Let `hata, hatb and hatc` be three unit vectors such that `hata=hatb+(hatb xx hatc)`, then the possible value(s) of `|hata+hatb+hatc|^(2)` can be :

Options

A. 1
B. 4
C. 16
D. 9

Answer

To solve the problem, we start with the given equation involving three unit vectors \(\hat{a}\), \(\hat{b}\), and \(\hat{c}\):

\[ \hat{a} = \hat{b} + (\hat{b} \times \hat{c}) \]

### Step 1: Dot Product with \(\hat{b}\)

Taking the dot product of both sides with \(\hat{b}\):

\[ \hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{b} + (\hat{b} \times \hat{c}) \cdot \hat{b} \]

Since \(\hat{b} \cdot \hat{b} = 1\) (as it is a unit vector) and \((\hat{b} \times \hat{c}) \cdot \hat{b} = 0\) (the dot product of a vector with itself crossed with another vector is zero), we have:

\[ \hat{a} \cdot \hat{b} = 1 \]

### Step 2: Dot Product with \(\hat{c}\)

Next, we take the dot product of both sides with \(\hat{c}\):

\[ \hat{a} \cdot \hat{c} = \hat{b} \cdot \hat{c} + (\hat{b} \times \hat{c}) \cdot \hat{c} \]

Again, since \((\hat{b} \times \hat{c}) \cdot \hat{c} = 0\), we have:

\[ \hat{a} \cdot \hat{c} = \hat{b} \cdot \hat{c} \]

### Step 3: Finding \(|\hat{a} + \hat{b} + \hat{c}|^2\)

Now, we need to compute \(|\hat{a} + \hat{b} + \hat{c}|^2\):

\[ |\hat{a} + \hat{b} + \hat{c}|^2 = |\hat{a}|^2 + |\hat{b}|^2 + |\hat{c}|^2 + 2(\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a}) \]

Since \(|\hat{a}|^2 = |\hat{b}|^2 = |\hat{c}|^2 = 1\):

\[ |\hat{a} + \hat{b} + \hat{c}|^2 = 1 + 1 + 1 + 2(\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a}) \]

This simplifies to:

\[ |\hat{a} + \hat{b} + \hat{c}|^2 = 3 + 2(1 + \hat{b} \cdot \hat{c} + \hat{b} \cdot \hat{c}) \]

Let \(x = \hat{b} \cdot \hat{c}\), then:

\[ |\hat{a} + \hat{b} + \hat{c}|^2 = 3 + 2(1 + 2x) = 5 + 4x \]

### Step 4: Finding the Range of \(x\)

The dot product \(\hat{b} \cdot \hat{c} = x\) can take values from \(-1\) to \(1\) (since both are unit vectors). Therefore, we evaluate:

- When \(x = -1\):
\[ |\hat{a} + \hat{b} + \hat{c}|^2 = 5 + 4(-1) = 1 \]

- When \(x = 1\):
\[ |\hat{a} + \hat{b} + \hat{c}|^2 = 5 + 4(1) = 9 \]

### Conclusion: Possible Values

Thus, the possible values of \(|\hat{a} + \hat{b} + \hat{c}|^2\) range from \(1\) to \(9\). The possible values can be \(1\), \(4\), and \(9\).