r}|^2 = 25 \). Now we need to find \( |\vec{m} \times \vec{n}|^2 \):

Using the identity \( |\vec{m} \times \vec{n}|^2 = \sin^2 \theta \):

From \( \tan \theta = \frac{1}{2} \), we can find \( \sin^2 \theta \):

\[ \sin^2 \theta = \frac{1}{1 + \tan^2 \theta} = \frac{1}{1 + \left(\frac{1}{2}\right)^2} = \frac{1}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5} \]

### Step 8: Final calculation
Thus,

\[ |\vec{m} \times \vec{n}|^2 = \frac{4}{5} \]

Now substituting back:

\[ |[\vec{m}, \vec{n}, \vec{r}]|^2 = 25 \cdot \frac{4}{5} = 20 \]

### Final Answer
The value of \( |[\vec{m}, \vec{n}, \vec{r}]|^2 \) is \( 20 \).

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