<p>The lines with vector equations \(\vec{r_1} = -3\vec{i} + 6\vec{j} + \lambda(-4\vec{i} + 3\vec{j} + 2\vec{k})\) and \(\vec{r_2} = -2\vec{i} + 7\vec{j} + \mu(-4\vec{i} + \vec{j} + \vec{k})\) are such that:</p>
<p>(a) they are coplanar</p>
<p>(b) they do not intersect</p>
<p>(c) they are skew</p>
<p>(d) the angle between them is \(\tan^{-1}(3/7)\)</p>
Step-by-Step Solution
Key Concept: Use scalar triple product to determine if lines are coplanar; use dot product to find the angle between direction vectors.
Point on first line: \(A_1 = (-3, 6, 0)\); Direction: \(\vec{d_1} = (-4, 3, 2)\) Point on second line: \(A_2 = (-2, 7, 0)\); Direction: \(\vec{d_2} = (-4, 1, 1)\) \(\vec{A_1A_2} = (1, 1, 0)\); Scalar triple product: \((1)(3 \cdot 1 - 2 \cdot 1) - (1)(-4 \cdot 1 - 2 \cdot (-4)) + 0 = 1 - 4 = -3 \neq 0\). Lines are skew (non-coplanar and non-intersecting). \(\cos\theta = \frac{|16 + 3 + 2|}{\sqrt{29}\sqrt{18}} = \frac{21}{\sqrt{522}} = \frac{7}{\sqrt{58}}\), giving \(\tan\theta = 3/7\). Correct.
Correct Answer: b, c, d