7. State whether the following pairs of lines are par- allel. If they are not parallel determine the point of intersection. (a) \( \underline{r}=i+j-k+\lambda(2 i-3 j+k) \) \( \underline{r}=3 i-2 j+\mu(i+j-k) \) (b) \( \frac{x-1}{4}=\frac{y-4}{3}=\frac{z+1}{2} \) \( \frac{x}{4}=\frac{y+5}{3}=\frac{z-3}{2} \)

Let's analyze each pair of lines to determine whether they are parallel. If they are not parallel, we'll find their point of intersection. --- ### **Problem 7(a)** **Given Lines:** 1. **Line 1:** \(\underline{r} = \mathbf{i} + \mathbf{j} - \mathbf{k} + \lambda(2\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) 2. **Line 2:** \(\underline{r} = 3\mathbf{i} - 2\mathbf{j} + \mu(\mathbf{i} + \mathbf{j} - \mathbf{k})\) **Step 1: Check for Parallelism** - **Direction Vector of Line 1:** \( \mathbf{d}_1 = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k} \) - **Direction Vector of Line 2:** \( \mathbf{d}_2 = \mathbf{i} + \mathbf{j} - \mathbf{k} \) To determine if the lines are parallel, check if the direction vectors are scalar multiples of each other: \[ 2\mathbf{i} - 3\mathbf{j} + \mathbf{k} = c(\mathbf{i} + \mathbf{j} - \mathbf{k}) \] Comparing components: - \( 2 = c \) - \( -3 = c \) - \( 1 = -c \) This leads to a contradiction (e.g., \( c = 2 \) and \( c = -3 \)), so **the lines are not parallel**. **Step 2: Find the Point of Intersection** Set the parametric equations equal to each other: \[ \mathbf{i} + \mathbf{j} - \mathbf{k} + \lambda(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}) = 3\mathbf{i} - 2\mathbf{j} + \mu(\mathbf{i} + \mathbf{j} - \mathbf{k}) \] Equate the corresponding components: \[ \begin{cases} 1 + 2\lambda = 3 + \mu \quad \text{(x-component)} \\ 1 - 3\lambda = -2 + \mu \quad \text{(y-component)} \\ -1 + \lambda = -\mu \quad \text{(z-component)} \end{cases} \] From the z-component: \[ -1 + \lambda = -\mu \Rightarrow \mu = 1 - \lambda \] Substitute \(\mu = 1 - \lambda\) into the x-component: \[ 1 + 2\lambda - (1 - \lambda) = 3 \Rightarrow 3\lambda = 3 \Rightarrow \lambda = 1 \] Then, \(\mu = 1 - 1 = 0\). **Intersection Point:** Substitute \(\lambda = 1\) into Line 1's equation: \[ \underline{r} = \mathbf{i} + \mathbf{j} - \mathbf{k} + 1(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}) = 3\mathbf{i} - 2\mathbf{j} + 0\mathbf{k} \] **Therefore, the lines intersect at the point \((3, -2, 0)\).** --- ### **Problem 7(b)** **Given Lines:** 1. **Line 1:** \(\frac{x - 1}{4} = \frac{y - 4}{3} = \frac{z + 1}{2}\) 2. **Line 2:** \(\frac{x}{4} = \frac{y + 5}{3} = \frac{z - 3}{2}\) **Step 1: Check for Parallelism** Express both lines in parametric form: - **Line 1:** \[ \begin{cases} x = 1 + 4\lambda \\ y = 4 + 3\lambda \\ z = -1 + 2\lambda \end{cases} \] - **Line 2:** \[ \begin{cases} x = 4\mu \\ y = -5 + 3\mu \\ z = 3 + 2\mu \end{cases} \] - **Direction Vector for Both Lines:** \( \mathbf{d} = 4\mathbf{i} + 3\mathbf{j} + 2\mathbf{k} \) Since both lines share the same direction vector, **they are parallel**. **Step 2: Determine if They are Distinct or Coincident** To check if the lines coincide, see if a point from one line lies on the other. - **Point on Line 1:** \((1, 4, -1)\) Check if this point satisfies Line 2's equations: \[ \frac{1}{4} = \frac{4 + 5}{3} = \frac{-1 - 3}{2} \Rightarrow \frac{1}{4} \neq 3 \neq -2 \] Since the point does not satisfy Line 2's equations, **the lines are distinct and parallel**. --- ### **Final Answer** **Problem 7(a) Answer:** The lines are not parallel and intersect at the point (3, –2, 0). **Problem 7(b) Answer:** The lines are parallel and do not intersect.