If `hati xx [ (veca-hatj) xxhati]+ hatj xx [(veca - hatk)xx hatj] +hatk xx [(veca-hati) xx hatk]=0 and veca=xhati+y hatj+z hatk`, then :

Options

A. `x+y=1`
B. `y+z=(1)/(2)`
C. `x+z=1`
D. None of these

Answer

To solve the given problem, we need to analyze the expression step by step. The expression is:

\[ \hat{i} \times (\vec{a} - \hat{j}) \times \hat{i} + \hat{j} \times (\vec{a} - \hat{k}) \times \hat{j} + \hat{k} \times (\vec{a} - \hat{i}) \times \hat{k} = 0 \]

where \(\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}\).

### Step 1: Expand the first term

The first term is:

\[ \hat{i} \times (\vec{a} - \hat{j}) \times \hat{i} \]

Substituting \(\vec{a}\):

\[ = \hat{i} \times ((x \hat{i} + y \hat{j} + z \hat{k}) - \hat{j}) \times \hat{i} \]

This simplifies to:

\[ = \hat{i} \times (x \hat{i} + (y - 1) \hat{j} + z \hat{k}) \times \hat{i} \]

### Step 2: Calculate the cross product

Using the properties of cross products:

\[ \hat{i} \times \hat{i} = 0 \]
\[ \hat{i} \times \hat{j} = \hat{k} \]
\[ \hat{i} \times \hat{k} = -\hat{j} \]

We can calculate:

\[ = \hat{i} \times ((y - 1) \hat{j}) + \hat{i} \times (z \hat{k}) = (y - 1)(\hat{i} \times \hat{j}) + z(\hat{i} \times \hat{k}) = (y - 1) \hat{k} - z \hat{j} \]

So the first term simplifies to:

\[ (y - 1) \hat{k} - z \hat{j} \]

### Step 3: Expand the second term

The second term is:

\[ \hat{j} \times (\vec{a} - \hat{k}) \times \hat{j} \]

Substituting \(\vec{a}\):

\[ = \hat{j} \times ((x \hat{i} + y \hat{j} + z \hat{k}) - \hat{k}) \times \hat{j} \]

This simplifies to:

\[ = \hat{j} \times (x \hat{i} + y \hat{j} + (z - 1) \hat{k}) \times \hat{j} \]

Calculating the cross product:

\[ = \hat{j} \times (x \hat{i}) + \hat{j} \times ((z - 1) \hat{k}) = x(\hat{j} \times \hat{i}) + (z - 1)(\hat{j} \times \hat{k}) = -x \hat{k} + (z - 1) \hat{i} \]

So the second term simplifies to:

\[ (z - 1) \hat{i} - x \hat{k} \]

### Step 4: Expand the third term

The third term is:

\[ \hat{k} \times (\vec{a} - \hat{i}) \times \hat{k} \]

Substituting \(\vec{a}\):

\[ = \hat{k} \times ((x \hat{i} + y \hat{j} + z \hat{k}) - \hat{i}) \times \hat{k} \]

This simplifies to:

\[ = \hat{k} \times ((x - 1) \hat{i} + y \hat{j}) \times \hat{k} \]

Calculating the cross product:

\[ = \hat{k} \times ((x - 1) \hat{i}) + \hat{k} \times (y \hat{j}) = (x - 1)(\hat{k} \times \hat{i}) + y(\hat{k} \times \hat{j}) = -(x - 1) \hat{j} + y \hat{i} \]

So the third term simplifies to:

\[ y \hat{i} - (x - 1) \hat{j} \]

### Step 5: Combine all terms

Now we combine all three simplified terms:

\[ [(y - 1) \hat{k} - z \hat{j}] + [(z - 1) \hat{i} - x \hat{k}] + [y \hat{i} - (x - 1) \hat{j}] = 0 \]

Combining like terms:

\[ (z - 1 + y) \hat{i} + [-(z + (x - 1))] \hat{j} + [(y - x)] \hat{k} = 0 \]

### Step 6: Set coefficients to zero

Setting each coefficient to zero gives us the following equations:

1. \(z - 1 + y = 0\)
2. \(-z - (x - 1) = 0\)
3. \(y - x = 0\)

### Step 7: Solve the equations

From equation 3, we have:

\[ y = x \]

Substituting \(y\) in equation 1:

\[ z - 1 + x = 0 \implies z = 1 - x \]

Substituting \(z\) in equation 2:

\[ -(1 - x) - (x - 1) = 0 \implies -1 + x - x + 1 = 0 \]

This holds true for any \(x\).

### Conclusion

Thus, we have:

\[ x + y + z = x + x + (1 - x) = 1 + x \]

If \(x = \frac{1}{2}\), then \(y = \frac{1}{2}\) and \(z = 0\).

### Final Result

The values of \(x\), \(y\), and \(z\) can be \( \frac{1}{2}, \frac{1}{2}, 0 \) respectively, leading to:

\[ x + y = 1 \]