Let's break it down step-by-step as requested: --- ### **2(a) Cartesian Equation for the Plane** **Step 1: Find a point and two direction vectors in the plane.** - **Point on \( L_1 \):** \( L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-1}{1} = t \) So, a point on \( L_1 \) (when \( t=0 \)): \( (x,y,z) = (1,2,1) \). - **Direction vector of \( L_1 \):** From the denominators: \( \vec{d_1} = (2,3,1) \). - **Direction vector of \( L_2 \):** \( L_2: \frac{x-2}{1} = \frac{y+1}{-2} = \frac{z-4}{-2} \) Direction vector: \( \vec{d_2} = (1,-2,-2) \). - **The plane contains \( L_1 \) (so it contains \( (1,2,1) \) and direction \( (2,3,1) \)), and is parallel to \( L_2 \) (so direction \( (1,-2,-2) \) is also in the plane).** **Step 1 Explanation:** We need a plane containing a given point and two direction vectors (one from \( L_1 \), one from \( L_2 \)), because the plane must contain \( L_1 \) and be parallel to \( L_2 \). --- **Step 2: Find the normal vector and write the equation.** - Two direction vectors in the plane: \( \vec{d_1} = (2,3,1) \), \( \vec{d_2} = (1,-2,-2) \). - Normal vector \( \vec{n} = \vec{d_1} \times \vec{d_2} \): \[ \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 1 \\ 1 & -2 & -2 \\ \end{vmatrix} = \mathbf{i}(3 \cdot -2 - 1 \cdot -2) - \mathbf{j}(2 \cdot -2 - 1 \cdot 1) + \mathbf{k}(2 \cdot -2 - 3 \cdot 1) \] \[ = \mathbf{i}(-6 + 2) - \mathbf{j}(-4 - 1) + \mathbf{k}(-4 - 3) = \mathbf{i}(-4) - \mathbf{j}(-5) + \mathbf{k}(-7) = (-4, 5, -7) \] - Equation of the plane through \( (1,2,1) \) with normal \( (-4,5,-7) \): \[ -4(x-1) + 5(y-2) - 7(z-1) = 0 \] or \[ -4x + 5y - 7z + 4 - 10 + 7 = 0 \] \[ -4x + 5y - 7z + 1 = 0 \] **Step 2 Explanation:** The normal vector is found using the cross product of the two direction vectors. The Cartesian equation is derived from the point-normal form of a plane. --- **Final Answer for 2(a):** \[ \boxed{-4x + 5y - 7z + 1 = 0} \] --- ## **2(b) True/False Statements** ### i. For \( a, b, c \in \mathbb{R}^n \), if \( a-b=0 \) and \( a-c=0 \), then \( b-c=0 \). **Step 1:** If \( a-b=0 \Rightarrow a = b \) and \( a-c=0 \Rightarrow a = c \). **Step 1 Explanation:** Both \( b \) and \( c \) are equal to \( a \). **Step 2:** Since \( b = c \), \( b-c = 0 \). **Step 2 Explanation:** Subtracting equal vectors gives the zero vector. **Final Answer for (i):** **True.** --- ### ii. For two planes in \( \mathbb{R}^3 \) with normals \( n_1, n_2 \) that intersect in a line with direction vector \( d \), \( d \) is always parallel to \( n_1 \times n_2 \). **Step 1:** The direction vector \( d \) of the intersection line of two planes is given by the cross product of their normals: \( d = n_1 \times n_2 \). **Step 1 Explanation:** The intersection line is perpendicular to both normal vectors, and the cross product gives such a direction. **Step 2:** Therefore, \( d \) is always parallel to \( n_1 \times n_2 \). **Step 2 Explanation:** This is a standard result in 3D analytic geometry. **Final Answer for (ii):** **True.**

Let's carefully solve each part step-by-step. --- ## **2(a) Cartesian Equation of the Plane** ### **Step 1: Identify a point and direction vectors** Given lines: \[ L_1: \frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z - 1}{1} \] - **Point on \( L_1 \):** When the parameter \( t=0 \): \[ x=1, \quad y=2, \quad z=1 \] - **Direction vector of \( L_1 \):** From the denominators: \[ \vec{d}_1 = (2, -1, 1) \] --- Given lines: \[ L_2: \frac{x - 3}{1} = \frac{y - 1}{1} = \frac{z - 4}{-2} \] - **Direction vector of \( L_2 \):** \[ \vec{d}_2 = (1, 1, -2) \] --- ### **Step 2: Find the plane containing \( L_1 \) and parallel to \( L_2 \)** - The plane **must contain** \( L_1 \), so it contains point \( P=(1,2,1) \). - The plane is **parallel** to \( L_2 \), meaning the direction vector \( \vec{d}_2 \) is **parallel to** the plane (but not necessarily in the plane). **Key insight:** The plane contains \( L_1 \), so it contains \( \vec{d}_1 \). It is **parallel** to \( L_2 \), so the vector \( \vec{d}_2 \) is **parallel** to the plane (but not necessarily in it). This implies the plane contains \( \vec{d}_1 \) and is **parallel** to \( \vec{d}_2 \). - The **normal vector** \( \vec{n} \) of the plane should be perpendicular to both \( \vec{d}_1 \) and \( \vec{d}_2 \). ### **Step 3: Calculate the normal vector** \[ \vec{n} = \vec{d}_1 \times \vec{d}_2 \] \[ \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 1 & 1 & -2 \\ \end{vmatrix} \] Calculate the cross product: \[ \vec{n} = \mathbf{i}((-1)\times(-2) - 1 \times 1) - \mathbf{j}(2 \times (-2) - 1 \times 1) + \mathbf{k}(2 \times 1 - (-1) \times 1) \] \[ = \mathbf{i} (2 - 1) - \mathbf{j} (-4 - 1) + \mathbf{k} (2 + 1) \] \[ = \mathbf{i} (1) - \mathbf{j} (-5) + \mathbf{k} (3) \] \[ \Rightarrow \vec{n} = (1, 5, 3) \] ### **Step 4: Write the Cartesian equation** Using point \( (1, 2, 1) \): \[ 1(x - 1) + 5(y - 2) + 3(z - 1) = 0 \] \[ (x - 1) + 5(y - 2) + 3(z - 1) = 0 \] Expand: \[ x - 1 + 5y - 10 + 3z - 3 = 0 \] \[ x + 5y + 3z - 14 = 0 \] --- ## **Final answer for 2(a):** \[ \boxed{x + 5y + 3z = 14} \] --- ## **2(b) True/False Statements** ### **i. If \( a, b, c \in \mathbb{R}^3 \) and \( a \times b = 0 \) and \( a \times c = 0 \), then \( b \times c = 0 \).** **Step 1:** Recall that \( a \times b = 0 \) implies \( a \) and \( b \) are **parallel** (or one is zero). Similarly, \( a \times c = 0 \) implies \( a \) and \( c \) are **parallel** (or one is zero). - So, \( b \) and \( c \) are both parallel to \( a \) (or \( a \) is zero). **Step 2:** - If \( a \neq 0 \), then \( b = \lambda a \) and \( c = \mu a \): \[ b \times c = (\lambda a) \times (\mu a) = \lambda \mu (a \times a) = 0 \] since the cross product of any vector with itself is zero. - If \( a=0 \), then \( a \times b=0 \) and \( a \times c=0 \) hold trivially, and \( b, c \) can be arbitrary. In that case, \( b \) and \( c \) are not necessarily parallel, so \( b \times c \) may **not** be zero. **Counterexample:** Let \( a=0 \), \( b=(1,0,0) \), \( c=(0,1,0) \): \[ a \times b = 0, \quad a \times c = 0, \quad \text{but} \quad b \times c = (0,0,1) \neq 0 \] Thus, the statement is **false**. --- ### **ii. For two planes in \( \mathbb{R}^3 \) with normals \( n_1, n_2 \), if they intersect in a line with direction vector \( d \), then \( d \) is always parallel to \( n_1 \times n_2 \).** **Step 1:** The line of intersection of two planes is perpendicular to both normals, so its direction vector is proportional to the cross product: \[ d \parallel n_1 \times n_2 \] **Step 2:** Since the cross product gives a vector perpendicular to both \( n_1 \) and \( n_2 \), \( d \) must be parallel to \( n_1 \times n_2 \). **Final conclusion:** The statement is **true**. --- ## **Summary:** ### **Final answers:** **2(a):** \[ \boxed{x + 5y + 3z = 14} \] **2(b):** - (i) **False** (counterexample: \( a=0 \), \( b=(1,0,0) \), \( c=(0,1,0) \)) - (ii) **True**