If `veca, vecb, vecc` are non coplanar vectors such that `vecbxxvecc=veca, vecaxxvecb=vecc and veccxxveca=vecb` then (A) `|veca|+|vecb|+|vecc|=3` (B) `|vecb|=1` (C) `|veca|=1` (D) none of these

Explanation

To solve the problem, we start with the given vector equations:

1. \(\vec{b} \times \vec{c} = \vec{a}\)
2. \(\vec{a} \times \vec{b} = \vec{c}\)
3. \(\vec{c} \times \vec{a} = \vec{b}\)

### Step 1: Taking the dot product of the first equation with \(\vec{a}\)

We take the dot product of the first equation with \(\vec{a}\):

\[ \vec{b} \times \vec{c} \cdot \vec{a} = \vec{a} \cdot \vec{a} \]

Using the property of the dot product and cross product, we know:

\[ \vec{b} \times \vec{c} \cdot \vec{a} = \det(\vec{a}, \vec{b}, \vec{c}) = |\vec{a}|^2 \]

Thus, we have:

\[ |\vec{a}|^2 = \det(\vec{a}, \vec{b}, \vec{c}) \]

### Step 2: Taking the dot product of the second equation with \(\vec{c}\)

Next, we take the dot product of the second equation with \(\vec{c}\):

\[ \vec{a} \times \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{c} \]

Again using the property of the dot product and cross product, we have:

\[ \vec{a} \times \vec{b} \cdot \vec{c} = \det(\vec{a}, \vec{b}, \vec{c}) = |\vec{c}|^2 \]

So, we get:

\[ |\vec{c}|^2 = \det(\vec{a}, \vec{b}, \vec{c}) \]

### Step 3: Taking the dot product of the third equation with \(\vec{b}\)

Now, we take the dot product of the third equation with \(\vec{b}\):

\[ \vec{c} \times \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{b} \]

Using the same property, we have:

\[ \vec{c} \times \vec{a} \cdot \vec{b} = \det(\vec{a}, \vec{b}, \vec{c}) = |\vec{b}|^2 \]

Thus, we conclude:

\[ |\vec{b}|^2 = \det(\vec{a}, \vec{b}, \vec{c}) \]

### Step 4: Equating the magnitudes

From the above equations, we have:

\[ |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 \]

Let \(k = |\vec{a}| = |\vec{b}| = |\vec{c}|\). Therefore, we can write:

\[ |\vec{a}|^2 = k^2, \quad |\vec{b}|^2 = k^2, \quad |\vec{c}|^2 = k^2 \]

### Step 5: Summing the magnitudes

Now, we can sum the magnitudes:

\[ |\vec{a}| + |\vec{b}| + |\vec{c}| = k + k + k = 3k \]

### Step 6: Finding the value of \(k\)

Given that the vectors are non-coplanar and the conditions imply that each vector has the same magnitude, we can assume \(k = 1\) (as a unit vector). Therefore:

\[ |\vec{a}| + |\vec{b}| + |\vec{c}| = 3 \cdot 1 = 3 \]

### Conclusion

Thus, we conclude that:

- (A) \(|\vec{a}| + |\vec{b}| + |\vec{c}| = 3\) is correct.
- (B) \(|\vec{b}| = 1\) is correct.
- (C) \(|\vec{a}| = 1\) is correct.
- (D) None of these is incorrect.

### Final Answer

The correct options are (A), (B), and (C).

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