A non-zero vecto `veca` is such tha its projections along vectors `(hati + hatj)/sqrt2, (-hati + hatj)/sqrt2 and hatk` are equal , then unit vector along `veca` us

Options

A. `(sqrt2hatj-hatk)/sqrt3`
B. `(hatj-sqrt2hatk)/sqrt3`
C. `sqrt2/sqrt3hatj+hatk/sqrt3`
D. `(hatj - hatk)/(sqrt2)`

Explanation

To solve the problem, we need to find the unit vector along a non-zero vector \(\vec{a}\) such that its projections along the vectors \(\frac{\hat{i} + \hat{j}}{\sqrt{2}}, \frac{-\hat{i} + \hat{j}}{\sqrt{2}},\) and \(\hat{k}\) are equal.

### Step-by-step Solution:

1. **Define the Vector \(\vec{a}\)**:
Let \(\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}\).

2. **Calculate Projections**:
The projection of \(\vec{a}\) along a unit vector \(\hat{u}\) is given by:
\[ \text{Projection of } \vec{a} \text{ along } \hat{u} = \vec{a} \cdot \hat{u} \]

- For the first vector \(\hat{u}_1 = \frac{\hat{i} + \hat{j}}{\sqrt{2}}\):
\[ \text{Projection}_1 = \vec{a} \cdot \hat{u}_1 = \frac{x + y}{\sqrt{2}} \]

- For the second vector \(\hat{u}_2 = \frac{-\hat{i} + \hat{j}}{\sqrt{2}}\):
\[ \text{Projection}_2 = \vec{a} \cdot \hat{u}_2 = \frac{-x + y}{\sqrt{2}} \]

- For the third vector \(\hat{u}_3 = \hat{k}\):
\[ \text{Projection}_3 = \vec{a} \cdot \hat{u}_3 = z \]

3. **Set Projections Equal**:
Since the projections are equal, we can set them equal to a common variable \(p\):
\[ \frac{x + y}{\sqrt{2}} = p \quad (1) \]
\[ \frac{-x + y}{\sqrt{2}} = p \quad (2) \]
\[ z = p \quad (3) \]

4. **Solve the Equations**:
From equations (1) and (2):
- From (1): \(x + y = p\sqrt{2}\)
- From (2): \(-x + y = p\sqrt{2}\)

Adding these two equations:
\[ (x + y) + (-x + y) = 2y = 2p\sqrt{2} \implies y = p\sqrt{2} \]

Substituting \(y\) back into equation (1):
\[ x + p\sqrt{2} = p\sqrt{2} \implies x = 0 \]

Now substituting \(y\) into equation (3):
\[ z = p \]

5. **Express \(\vec{a}\)**:
Now we have:
\[ \vec{a} = 0 \hat{i} + p\sqrt{2} \hat{j} + p \hat{k} = p\sqrt{2} \hat{j} + p \hat{k} \]

6. **Find the Unit Vector**:
The unit vector \(\hat{a}\) along \(\vec{a}\) is given by:
\[ \hat{a} = \frac{\vec{a}}{|\vec{a}|} \]

First, calculate the magnitude of \(\vec{a}\):
\[ |\vec{a}| = \sqrt{(0)^2 + (p\sqrt{2})^2 + (p)^2} = \sqrt{2p^2 + p^2} = \sqrt{3p^2} = p\sqrt{3} \]

Therefore, the unit vector is:
\[ \hat{a} = \frac{p\sqrt{2} \hat{j} + p \hat{k}}{p\sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \]

### Final Answer:
The unit vector along \(\vec{a}\) is:
\[ \hat{a} = \frac{\sqrt{2}}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \]