Let's proceed step by step for each part, with explanations after each step, and provide final answers at the end. **First, let's rewrite the coordinates for clarity:** - \( A = [-1, -1, -1] \) - \( B = [-1, -6, 0] \) - \( C = [-1, 17, 2] \) - \( D = [0, 2, 5] \) - \( P = [8, 0, -1] \) - \( Q = [0, 1, -1] \) - \( R = [-18, -1, -1] \) - \( S = [2, 3, 5] \) --- ## (a) Area of triangle \(ABC\) ### **Step 1: Find vectors \( \vec{AB} \) and \( \vec{AC} \):** \[ \vec{AB} = B - A = [-1, -6, 0] - [-1, -1, -1] = [0, -5, 1] \] \[ \vec{AC} = C - A = [-1, 17, 2] - [-1, -1, -1] = [0, 18, 3] \] #### **Explanation:** Subtracting coordinates gives the direction vectors from A to B and from A to C. --- ### **Step 2: Compute the cross product \( \vec{AB} \times \vec{AC} \):** \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & -5 & 1\\ 0 & 18 & 3 \end{vmatrix} \] Using the determinant: \[ = \mathbf{i}[(-5)(3) - (1)(18)] - \mathbf{j}[0 \cdot 3 - 1 \cdot 0] + \mathbf{k}[0 \cdot 18 - (-5) \cdot 0] \] \[ = \mathbf{i}[-15 - 18] - \mathbf{j}[0 - 0] + \mathbf{k}[0 - 0] \] \[ = \mathbf{i}[-33] - \mathbf{j}[0] + \mathbf{k}[0] = [-33, 0, 0] \] #### **Explanation:** The cross product gives a vector perpendicular to the triangle's plane; its magnitude is twice the triangle's area. --- ### **Step 3: Find the magnitude and area:** \[ |\vec{AB} \times \vec{AC}| = \sqrt{(-33)^2 + 0^2 + 0^2} = 33 \] Area \( = \frac{1}{2} \times 33 = 16.5 \) #### **Explanation:** The triangle's area is half the magnitude of the cross product. --- **Final answer (a):** \[ \boxed{16.5} \] The area of triangle \(ABC\) is 16.5 square units. --- ## (b) Do \(A,B,C,D\) lie in a plane in \(\mathbb{R}^3\)? ### **Step 1: Compute three vectors from point A:** \[ \vec{AB} = [0, -5, 1] \\ \vec{AC} = [0, 18, 3] \\ \vec{AD} = [1, 3, 6] \] #### **Explanation:** If vectors \( \vec{AB}, \vec{AC}, \vec{AD} \) are coplanar, \(A,B,C,D\) are coplanar. --- ### **Step 2: Check if \( \vec{AD} \) is a linear combination of \( \vec{AB} \) and \( \vec{AC} \):** Set \( \vec{AD} = x\vec{AB} + y\vec{AC} \): \[ [1,3,6] = x[0,-5,1] + y[0,18,3] \] So: \[ x \cdot 0 + y \cdot 0 = 1 \implies 0 = 1 \rightarrow \text{Contradiction} \] #### **Explanation:** Since the first component cannot be made to match, the vectors are not coplanar. --- ### **Step 3: Conclude.** #### **Explanation:** Since the 4 points are not coplanar, they do not all lie on the same plane. --- **Final answer (b):** \[ \boxed{\text{No}} \] The points \(A, B, C, D\) do **not** all lie on a plane in \(\mathbb{R}^3\). --- ## (c) Orthogonal and parallel vectors ### **Step 1: Find vectors \(\vec{PQ}, \vec{PR}, \vec{PS}\):** \[ \vec{PQ} = Q - P = [0,1,-1] - [8,0,-1] = [-8, 1, 0] \] \[ \vec{PR} = R - P = [-18, -1, -1] - [8,0,-1] = [-26, -1, 0] \] \[ \vec{PS} = S - P = [2,3,5] - [8,0,-1] = [-6, 3, 6] \] #### **Explanation:** Subtract to get direction vectors from P to Q, R, S. --- ### **Step 2: Find a vector orthogonal to both \(\vec{PQ}\) and \(\vec{PR}\) (cross product):** \[ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ -8 & 1 & 0\\ -26 & -1 & 0 \end{vmatrix} \] \[ = \mathbf{i}(1\cdot0 - 0\cdot(-1)) - \mathbf{j}((-8)\cdot0 - 0\cdot(-26)) + \mathbf{k}((-8)\cdot(-1) - 1\cdot(-26)) \] \[ = \mathbf{i}(0) - \mathbf{j}(0) + \mathbf{k}(8 + 26) \] \[ = [0, 0, 34] \] #### **Explanation:** The cross product gives a vector orthogonal to both. --- ### **Step 3: Find a vector parallel to \(\vec{PS}\):** Any scalar multiple of \(\vec{PS}\) is parallel to it. The simplest is itself: \[ \vec{PS} = [-6, 3, 6] \] #### **Explanation:** A vector is always parallel to itself. --- **Final answer (c):** - A vector orthogonal to both \(\vec{PQ}\) and \(\vec{PR}\) is \(\boxed{[0,0,34]}\) (or any nonzero scalar multiple). - A vector parallel to \(\vec{PS}\) is \(\boxed{[-6,3,6]}\) (or any nonzero scalar multiple). --- ## (d) Compute dot products ### **Step 1: Find the vectors** \[ \vec{QP} = P-Q = [8,0,-1] - [0,1,-1] = [8,-1,0] \] \[ \vec{QR} = R-Q = [-18,-1,-1] - [0,1,-1] = [-18,-2,0] \] \[ \vec{PQ} = [-8,1,0] \] \[ \vec{PR} = [-26,-1,0] \] \[ \vec{RQ} = Q-R = [0,1,-1] - [-18,-1,-1] = [18,2,0] \] \[ \vec{RP} = P-R = [8,0,-1] - [-18,-1,-1] = [26,1,0] \] #### **Explanation:** Compute all required vectors. --- ### **Step 2: Compute each dot product** **(i) \(\vec{QP} \cdot \vec{QR} = [8,-1,0] \cdot [-18,-2,0] = 8\cdot(-18) + (-1)\cdot(-2) + 0\cdot0 = -144 + 2 + 0 = -142\)** **(ii) \(\vec{PQ} \cdot \vec{PR} = [-8,1,0] \cdot [-26,-1,0] = (-8)\cdot(-26) + 1\cdot(-1) + 0\cdot0 = 208 - 1 = 207\)** **(iii) \(\vec{RQ} \cdot \vec{RP} = [18,2,0] \cdot [26,1,0] = 18\cdot26 + 2\cdot1 + 0\cdot0 = 468 + 2 = 470\)** #### **Explanation:** Dot product: \(a_1b_1 + a_2b_2 + a_3b_3\) --- ### **Step 3: List answers** #### **Explanation:** Final calculated values. --- **Final answer (d):** - \(\vec{QP} \cdot \vec{QR} = \boxed{-142}\) - \(\vec{PQ} \cdot \vec{PR} = \boxed{207}\) - \(\vec{RQ} \cdot \vec{RP} = \boxed{470}\) --- ## (e) Nature of triangle \(PQR\): acute, right, or obtuse? ### **Step 1: Use dot products from (d) to check for right/obtuse/acute** Recall: - If the dot product between two sides (from the same vertex) is 0, the triangle is right-angled at that vertex. - If the dot product is negative, the angle is obtuse. - If all are positive, it is acute. From above: - \(\vec{QP} \cdot \vec{QR} = -142\) (**negative**) - \(\vec{PQ} \cdot \vec{PR} = 207\) (**positive**) - \(\vec{RQ} \cdot \vec{RP} = 470\) (**positive**) #### **Explanation:** At vertex \(Q\), the dot product is negative, so angle at \(Q\) is obtuse. --- ### **Step 2: Conclusion** Since one angle is obtuse, triangle is obtuse. --- ### **Step 3: State the answer** #### **Explanation:** Only one angle is obtuse, so the triangle is obtuse. --- **Final answer (e):** \[ \boxed{\text{The triangle } PQR \text{ is obtuse (at vertex Q)}} \] --- ## **Summary of Answers** **(a)** Area of triangle \(ABC\): \(\boxed{16.5}\) **(b)** Are \(A,B,C,D\) coplanar? \(\boxed{\text{No}}\) **(c)** Orthogonal vector: \(\boxed{[0,0,34]}\), Parallel vector to \(\vec{PS}\): \(\boxed{[-6,3,6]}\) **(d)** Dot products: \(\boxed{-142}, \boxed{207}, \boxed{470}\) **(e)** Triangle \(PQR\) is \(\boxed{\text{obtuse}}\) at \(Q\).