Let's break the problem into the required sections step-by-step, with explanations and final answers. --- ## **Given Points (assumed column vectors in \(\mathbb{R}^3\)):** - \(A = \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}\) - \(B = \begin{bmatrix} -1 \\ -6 \\ 0 \end{bmatrix}\) - \(C = \begin{bmatrix} -1 \\ 17 \\ 2 \end{bmatrix}\) - \(D = \begin{bmatrix} 0 \\ 2 \\ 5 \\ -2 \end{bmatrix}\) (however, this has four coordinates; probably a typo, see below) - \(P = \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix}\) - \(Q = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}\) - \(R = \begin{bmatrix} -1 \\ 8 \\ -1 \end{bmatrix}\) - \(S = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix}\) --- ### **(a) Area of Triangle ABC** #### **Step 1: Find vectors AB and AC** \[ \vec{AB} = B - A = \begin{bmatrix} -1 \\ -6 \\ 0 \end{bmatrix} - \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ -5 \\ 1 \end{bmatrix} \] \[ \vec{AC} = C - A = \begin{bmatrix} -1 \\ 17 \\ 2 \end{bmatrix} - \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 18 \\ 3 \end{bmatrix} \] #### **Step 2: Compute the cross product \(\vec{AB} \times \vec{AC}\)** \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -5 & 1 \\ 0 & 18 & 3 \\ \end{vmatrix} \] Expand: - \(i\) component: \((-5)(3) - (1)(18) = -15 - 18 = -33\) - \(j\) component: \,- [ (0)(3) - (1)(0) ] = -(0 - 0) = 0 - \(k\) component: \((0)(18) - (-5)(0) = 0 - 0 = 0\) So, \[ \vec{AB} \times \vec{AC} = \begin{bmatrix} -33 \\ 0 \\ 0 \end{bmatrix} \] #### **Step 3: The Area Formula** \[ \text{Area} = \frac{1}{2} \left\| \vec{AB} \times \vec{AC} \right\| = \frac{1}{2} \times | -33 | = 16.5 \] #### **Step 4: Justification** The cross product gives a nonzero vector, so the points are not collinear. **Final Answer for (a):** \[ \boxed{16.5} \] --- ### **(b) Do A, B, C, D lie on a plane in \( \mathbb{R}^3 \)?** However, D is given as \( D = [0, 2, 5, -2] \), which is a 4D vector, but all others are 3D. Assuming a typo, perhaps D is \( D = [0, 2, 5] \). #### **Step 1: Find if D is coplanar with A, B, C** Points in \( \mathbb{R}^3 \) are coplanar if the vectors \( \vec{AB}, \vec{AC}, \vec{AD} \) are linearly dependent (i.e., the scalar triple product is zero). \[ \vec{AD} = D - A = \begin{bmatrix} 0 \\ 2 \\ 5 \end{bmatrix} - \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 6 \end{bmatrix} \] \[ \text{Scalar triple product} = (\vec{AB} \times \vec{AC}) \cdot \vec{AD} \] \[ \vec{AB} \times \vec{AC} = \begin{bmatrix} -33 \\ 0 \\ 0 \end{bmatrix} \] \[ (\vec{AB} \times \vec{AC}) \cdot \vec{AD} = (-33)(1) + 0(3) + 0(6) = -33 \] Since the scalar triple product is **not zero**, the points are **not coplanar**. **Final Answer for (b):** \[ \boxed{\text{No, they do not lie on a plane in } \mathbb{R}^3} \] --- ### **(c) Orthogonal and Parallel Vectors** #### **Step 1: Vectors \( \vec{PQ} \) and \( \vec{PR} \)** \[ \vec{PQ} = Q - P = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} - \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} -8 \\ 1 \\ 0 \end{bmatrix} \] \[ \vec{PR} = R - P = \begin{bmatrix} -1 \\ 8 \\ -1 \end{bmatrix} - \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} -9 \\ 8 \\ 0 \end{bmatrix} \] #### **Step 2: Vector Orthogonal to Both (\( \vec{PQ} \times \vec{PR} \))** \[ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & 1 & 0 \\ -9 & 8 & 0 \\ \end{vmatrix} \] - \(i\) component: \( (1)(0) - (0)(8) = 0 \) - \(j\) component: \(- [ (-8)(0) - (0)(-9) ] = 0 \) - \(k\) component: \( (-8)(8) - (1)(-9) = -64 + 9 = -55 \) So, \[ \vec{PQ} \times \vec{PR} = \begin{bmatrix} 0 \\ 0 \\ -55 \end{bmatrix} \] Any scalar multiple of this (e.g., \( [0,0,1] \)) is also orthogonal. #### **Step 3: Vector Parallel to \( \vec{PS} \)** \[ \vec{PS} = S - P = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix} - \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} -6 \\ 3 \\ 6 \end{bmatrix} \] Any scalar multiple of this vector is parallel. **Final Answer for (c):** - Orthogonal vector: \( \boxed{\begin{bmatrix} 0 \\ 0 \\ -55 \end{bmatrix}} \) (or any nonzero multiple) - Parallel vector: \( \boxed{\begin{bmatrix} -6 \\ 3 \\ 6 \end{bmatrix}} \) (or any nonzero multiple) --- ### **(d) Compute Dot Products** Let’s clarify the notation: - \( \vec{QP} = P - Q \) - \( \vec{QR} = R - Q \) - \( \vec{PQ} = Q - P \) - \( \vec{PR} = R - P \) - \( \vec{RQ} = Q - R \) - \( \vec{RP} = P - R \) #### **First pair: \( (\vec{QP}, \vec{QR}) \)** \[ \vec{QP} = P - Q = \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 8 \\ -1 \\ 0 \end{bmatrix} \] \[ \vec{QR} = R - Q = \begin{bmatrix} -1 \\ 8 \\ -1 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} -1 \\ 7 \\ 0 \end{bmatrix} \] \[ \vec{QP} \cdot \vec{QR} = (8)(-1) + (-1)(7) + 0(0) = -8 - 7 = -15 \] #### **Second pair: \( (\vec{PQ}, \vec{PR}) \)** \[ \vec{PQ} = Q - P = \begin{bmatrix} -8 \\ 1 \\ 0 \end{bmatrix} \] \[ \vec{PR} = R - P = \begin{bmatrix} -9 \\ 8 \\ 0 \end{bmatrix} \] \[ \vec{PQ} \cdot \vec{PR} = (-8)(-9) + (1)(8) + (0)(0) = 72 + 8 = 80 \] #### **Third pair: \( (\vec{RQ}, \vec{RP}) \)** \[ \vec{RQ} = Q - R = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} - \begin{bmatrix} -1 \\ 8 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ -7 \\ 0 \end{bmatrix} \] \[ \vec{RP} = P - R = \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix} - \begin{bmatrix} -1 \\ 8 \\ -1 \end{bmatrix} = \begin{bmatrix} 9 \\ -8 \\ 0 \end{bmatrix} \] \[ \vec{RQ} \cdot \vec{RP} = (1)(9) + (-7)(-8) + (0)(0) = 9 + 56 = 65 \] **Final Answer for (d):** - \( \vec{QP} \cdot \vec{QR} = \boxed{-15} \) - \( \vec{PQ} \cdot \vec{PR} = \boxed{80} \) - \( \vec{RQ} \cdot \vec{RP} = \boxed{65} \) --- ### **(e) Nature of Triangle \(PQR\) (Acute, Right, or Obtuse)** A triangle is: - **Acute**: all angles < 90° - **Right**: one angle = 90° - **Obtuse**: one angle > 90° #### **Step 1: Use Dot Products** Recall: - If the dot product between two sides at a vertex is positive, the angle is < 90° - If zero, angle = 90° - If negative, angle > 90° From above: - At Q: \( \vec{QP} \cdot \vec{QR} = -15 \) (angle > 90°, i.e., obtuse) - At P: \( \vec{PQ} \cdot \vec{PR} = 80 \) (angle < 90°) - At R: \( \vec{RQ} \cdot \vec{RP} = 65 \) (angle < 90°) So, angle at Q is **obtuse**. **Final Answer for (e):** \[ \boxed{\text{Triangle } PQR \text{ is obtuse, since the angle at Q is greater than 90°.}} \] --- ## **Summary Table** | Part | Final Answer | |------|----------------------------------------------------------------------------------------------| | (a) | \( \boxed{16.5} \) area units | | (b) | \( \boxed{\text{No, A, B, C, D do not lie on a plane in } \mathbb{R}^3} \) | | (c) | Orthogonal: \( \boxed{[0, 0, -55]} \), Parallel: \( \boxed{[-6, 3, 6]} \) | | (d) | \( \vec{QP} \cdot \vec{QR} = \boxed{-15} \), \( \vec{PQ} \cdot \vec{PR} = \boxed{80} \), \( \vec{RQ} \cdot \vec{RP} = \boxed{65

