A vectors which makes equal angles with the vectors ` 1/3(hati - 2hatj + 2 hatk ) , 1/5(-4hati - 3hatk) , hatj ` is:

Options

A. a) `5hat(i)+5hat(j)+hat(k)`
B. b) `5hat(i)+hat(j)-5hat(k)`
C. c) `5hat(i)+hat(j)+5hat(k)`
D. d) `pm(5hat(i)-hat(j)-5hat(k))`

Answer

To find a vector \( \mathbf{A} \) that makes equal angles with the given vectors, we can follow these steps:

### Step 1: Define the vectors
Let:
- \( \mathbf{B} = \frac{1}{3} \mathbf{i} - 2 \mathbf{j} + 2 \mathbf{k} \)
- \( \mathbf{C} = \frac{1}{5} (-4 \mathbf{i} - 3 \mathbf{k}) \)
- \( \mathbf{D} = \mathbf{j} \)

### Step 2: Normalize the vectors
First, we need to check if these vectors are unit vectors. If they are not, we can normalize them.

1. **Magnitude of \( \mathbf{B} \)**:
\[ |\mathbf{B}| = \sqrt{\left(\frac{1}{3}\right)^2 + (-2)^2 + 2^2} = \sqrt{\frac{1}{9} + 4 + 4} = \sqrt{\frac{1}{9} + \frac{36}{9} + \frac{36}{9}} = \sqrt{\frac{73}{9}} = \frac{\sqrt{73}}{3} \]

2. **Magnitude of \( \mathbf{C} \)**:
\[ |\mathbf{C}| = \sqrt{\left(-\frac{4}{5}\right)^2 + 0^2 + \left(-\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25} + \frac{9}{25}} = \sqrt{\frac{25}{25}} = 1 \]

3. **Magnitude of \( \mathbf{D} \)**:
\[ |\mathbf{D}| = \sqrt{0^2 + 1^2 + 0^2} = 1 \]

### Step 3: Set up the equations
Since \( \mathbf{A} \) makes equal angles with \( \mathbf{B} \), \( \mathbf{C} \), and \( \mathbf{D} \), we can set up the following equations based on the dot product:

1. **Equation from \( \mathbf{B} \)**:
\[ \mathbf{A} \cdot \mathbf{B} = k \quad \text{(for some scalar \( k \))} \]
\[ \mathbf{A} \cdot \left(\frac{1}{3} \mathbf{i} - 2 \mathbf{j} + 2 \mathbf{k}\right) = k \]
\[ \Rightarrow \frac{1}{3} X - 2Y + 2Z = k \]

2. **Equation from \( \mathbf{C} \)**:
\[ \mathbf{A} \cdot \mathbf{C} = k \]
\[ \mathbf{A} \cdot \left(-\frac{4}{5} \mathbf{i} - \frac{3}{5} \mathbf{k}\right) = k \]
\[ \Rightarrow -\frac{4}{5} X - \frac{3}{5} Z = k \]

3. **Equation from \( \mathbf{D} \)**:
\[ \mathbf{A} \cdot \mathbf{D} = k \]
\[ \Rightarrow Y = k \]

### Step 4: Equate the equations
From the equations we have:
1. \( \frac{1}{3} X - 2Y + 2Z = k \)
2. \( -\frac{4}{5} X - \frac{3}{5} Z = k \)
3. \( Y = k \)

Substituting \( Y = k \) into the first equation:
\[ \frac{1}{3} X - 2k + 2Z = k \implies \frac{1}{3} X + 2Z = 3k \quad \text{(1)} \]

Substituting \( Y = k \) into the second equation:
\[ -\frac{4}{5} X - \frac{3}{5} Z = k \quad \text{(2)} \]

### Step 5: Solve the equations
From equation (1):
\[ \frac{1}{3} X + 2Z = 3k \implies X = 9k - 6Z \quad \text{(3)} \]

Substituting (3) into (2):
\[ -\frac{4}{5}(9k - 6Z) - \frac{3}{5}Z = k \]
\[ -\frac{36}{5}k + \frac{24}{5}Z - \frac{3}{5}Z = k \]
\[ -\frac{36}{5}k + \frac{21}{5}Z = k \]
Multiplying through by 5:
\[ -36k + 21Z = 5k \implies 21Z = 41k \implies Z = \frac{41}{21}k \]

Substituting \( Z \) back into (3):
\[ X = 9k - 6\left(\frac{41}{21}k\right) = 9k - \frac{246}{21}k = \frac{189}{21}k - \frac{246}{21}k = -\frac{57}{21}k \]

### Step 6: Write the vector
Thus, we can express the vector \( \mathbf{A} \) as:
\[ \mathbf{A} = X \mathbf{i} + Y \mathbf{j} + Z \mathbf{k} = -\frac{57}{21}k \mathbf{i} + k \mathbf{j} + \frac{41}{21}k \mathbf{k} \]

### Final Step: Simplify and check options
From the calculations, we can express \( \mathbf{A} \) in terms of \( k \) and check against the options provided.