Find the angle between the two lines `2x=3y=-z and 6x=-y=-4z`.

Answer

To find the angle between the two lines given by the equations \(2x = 3y = -z\) and \(6x = -y = -4z\), we can follow these steps:

### Step 1: Convert the equations into standard form
The equations of the lines can be expressed in the form of direction ratios.

For the first line:
\[ 2x = 3y = -z \]
This can be rewritten as:
\[ \frac{x}{3} = \frac{y}{2} = \frac{-z}{1} \]
Thus, the direction ratios for the first line are \( (3, 2, -1) \).

For the second line:
\[ 6x = -y = -4z \]
This can be rewritten as:
\[ \frac{x}{1} = \frac{y}{-6} = \frac{z}{-1.5} \]
Thus, the direction ratios for the second line are \( (1, -6, -1.5) \).

### Step 2: Identify the direction ratios
Let:
- For the first line, \(a_1 = 3\), \(b_1 = 2\), \(c_1 = -1\)
- For the second line, \(a_2 = 1\), \(b_2 = -6\), \(c_2 = -1.5\)

### Step 3: Use the formula to find the angle between the lines
The formula for the cosine of the angle \(\theta\) between two lines is given by:
\[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \]

### Step 4: Calculate the dot product and magnitudes
1. Calculate the dot product:
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = 3 \cdot 1 + 2 \cdot (-6) + (-1) \cdot (-1.5) \]
\[ = 3 - 12 + 1.5 = -7.5 \]

2. Calculate the magnitudes:
\[ \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \]
\[ \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{1^2 + (-6)^2 + (-1.5)^2} = \sqrt{1 + 36 + 2.25} = \sqrt{39.25} \]

### Step 5: Substitute into the formula
Now substituting the values into the formula:
\[ \cos \theta = \frac{-7.5}{\sqrt{14} \cdot \sqrt{39.25}} \]

### Step 6: Calculate \(\theta\)
Since the cosine value is negative, we can find the angle:
\[ \theta = \cos^{-1}(-1) = 90^\circ \]

### Conclusion
Thus, the angle between the two lines is \(90^\circ\).

---