The vectors `veca , vecb and vecu xx vecu` are of the same length and taken pairwise they form equal angles. If ` veca = hati +hatj and vecb =hatj +hatk` then `vecc` is equal to

Options

A. ` hati +hatk, 1/3 (-hati +4hatj-hatk)`
B. ` hati +2hatj +3hatk,hati +hatj`
C. ` -hati +hatj +2hatk, hati +hatk`
D. `1/3 (-hati +4hatj -hatk) , hatj +hatk`

Explanation

To solve the problem, we need to find the vector \( \vec{c} \) given that the vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are of the same length and form equal angles with each other. We are given:

\[ \vec{a} = \hat{i} + \hat{j} \]
\[ \vec{b} = \hat{j} + \hat{k} \]

### Step 1: Find the magnitude of vectors \( \vec{a} \) and \( \vec{b} \)

The magnitude of \( \vec{a} \) is calculated as follows:

\[ |\vec{a}| = \sqrt{(1)^2 + (1)^2} = \sqrt{2} \]

The magnitude of \( \vec{b} \) is calculated as follows:

\[ |\vec{b}| = \sqrt{(0)^2 + (1)^2 + (1)^2} = \sqrt{2} \]

Since \( |\vec{a}| = |\vec{b}| \), we can conclude that the magnitudes are equal.

### Step 2: Set up the vector \( \vec{c} \)

Let \( \vec{c} = C_1 \hat{i} + C_2 \hat{j} + C_3 \hat{k} \).

### Step 3: Use the condition that \( |\vec{c}| = |\vec{a}| = |\vec{b}| \)

The magnitude of \( \vec{c} \) is given by:

\[ |\vec{c}| = \sqrt{C_1^2 + C_2^2 + C_3^2} \]

Setting this equal to \( \sqrt{2} \):

\[ \sqrt{C_1^2 + C_2^2 + C_3^2} = \sqrt{2} \]

Squaring both sides gives:

\[ C_1^2 + C_2^2 + C_3^2 = 2 \quad \text{(1)} \]

### Step 4: Find the angles between the vectors

The angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \) can be found using the dot product:

\[ \vec{a} \cdot \vec{b} = (1)(0) + (1)(1) + (0)(1) = 1 \]

Using the formula for the cosine of the angle:

\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{2} \]

### Step 5: Set up equations for \( \vec{c} \)

Using the same angle condition for \( \vec{a} \) and \( \vec{c} \):

\[ \vec{a} \cdot \vec{c} = C_1 + C_2 = 1 \quad \text{(2)} \]

Using the angle condition for \( \vec{b} \) and \( \vec{c} \):

\[ \vec{b} \cdot \vec{c} = C_2 + C_3 = 1 \quad \text{(3)} \]

### Step 6: Solve the equations

From equation (2), we have:

\[ C_1 + C_2 = 1 \quad \text{(4)} \]

From equation (3):

\[ C_2 + C_3 = 1 \quad \text{(5)} \]

Now we can express \( C_1 \) and \( C_3 \) in terms of \( C_2 \):

From (4):

\[ C_1 = 1 - C_2 \quad \text{(6)} \]

From (5):

\[ C_3 = 1 - C_2 \quad \text{(7)} \]

### Step 7: Substitute into equation (1)

Substituting equations (6) and (7) into equation (1):

\[ (1 - C_2)^2 + C_2^2 + (1 - C_2)^2 = 2 \]

Expanding this gives:

\[ (1 - 2C_2 + C_2^2) + C_2^2 + (1 - 2C_2 + C_2^2) = 2 \]

Combining like terms:

\[ 2 - 4C_2 + 3C_2^2 = 2 \]

Simplifying:

\[ 3C_2^2 - 4C_2 = 0 \]

Factoring out \( C_2 \):

\[ C_2(3C_2 - 4) = 0 \]

Thus, \( C_2 = 0 \) or \( C_2 = \frac{4}{3} \).

### Step 8: Find \( C_1 \) and \( C_3 \)

If \( C_2 = 0 \):

From (6):

\[ C_1 = 1 \]

From (7):

\[ C_3 = 1 \]

Thus, \( \vec{c} = 1\hat{i} + 0\hat{j} + 1\hat{k} = \hat{i} + \hat{k} \).

If \( C_2 = \frac{4}{3} \):

From (6):

\[ C_1 = 1 - \frac{4}{3} = -\frac{1}{3} \]

From (7):

\[ C_3 = 1 - \frac{4}{3} = -\frac{1}{3} \]

Thus, \( \vec{c} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} - \frac{1}{3}\hat{k} \).

### Final Answer

The vector \( \vec{c} \) is:

\[ \vec{c} = -\frac{1}{3}\hat{i} + \frac{4}{3}\hat{j} - \frac{1}{3}\hat{k} \]