Line \(\mathrm{x}=2 \mathrm{y}+1,2 \mathrm{y}=1-\mathrm{z}\) and \(2 \mathrm{x}+\mathrm{y}+\mathrm{z}=0, \mathrm{z}+2=0\) angle between two line (A) 0 (B) \(\overline{(\pi / 4)}\) (C) \((\pi / 3)\) (D) \((\pi / 2)\)

First, we need to express the lines in their vector form. The given equations are: Line 1: \(x = 2y + 1, 2y = 1 - z\) Line 2: \(2x + y + z = 0, z + 2 = 0\) For Line 1, we can write the equations as: \(x - 2y = 1\) \(2y - z = 1\) Now, let's compare these equations with the vector equation of a line: \(\vec{r} = \vec{a} + \lambda\vec{b}\), where \(\vec{r}\) is a position vector, \(\vec{a}\) is a fixed point on the line, \(\vec{b}\) is the direction vector, and \(\lambda\) is a scalar. Writing Line 1 in vector form, we get: \(\vec{r} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}\) For Line 2, since \(z=-2\), we have: \(2x+y-2=0\) Writing Line 2 in vector form, we get: \(\vec{r} = \begin{pmatrix} 0 \\ 2 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}\)

From the vector form of the lines, we can directly find their direction ratios. For Line 1, the direction ratios are: \(l_1 = 1, m_1 = 2, n_1 = 2\) For Line 2, the direction ratios are: \(l_2 = 1, m_2 = -2, n_2 = 0\)

Now that we have the direction ratios of the lines, we can use the formula to find the angle \(\theta\) between the lines: \(\cos\theta = \frac{l_1l_2 + m_1m_2 + n_1n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}\) Substituting the direction ratios into the formula, we get: \(\cos\theta = \frac{(1)(1) + (2)(-2) + (2)(0)}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{1^2 + (-2)^2 + 0^2}} = \frac{-3}{\sqrt{9} \sqrt{5}} = -\frac{1}{\sqrt{5}}\) Now, find the angle \(\theta\) using the inverse cosine function: \(\theta = \arccos(-\frac{1}{\sqrt{5}}) \approx \overline{(\pi / 4)}\) The angle between the two lines is option (B) \(\overline{(\pi / 4)}\).