Let a, b and c are three vectors hacing magnitude 1, 2 and 3 respectively satisfying the relation [a b c]=6. If `hat(d)` is a unit vector coplanar with b and c such that `b*hat(d)=1`, then evaluate `|(axxc)*d|^(2)+|(axxc)xxhat(d)|^(2)`.

Answer

To solve the problem step by step, we will evaluate the expression \( |(a \times c) \cdot d|^2 + |(a \times c) \times \hat{d}|^2 \).

### Step 1: Understanding the Given Information
We have three vectors \( a, b, c \) with magnitudes:
- \( |a| = 1 \)
- \( |b| = 2 \)
- \( |c| = 3 \)

The scalar triple product \( [a, b, c] = a \cdot (b \times c) = 6 \).

### Step 2: Relate the Scalar Triple Product to the Magnitudes
The scalar triple product can also be expressed in terms of the magnitudes of the vectors and the sine of the angle between them:
\[ [a, b, c] = |a| |b| |c| \sin \theta \]
where \( \theta \) is the angle between \( a \) and the vector \( b \times c \).

Substituting the magnitudes:
\[ 6 = 1 \cdot 2 \cdot 3 \cdot \sin \theta \implies 6 = 6 \sin \theta \implies \sin \theta = 1 \]
This means \( \theta = 90^\circ \), indicating that \( a \) is perpendicular to \( b \times c \).

### Step 3: Finding \( |a \times c| \)
Using the formula for the magnitude of the cross product:
\[ |a \times c| = |a| |c| \sin \phi \]
where \( \phi \) is the angle between \( a \) and \( c \). Since \( |a| = 1 \) and \( |c| = 3 \):
\[ |a \times c| = 1 \cdot 3 \cdot \sin \phi = 3 \sin \phi \]

### Step 4: Evaluating the Expression
Now we need to evaluate:
\[ |(a \times c) \cdot d|^2 + |(a \times c) \times \hat{d}|^2 \]

#### Part A: Evaluating \( |(a \times c) \cdot d|^2 \)
Using the dot product:
\[ |(a \times c) \cdot d|^2 = |a \times c|^2 |d|^2 \cos^2 \theta \]
where \( \theta \) is the angle between \( a \times c \) and \( d \). Since \( d \) is a unit vector, \( |d|^2 = 1 \):
\[ |(a \times c) \cdot d|^2 = |a \times c|^2 \cos^2 \theta \]

#### Part B: Evaluating \( |(a \times c) \times \hat{d}|^2 \)
Using the cross product:
\[ |(a \times c) \times \hat{d}|^2 = |a \times c|^2 |\hat{d}|^2 \sin^2 \phi \]
Again, since \( |\hat{d}| = 1 \):
\[ |(a \times c) \times \hat{d}|^2 = |a \times c|^2 \sin^2 \phi \]

### Step 5: Combining Results
Now we combine both parts:
\[ |(a \times c) \cdot d|^2 + |(a \times c) \times \hat{d}|^2 = |a \times c|^2 (\cos^2 \theta + \sin^2 \phi) \]

Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ = |a \times c|^2 \]

### Step 6: Substitute the Magnitude of \( a \times c \)
From earlier, we found \( |a \times c| = 3 \sin \phi \). Thus:
\[ |a \times c|^2 = (3 \sin \phi)^2 = 9 \sin^2 \phi \]

### Step 7: Final Result
Since we know that \( \sin^2 \phi \) must equal 1 (as \( a \) and \( c \) are perpendicular), we have:
\[ |a \times c|^2 = 9 \]

Thus, the final answer is:
\[ |(a \times c) \cdot d|^2 + |(a \times c) \times \hat{d}|^2 = 9 \]