The vectors a, b and c are of the same length and taken pairwise, they form equal angles. If `a = i+j` and `b =j+k`, then c is equal to

Options

A. `i+k`
B. `i+2j+3k`
C. `-i+j+2k`
D. `(1)/(3)(-i+4j-k)`

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 of the same length and taken pairwise, they form equal angles. We are given:

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

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

First, we calculate the magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \):

\[ |\mathbf{a}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \]

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

Since \( \mathbf{a} \) and \( \mathbf{b} \) have the same magnitude, we can denote the magnitude of \( \mathbf{c} \) as:

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

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

From the magnitude of \( \mathbf{c} \):

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

### Step 3: Calculate the Dot Products to Find Angles

Next, we need to use the condition that the angles between the vectors are equal. We start with the dot product of \( \mathbf{a} \) and \( \mathbf{b} \):

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

Using the formula for the cosine of the angle between two vectors:

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

This means \( \theta = 60^\circ \).

### Step 4: Set Up the Dot Product Equations for \( \mathbf{c} \)

Now, we use the same cosine value for the angles between \( \mathbf{a} \) and \( \mathbf{c} \), and \( \mathbf{b} \) and \( \mathbf{c} \):

1. For \( \mathbf{a} \) and \( \mathbf{c} \):

\[ \mathbf{a} \cdot \mathbf{c} = c_1 + c_2 = |\mathbf{a}| |\mathbf{c}| \cos \theta = \sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2} = 1 \quad \text{(Equation 2)} \]

2. For \( \mathbf{b} \) and \( \mathbf{c} \):

\[ \mathbf{b} \cdot \mathbf{c} = c_2 + c_3 = |\mathbf{b}| |\mathbf{c}| \cos \theta = \sqrt{2} \cdot \sqrt{2} \cdot \frac{1}{2} = 1 \quad \text{(Equation 3)} \]

### Step 5: 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 \( c_1 \) and \( c_3 \) into Equation 1:

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

Expanding this:

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

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 6: Find Corresponding Values of \( c_1 \) and \( c_3 \)

1. If \( c_2 = 0 \):
- \( c_1 = 1 \)
- \( c_3 = 1 \)
- Thus, \( \mathbf{c} = 1\mathbf{i} + 0\mathbf{j} + 1\mathbf{k} = \mathbf{i} + \mathbf{k} \).

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

### Final Answers

The two possible vectors \( \mathbf{c} \) are:

1. \( \mathbf{c} = \mathbf{i} + \mathbf{k} \)
2. \( \mathbf{c} = -\frac{1}{3}\mathbf{i} + \frac{4}{3}\mathbf{j} - \frac{1}{3}\mathbf{k} \)