Question

If \(\theta\) is angle between the lines \(\frac{x}{1}=\frac{y+1}{2}=\frac{z-1}{3}\) and \(\frac{x+1}{3}=\frac{y}{2}=\frac{z}{1}\), then \(\cos\theta=\)

• A. \(\frac{5}{9}\)
• B. \(\frac{5}{8}\)
• C. \(\frac{5}{6}\)
• D. \(\frac{5}{7}\)
• E. \(\frac{6}{7}\)

Answer 1

The Correct Option is D

The equations of the two lines are given as symmetric equations. Let's first convert the symmetric form into parametric form. For the first line: \[ \frac{x}{1} = \frac{y + 1}{2} = \frac{z - 1}{3} \] This can be written as: \[ x = t, \quad y = 2t - 1, \quad z = 3t + 1 \] So, the direction ratios of the first line are \( \vec{l_1} = (1, 2, 3) \). For the second line: \[ \frac{x + 1}{3} = \frac{y}{2} = \frac{z - 1}{1} \] This can be written as: \[ x = 3s - 1, \quad y = 2s, \quad z = s + 1 \] So, the direction ratios of the second line are \( \vec{l_2} = (3, 2, 1) \). The formula for the cosine of the angle between two lines with direction ratios \( \vec{l_1} = (l_1, m_1, n_1) \) and \( \vec{l_2} = (l_2, m_2, n_2) \) is: \[ \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \cdot \sqrt{l_2^2 + m_2^2 + n_2^2}} \] Substitute \( \vec{l_1} = (1, 2, 3) \) and \( \vec{l_2} = (3, 2, 1) \) into the formula: \[ \cos \theta = \frac{(1)(3) + (2)(2) + (3)(1)}{\sqrt{1^2 + 2^2 + 3^2} \cdot \sqrt{3^2 + 2^2 + 1^2}} \] \[ = \frac{3 + 4 + 3}{\sqrt{1 + 4 + 9} \cdot \sqrt{9 + 4 + 1}} = \frac{10}{\sqrt{14} \cdot \sqrt{14}} = \frac{10}{14} = \frac{5}{7} \] Thus, \( \cos \theta = \frac{5}{7} \).

The correct option is (D) : \(\frac{5}{7}\)

Answer 2

The direction ratios of the first line \(\frac{x}{1} = \frac{y + 1}{2} = \frac{z - 1}{3}\) are (1, 2, 3). Let this vector be \(\bar{a} = \hat{i} + 2\hat{j} + 3\hat{k}\).

The direction ratios of the second line \(\frac{x + 1}{3} = \frac{y}{2} = \frac{z}{1}\) are (3, 2, 1). Let this vector be \(\bar{b} = 3\hat{i} + 2\hat{j} + \hat{k}\).

If \(\theta\) is the angle between the lines, then

\(\cos\theta = \frac{|\bar{a} \cdot \bar{b}|}{|\bar{a}| |\bar{b}|}\)

We have:

\(\bar{a} \cdot \bar{b} = (1)(3) + (2)(2) + (3)(1) = 3 + 4 + 3 = 10\)

\(|\bar{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}\)

\(|\bar{b}| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}\)

Therefore:

\(\cos\theta = \frac{|10|}{\sqrt{14}\sqrt{14}} = \frac{10}{14} = \frac{5}{7}\)