Which one of the following vectors of magnitude `sqrt(51)` makes equal angles with three vectors
`vec(a)=(hat(i)-2hat(j)+2hat(k))/(3), vec(b)=(-4hat(i)-3hat(k))/(5) and vec(c)=hat(j)`?

Options

A. `5hat(i)-hat(j)-5hat(k)`
B. `5hat(i)+hat(j)+5hat(k)`
C. `-5hat(i)-hat(j)+5hat(k)`
D. `5hat(i)+5hat(j)-k`

Explanation

To solve the problem of finding a vector \( \vec{P} \) of magnitude \( \sqrt{51} \) that makes equal angles with the vectors \( \vec{A} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} \), \( \vec{B} = \frac{-4\hat{i} - 3\hat{k}}{5} \), and \( \vec{C} = \hat{j} \), we will follow these steps:

### Step 1: Define the vector \( \vec{P} \)
Let \( \vec{P} = x\hat{i} + y\hat{j} + z\hat{k} \).

### Step 2: Set the magnitude of \( \vec{P} \)
The magnitude of \( \vec{P} \) is given by:
\[ |\vec{P}| = \sqrt{x^2 + y^2 + z^2} = \sqrt{51} \]
Squaring both sides, we get:
\[ x^2 + y^2 + z^2 = 51 \tag{1} \]

### Step 3: Use the condition of equal angles
The condition that \( \vec{P} \) makes equal angles with \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) implies:
\[ \frac{\vec{P} \cdot \vec{A}}{|\vec{P}| |\vec{A}|} = \frac{\vec{P} \cdot \vec{B}}{|\vec{P}| |\vec{B}|} = \frac{\vec{P} \cdot \vec{C}}{|\vec{P}| |\vec{C}|} = \cos \theta \]

### Step 4: Calculate \( |\vec{A}| \), \( |\vec{B}| \), and \( |\vec{C}| \)
1. For \( \vec{A} \):
\[ |\vec{A}| = \left|\frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}\right| = \frac{1}{3} \sqrt{1^2 + (-2)^2 + 2^2} = \frac{1}{3} \sqrt{1 + 4 + 4} = \frac{1}{3} \sqrt{9} = 1 \]

2. For \( \vec{B} \):
\[ |\vec{B}| = \left|\frac{-4\hat{i} - 3\hat{k}}{5}\right| = \frac{1}{5} \sqrt{(-4)^2 + 0^2 + (-3)^2} = \frac{1}{5} \sqrt{16 + 9} = \frac{1}{5} \sqrt{25} = 1 \]

3. For \( \vec{C} \):
\[ |\vec{C}| = |\hat{j}| = 1 \]

### Step 5: Set up the dot product equations
Now, we can write the dot product equations:
1. For \( \vec{A} \):
\[ \vec{P} \cdot \vec{A} = x \cdot \frac{1}{3} + y \cdot \frac{-2}{3} + z \cdot \frac{2}{3} = \frac{x - 2y + 2z}{3} \]

2. For \( \vec{B} \):
\[ \vec{P} \cdot \vec{B} = x \cdot \frac{-4}{5} + y \cdot 0 + z \cdot \frac{-3}{5} = \frac{-4x - 3z}{5} \]

3. For \( \vec{C} \):
\[ \vec{P} \cdot \vec{C} = 0 + y + 0 = y \]

### Step 6: Set the equations equal to each other
Since all three dot products are equal:
\[ \frac{x - 2y + 2z}{3} = \frac{-4x - 3z}{5} = y \]

### Step 7: Solve the equations
From the first two equations:
\[ 5(x - 2y + 2z) = 3(-4x - 3z) \]
Expanding gives:
\[ 5x - 10y + 10z = -12x - 9z \implies 17x - 10y + 19z = 0 \tag{2} \]

From the second and third equations:
\[ \frac{-4x - 3z}{5} = y \implies -4x - 3z = 5y \implies 4x + 5y + 3z = 0 \tag{3} \]

### Step 8: Solve the system of equations
We now have two equations (2) and (3):
1. \( 17x - 10y + 19z = 0 \)
2. \( 4x + 5y + 3z = 0 \)

We can solve this system using substitution or elimination methods.

### Step 9: Substitute and solve
From equation (3), express \( z \):
\[ z = -\frac{4x + 5y}{3} \]
Substituting into equation (2):
\[ 17x - 10y + 19\left(-\frac{4x + 5y}{3}\right) = 0 \]
Multiply through by 3 to eliminate the fraction:
\[ 51x - 30y - 76x - 95y = 0 \implies -25x - 125y = 0 \implies x + 5y = 0 \implies x = -5y \]

### Step 10: Substitute back to find \( z \)
Substituting \( x = -5y \) into (3):
\[ 4(-5y) + 5y + 3z = 0 \implies -20y + 5y + 3z = 0 \implies 3z = 15y \implies z = 5y \]

### Step 11: Substitute into the magnitude equation
Now substituting \( x = -5y \) and \( z = 5y \) into equation (1):
\[ (-5y)^2 + y^2 + (5y)^2 = 51 \implies 25y^2 + y^2 + 25y^2 = 51 \implies 51y^2 = 51 \implies y^2 = 1 \implies y = \pm 1 \]

### Step 12: Find \( x \) and \( z \)
If \( y = 1 \):
\[ x = -5(1) = -5, \quad z = 5(1) = 5 \quad \Rightarrow \quad \vec{P} = -5\hat{i} + 1\hat{j} + 5\hat{k} \]
If \( y = -1 \):
\[ x = -5(-1) = 5, \quad z = 5(-1) = -5 \quad \Rightarrow \quad \vec{P} = 5\hat{i} - 1\hat{j} - 5\hat{k} \]

### Conclusion
The two vectors \( \vec{P} = -5\hat{i} + 1\hat{j} + 5\hat{k} \) and \( \vec{P} = 5\hat{i} - 1\hat{j} - 5\hat{k} \) both have a magnitude of \( \sqrt{51} \) and make equal angles with the given vectors.