To find the angle between the lines with direction cosines \(\left(\frac{1}{2}, -\frac{1}{3}, \frac{1}{4}\right)\) and \(\left(\frac{1}{3}, \frac{2}{3}, 0\right)\), we can use the formula for the cosine of the angle \(\theta\) between two lines given their direction cosines:
\[
\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}
\]
### Step 1: Identify the direction cosines
Let:
- Line 1: \((a_1, b_1, c_1) = \left(\frac{1}{2}, -\frac{1}{3}, \frac{1}{4}\right)\)
- Line 2: \((a_2, b_2, c_2) = \left(\frac{1}{3}, \frac{2}{3}, 0\right)\)
### Step 2: Calculate the numerator
Calculate \(a_1 a_2 + b_1 b_2 + c_1 c_2\):
\[
a_1 a_2 = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}
\]
\[
b_1 b_2 = -\frac{1}{3} \cdot \frac{2}{3} = -\frac{2}{9}
\]
\[
c_1 c_2 = \frac{1}{4} \cdot 0 = 0
\]
Now sum these values:
\[
a_1 a_2 + b_1 b_2 + c_1 c_2 = \frac{1}{6} - \frac{2}{9} + 0
\]
To add \(\frac{1}{6}\) and \(-\frac{2}{9}\), we need a common denominator, which is 18:
\[
\frac{1}{6} = \frac{3}{18}, \quad -\frac{2}{9} = -\frac{4}{18}
\]
Thus,
\[
\frac{1}{6} - \frac{2}{9} = \frac{3}{18} - \frac{4}{18} = -\frac{1}{18}
\]
### Step 3: Calculate the denominator
Calculate \(\sqrt{a_1^2 + b_1^2 + c_1^2}\):
\[
a_1^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}, \quad b_1^2 = \left(-\frac{1}{3}\right)^2 = \frac{1}{9}, \quad c_1^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}
\]
Now sum these values:
\[
a_1^2 + b_1^2 + c_1^2 = \frac{1}{4} + \frac{1}{9} + \frac{1}{16}
\]
Finding a common denominator (144):
\[
\frac{1}{4} = \frac{36}{144}, \quad \frac{1}{9} = \frac{16}{144}, \quad \frac{1}{16} = \frac{9}{144}
\]
Thus,
\[
a_1^2 + b_1^2 + c_1^2 = \frac{36}{144} + \frac{16}{144} + \frac{9}{144} = \frac{61}{144}
\]
Now calculate \(\sqrt{a_1^2 + b_1^2 + c_1^2}\):
\[
\sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{\frac{61}{144}} = \frac{\sqrt{61}}{12}
\]
Next, calculate \(\sqrt{a_2^2 + b_2^2 + c_2^2}\):
\[
a_2^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}, \quad b_2^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}, \quad c_2^2 = 0^2 = 0
\]
Sum these values:
\[
a_2^2 + b_2^2 + c_2^2 = \frac{1}{9} + \frac{4}{9} + 0 = \frac{5}{9}
\]
Now calculate \(\sqrt{a_2^2 + b_2^2 + c_2^2}\):
\[
\sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}
\]
### Step 4: Combine to find \(\cos \theta\)
Now substitute back into the cosine formula:
\[
\cos \theta = \frac{-\frac{1}{18}}{\left(\frac{\sqrt{61}}{12}\right) \cdot \left(\frac{\sqrt{5}}{3}\right)} = \frac{-\frac{1}{18}}{\frac{\sqrt{305}}{36}} = -\frac{2}{\sqrt{305}}
\]
### Step 5: Find the angle \(\theta\)
To find \(\theta\), we take the inverse cosine:
\[
\theta = \cos^{-1}\left(-\frac{2}{\sqrt{305}}\right)
\]
### Conclusion
The angle between the two lines is given by \(\theta = \cos^{-1}\left(-\frac{2}{\sqrt{305}}\right)\).
