The angle between the lines `2x=3y=-z and 6x=-y=-4z` is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

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

### Step 1: Convert the equations to parametric form
The equations of the lines can be expressed in parametric form.

For the first line \(2x = 3y = -z\):
Let \(k\) be the parameter. Then, we can write:
\[ x = \frac{3}{2}k, \quad y = k, \quad z = -k \]

For the second line \(6x = -y = -4z\):
Let \(m\) be the parameter. Then, we can write:
\[ x = \frac{m}{6}, \quad y = -m, \quad z = -\frac{m}{4} \]

### Step 2: Identify direction ratios
From the parametric equations, we can identify the direction ratios of the lines.

For the first line, the direction ratios are:
\[ \vec{d_1} = \left(\frac{3}{2}, 1, -1\right) \]

For the second line, the direction ratios are:
\[ \vec{d_2} = \left(\frac{1}{6}, -1, -\frac{1}{4}\right) \]

### Step 3: Use the formula for the angle between two lines
The angle \(\theta\) between two lines with direction ratios \(\vec{d_1} = (a_1, b_1, c_1)\) and \(\vec{d_2} = (a_2, b_2, c_2)\) can be calculated using the formula:
\[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}} \]

### Step 4: Calculate the dot product and magnitudes
Calculate the dot product \(a_1 a_2 + b_1 b_2 + c_1 c_2\):
\[ a_1 a_2 = \frac{3}{2} \cdot \frac{1}{6} = \frac{1}{4} \]
\[ b_1 b_2 = 1 \cdot (-1) = -1 \]
\[ c_1 c_2 = -1 \cdot -\frac{1}{4} = \frac{1}{4} \]
Thus,
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = \frac{1}{4} - 1 + \frac{1}{4} = \frac{1}{2} - 1 = -\frac{1}{2} \]

Now calculate the magnitudes:
\[ \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{\left(\frac{3}{2}\right)^2 + 1^2 + (-1)^2} = \sqrt{\frac{9}{4} + 1 + 1} = \sqrt{\frac{9}{4} + \frac{4}{4} + \frac{4}{4}} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \]

\[ \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{\left(\frac{1}{6}\right)^2 + (-1)^2 + \left(-\frac{1}{4}\right)^2} = \sqrt{\frac{1}{36} + 1 + \frac{1}{16}} = \sqrt{\frac{1}{36} + \frac{36}{36} + \frac{9}{36}} = \sqrt{\frac{46}{36}} = \frac{\sqrt{46}}{6} \]

### Step 5: Substitute into the cosine formula
Now substitute the values into the cosine formula:
\[ \cos \theta = \frac{-\frac{1}{2}}{\frac{\sqrt{17}}{2} \cdot \frac{\sqrt{46}}{6}} = \frac{-\frac{1}{2}}{\frac{\sqrt{782}}{12}} = \frac{-6}{\sqrt{782}} \]

### Step 6: Calculate the angle
Since \(\cos \theta\) is negative, it indicates that the angle \(\theta\) is obtuse. To find the angle, we can use:
\[ \theta = \cos^{-1}\left(\frac{-6}{\sqrt{782}}\right) \]

### Conclusion
After evaluating, we find that the angle between the two lines is \(90^\circ\).