The vector a, b, c are equal in length and taken pairwise they mak equal-angles.
If `a=i+j, b=j+k` and c makes obtuse angle with x-axis, then c =

Options

A. `-1,4,-1`
B. `1,0,1`
C. `-1//3,4//3,-1//3`
D. `1//3,-4//3,1//3`

Answer

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

\[ \mathbf{a} = \mathbf{i} + \mathbf{j}, \quad \mathbf{b} = \mathbf{j} + \mathbf{k} \]

### Step 1: Calculate the Magnitudes of Vectors \( \mathbf{a} \) and \( \mathbf{b} \)

The magnitude of vector \( \mathbf{a} \):

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

The magnitude of vector \( \mathbf{b} \):

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

### Step 2: Set Up the Magnitude Condition for Vector \( \mathbf{c} \)

Let \( \mathbf{c} = c_1 \mathbf{i} + c_2 \mathbf{j} + c_3 \mathbf{k} \). The magnitude of vector \( \mathbf{c} \) must also equal \( \sqrt{2} \):

\[ |\mathbf{c}| = \sqrt{c_1^2 + c_2^2 + c_3^2} = \sqrt{2} \]

Squaring both sides gives:

\[ c_1^2 + c_2^2 + c_3^2 = 2 \quad \text{(Equation 1)} \]

### Step 3: Calculate the Angle Between Vectors \( \mathbf{a} \) and \( \mathbf{b} \)

Using the dot product to find the angle \( \theta \) between \( \mathbf{a} \) and \( \mathbf{b} \):

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

The cosine of the angle is given by:

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

Thus, \( \theta = 60^\circ \).

### Step 4: Set Up the Angle Condition for Vectors \( \mathbf{a} \) and \( \mathbf{c} \)

The angle between \( \mathbf{a} \) and \( \mathbf{c} \) must also be \( 60^\circ \):

\[ \cos 60^\circ = \frac{1}{2} = \frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{a}| |\mathbf{c}|} \]

Substituting the values:

\[ \frac{(1)(c_1) + (1)(c_2) + (0)(c_3)}{\sqrt{2} \cdot \sqrt{2}} = \frac{c_1 + c_2}{2} = \frac{1}{2} \]

This simplifies to:

\[ c_1 + c_2 = 1 \quad \text{(Equation 2)} \]

### Step 5: Set Up the Angle Condition for Vectors \( \mathbf{b} \) and \( \mathbf{c} \)

Similarly, for the angle between \( \mathbf{b} \) and \( \mathbf{c} \):

\[ \cos 60^\circ = \frac{\mathbf{b} \cdot \mathbf{c}}{|\mathbf{b}| |\mathbf{c}|} \]

Calculating the dot product:

\[ \mathbf{b} \cdot \mathbf{c} = (0)(c_1) + (1)(c_2) + (1)(c_3) = c_2 + c_3 \]

Thus:

\[ \frac{c_2 + c_3}{2} = \frac{1}{2} \implies c_2 + c_3 = 1 \quad \text{(Equation 3)} \]

### Step 6: Solve the System of Equations

Now we have three equations:

1. \( c_1^2 + c_2^2 + c_3^2 = 2 \) (Equation 1)
2. \( c_1 + c_2 = 1 \) (Equation 2)
3. \( c_2 + c_3 = 1 \) (Equation 3)

From Equation 2, we can express \( c_1 \):

\[ c_1 = 1 - c_2 \]

From Equation 3, we can express \( c_3 \):

\[ c_3 = 1 - c_2 \]

Substituting these into Equation 1:

\[ (1 - c_2)^2 + c_2^2 + (1 - c_2)^2 = 2 \]

Expanding:

\[ (1 - 2c_2 + c_2^2) + c_2^2 + (1 - 2c_2 + c_2^2) = 2 \]

Combining like terms:

\[ 2 - 4c_2 + 3c_2^2 = 2 \]

This simplifies to:

\[ 3c_2^2 - 4c_2 = 0 \]

Factoring out \( c_2 \):

\[ c_2(3c_2 - 4) = 0 \]

Thus, \( c_2 = 0 \) or \( c_2 = \frac{4}{3} \).

### Step 7: Find Corresponding Values of \( c_1 \) and \( c_3 \)

1. If \( c_2 = 0 \):
- \( c_1 = 1 \)
- \( c_3 = 1 \)
- \( \mathbf{c} = \mathbf{i} + \mathbf{k} \) (which does not satisfy the obtuse angle condition).

2. If \( c_2 = \frac{4}{3} \):
- \( c_1 = 1 - \frac{4}{3} = -\frac{1}{3} \)
- \( c_3 = 1 - \frac{4}{3} = -\frac{1}{3} \)
- \( \mathbf{c} = -\frac{1}{3} \mathbf{i} + \frac{4}{3} \mathbf{j} - \frac{1}{3} \mathbf{k} \)

### Conclusion

The vector \( \mathbf{c} \) that makes an obtuse angle with the x-axis is:

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