Let's carefully analyze each part of the problem step-by-step, providing explanations at each stage and a final answer. --- ## **Given points in \(\mathbb{R}^3\):** - \(A = \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}\) - \(B = \begin{bmatrix} -1 \\ -6 \\ 0 \end{bmatrix}\) - \(C = \begin{bmatrix} -1 \\ 7 \\ 2 \end{bmatrix}\) - \(D = \begin{bmatrix} 0 \\ 2 \\ 5 \end{bmatrix}\) - \(P = \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix}\) - \(Q = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix}\) - \(R = \begin{bmatrix} -1 \\ 8 \\ -1 \end{bmatrix}\) - \(S = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix}\) --- ### **(a) Find the area of triangle \(ABC\).** **Step 1: Compute vectors \(\vec{AB}\) and \(\vec{AC}\).** \[ \vec{AB} = B - A = \begin{bmatrix} -1 \\ -6 \\ 0 \end{bmatrix} - \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ -5 \\ 1 \end{bmatrix} \] \[ \vec{AC} = C - A = \begin{bmatrix} -1 \\ 7 \\ 2 \end{bmatrix} - \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 8 \\ 3 \end{bmatrix} \] **Step 2: Compute the cross product \(\vec{AB} \times \vec{AC}\) to find the area.** \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & -5 & 1 \\ 0 & 8 & 3 \end{vmatrix} \] - \(i\) component: \((-5)(3) - (1)(8) = -15 - 8 = -23\) - \(j\) component: \(-[0 \times 3 - 1 \times 0] = -[0 - 0] = 0\) - \(k\) component: \(0 \times 8 - (-5) \times 0 = 0 - 0 = 0\) \[ \Rightarrow \vec{AB} \times \vec{AC} = \begin{bmatrix} -23 \\ 0 \\ 0 \end{bmatrix} \] **Step 3: Compute the magnitude of the cross product and find the area.** \[ \left\| \vec{AB} \times \vec{AC} \right\| = \sqrt{(-23)^2 + 0^2 + 0^2} = 23 \] \[ \text{Area of } \triangle ABC = \frac{1}{2} \times 23 = 11.5 \] --- **Final answer for (a):** \(\boxed{11.5}\) --- ### **(b) Do points \(A, B, C, D\) lie on a plane?** **Step 1: Find if \(A, B, C, D\) are coplanar.** Points are coplanar if the scalar triple product of vectors \(\vec{AB}\), \(\vec{AC}\), and \(\vec{AD}\) is zero. Calculate \(\vec{AD} = D - A\): \[ \vec{AD} = \begin{bmatrix} 0 \\ 2 \\ 5 \end{bmatrix} - \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 6 \end{bmatrix} \] Calculate the scalar triple product: \[ \text{Scalar triple} = (\vec{AB} \times \vec{AC}) \cdot \vec{AD} \] From part (a), \(\vec{AB} \times \vec{AC} = \begin{bmatrix} -23 \\ 0 \\ 0 \end{bmatrix}\). Now: \[ (-23) \times 1 + 0 \times 3 + 0 \times 6 = -23 \] Since the scalar triple product is **not zero**, the four points are **not coplanar**. --- **Final answer for (b):** \(\boxed{\text{No}}\) --- ### **(c) Find a vector orthogonal to both \(\vec{PQ}\) and \(\vec{PR}\), and a vector parallel to \(\vec{PS}\).** **Step 1: Compute \(\vec{PQ}\) and \(\vec{PR}\).** \[ \vec{PQ} = Q - P = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} - \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} -8 \\ 1 \\ 0 \end{bmatrix} \] \[ \vec{PR} = R - P = \begin{bmatrix} -1 \\ 8 \\ -1 \end{bmatrix} - \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} -9 \\ 8 \\ 0 \end{bmatrix} \] **Step 2: Find a vector orthogonal to both \(\vec{PQ}\) and \(\vec{PR}\).