A rigid body rotates about an axis through the origin with an angular velocity `10sqrt(3)`rad/s. If `omega` points in the direction of `hat(i)+hat(j)+hat(k)`, then the equation to the locus of the points having tangential speed 20m/s.

Options

A. `x^(2)+y^(2)+z^(2)-xy-yz-xz-1=0`
B. `x^(2)+y^(2)+z^(2)-2xy-2yz-2xz-1=0`
C. `x^(2)+y^(2)+z^(2)-xy-yz-xz-2=0`
D. `x^(2)+y^(2)+z^(2)-2xy-2yz-2xz-2=0`

Answer

To solve the problem step by step, we will follow the given information and derive the equation for the locus of points with a specific tangential speed.

### Step 1: Determine the Angular Velocity Vector
The angular velocity \( \omega \) is given as \( 10\sqrt{3} \) rad/s and points in the direction of \( \hat{i} + \hat{j} + \hat{k} \).

To express \( \omega \) as a vector:
1. Find the unit vector in the direction of \( \hat{i} + \hat{j} + \hat{k} \):
\[ \text{Magnitude of } (\hat{i} + \hat{j} + \hat{k}) = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
Thus, the unit vector \( \hat{n} \) is:
\[ \hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \]

2. Now, express \( \omega \):
\[ \omega = 10\sqrt{3} \cdot \hat{n} = 10\sqrt{3} \cdot \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = 10(\hat{i} + \hat{j} + \hat{k}) = 10\hat{i} + 10\hat{j} + 10\hat{k} \]

### Step 2: Relate Tangential Speed to Angular Velocity
The tangential speed \( v \) is given as 20 m/s. The relationship between tangential speed, angular velocity, and the position vector \( \mathbf{r} \) is given by:
\[ \mathbf{v} = \omega \times \mathbf{r} \]
where \( \mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k} \).

### Step 3: Compute the Cross Product
Using the values of \( \omega \) and \( \mathbf{r} \):
\[ \mathbf{v} = (10\hat{i} + 10\hat{j} + 10\hat{k}) \times (x\hat{i} + y\hat{j} + z\hat{k}) \]

Setting up the determinant for the cross product:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 10 & 10 & 10 \\ x & y & z \end{vmatrix} \]

Calculating this determinant:
1. For \( \hat{i} \):
\[ \hat{i}(10z - 10y) = 10(z - y)\hat{i} \]
2. For \( \hat{j} \):
\[ -\hat{j}(10z - 10x) = -10(z - x)\hat{j} \]
3. For \( \hat{k} \):
\[ \hat{k}(10y - 10x) = 10(y - x)\hat{k} \]

Thus, the velocity vector \( \mathbf{v} \) becomes:
\[ \mathbf{v} = 10(z - y)\hat{i} - 10(z - x)\hat{j} + 10(y - x)\hat{k} \]

### Step 4: Find the Magnitude of the Velocity Vector
The magnitude of \( \mathbf{v} \) is:
\[ |\mathbf{v}| = \sqrt{(10(z - y))^2 + (-10(z - x))^2 + (10(y - x))^2} \]
\[ = 10 \sqrt{(z - y)^2 + (z - x)^2 + (y - x)^2} \]

### Step 5: Set the Magnitude Equal to Tangential Speed
Given that the tangential speed \( v = 20 \) m/s:
\[ 20 = 10 \sqrt{(z - y)^2 + (z - x)^2 + (y - x)^2} \]
Dividing both sides by 10:
\[ 2 = \sqrt{(z - y)^2 + (z - x)^2 + (y - x)^2} \]

### Step 6: Square Both Sides
Squaring both sides gives:
\[ 4 = (z - y)^2 + (z - x)^2 + (y - x)^2 \]

### Step 7: Expand and Rearrange
Expanding the right-hand side:
\[ (z^2 - 2zy + y^2) + (z^2 - 2zx + x^2) + (y^2 - 2xy + x^2) = 4 \]
Combining like terms:
\[ 2z^2 + 2y^2 + 2x^2 - 2zy - 2zx - 2xy = 4 \]
Dividing through by 2:
\[ z^2 + y^2 + x^2 - zy - zx - xy = 2 \]

### Final Equation
Thus, the equation of the locus is:
\[ x^2 + y^2 + z^2 - xy - xz - yz - 2 = 0 \]