** Compute their cross product: \[ \vec{PQ} \times \vec{PR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & 1 & 0 \\ -9 & 8 & 0 \end{vmatrix} \] - \(i\): \((1)(0) - (0)(8) = 0\) - \(j\): \(-[(-8)(0) - (0)(-9)] = -[0 - 0] = 0\) - \(k\): \((-8)(8) - (1)(-9) = -64 + 9 = -55\) So, an orthogonal vector: \(\boxed{\begin{bmatrix} 0 \\ 0 \\ -55 \end{bmatrix}}\). **Step 3: Find a vector parallel to \(\vec{PS}\).** \[ \vec{PS} = S - P = \begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix} - \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix} = \begin{bmatrix} -6 \\ 3 \\ 6 \end{bmatrix} \] Any scalar multiple of this vector is parallel. --- **Final answers for (c):** - Orthogonal vector: \(\boxed{\begin{bmatrix} 0 \\ 0 \\ -55 \end{bmatrix}}\) (or scaled version) - Parallel vector: \(\boxed{\begin{bmatrix} -6 \\ 3 \\ 6 \end{bmatrix}}\) --- ### **(d) Dot products between each pair:** - \( \vec{QP} = P - Q = \begin{bmatrix} 8 \\ 0 \\ -1 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 8 \\ -1 \\ 0 \end{bmatrix} \) - \( \vec{QR} = R - Q = \begin{bmatrix} -1 \\ 8 \\ -1 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} -1 \\ 7 \\ 0 \end{bmatrix} \) - \( \vec{PQ} = Q - P = -\vec{QP} = \begin{bmatrix} -8 \\ 1 \\ 0 \end{bmatrix} \) - \( \vec{PR} = R - P = \begin{bmatrix} -9 \\ 8 \\ 0 \end{bmatrix} \) - \( \vec{RQ} = Q - R = -\vec{QR} = \begin{bmatrix} 1 \\ -7 \\ 0 \end{bmatrix} \) - \( \vec{RP} = P - R = \begin{bmatrix} 9 \\ -8 \\ 0 \end{bmatrix} \) Calculate dot products: 1. \( \vec{QP} \cdot \vec{QR} = \begin{bmatrix} 8 \\ -1 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} -1 \\ 7 \\ 0 \end{bmatrix} = (8)(-1) + (-1)(7) + 0 = -8 -7 = -15 \) 2. \( \vec{PQ} \cdot \vec{PR} = \begin{bmatrix} -8 \\ 1 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} -9 \\ 8 \\ 0 \end{bmatrix} = (-8)(-9) + 1 \times 8 + 0 = 72 + 8 = 80 \) 3. \( \vec{RQ} \cdot \vec{RP} = \begin{bmatrix} 1 \\ -7 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} 9 \\ -8 \\ 0 \end{bmatrix} = 1 \times 9 + (-7)(-8) + 0 = 9 + 56 = 65 \) --- ## **Final answers for (d):** | Pair | Dot product | |---------|--------------| | \((\vec{QP}, \vec{QR})\) | \(-15\) | | \((\vec{PQ}, \vec{PR})\) | \(80\) | | \((\vec{RQ}, \vec{RP})\) | \(65\) | --- ### **(e) Is triangle \(PQR\) acute, right, or obtuse?** **Step 1:** Use the dot products to determine angles: - If dot product > 0: angle < 90° - If dot product = 0: angle = 90° - If dot product < 0: angle > 90° From (d): - At \(Q\): \(\vec{QP} \cdot \vec{QR} = -15 < 0 \Rightarrow\) angle at \(Q\) is **obtuse**. - At \(P\): \(\vec{PQ} \cdot \vec{PR} = 80 > 0 \Rightarrow\) angle at \(P\) is **acute**. - At \(R\): \(\vec{RQ} \cdot \vec{RP} = 65 > 0 \Rightarrow\) angle at \(R\) is **acute**. Since at least one angle (at \(Q\)) is obtuse, **the triangle is obtuse**. --- ## **Final summary:** | Part | Final conclusion | |---------|-------------------| | (a) | Area = **11.5** | | (b) | **No**, points are not coplanar | | (c) | Orthogonal: \(\begin{bmatrix} 0 \\ 0 \\ -55 \end{bmatrix}\), Parallel: \(\begin{bmatrix} -6 \\ 3 \\ 6 \end{bmatrix}\) | | (d) | Dot products: \(-15\), \(80\), \(65\) | | (e) | Triangle \(PQR\) is **obtuse** | Let me know if you'd like further clarification!