Orthocenter Triangle B22B29

1. **Problem Statement:** Find the orthocenter of the triangle with vertices \(A(2,1,0)\), \(B(5,4,3)\), and \(C(1,6,2)\). 2. **Definition:** The orthocenter is the point where the three altitudes of a triangle intersect. An altitude is a perpendicular line from a vertex to the opposite side (or its extension). 3. **Step 1: Find vectors for sides** - Vector \(\overrightarrow{BC} = C - B = (1-5, 6-4, 2-3) = (-4, 2, -1)\) - Vector \(\overrightarrow{AC} = C - A = (1-2, 6-1, 2-0) = (-1, 5, 2)\) 4. **Step 2: Find direction vectors of altitudes** - Altitude from \(A\) is perpendicular to side \(BC\), so its direction vector is perpendicular to \(\overrightarrow{BC}\). - Altitude from \(B\) is perpendicular to side \(AC\), so its direction vector is perpendicular to \(\overrightarrow{AC}\). 5. **Step 3: Find equations of altitudes from vertices A and B** - Let \(\vec{r}_A = (2,1,0) + t\vec{d}_A\), where \(\vec{d}_A\) is perpendicular to \(\overrightarrow{BC} = (-4,2,-1)\). - To find \(\vec{d}_A\), find any vector perpendicular to \(\overrightarrow{BC}\). For example, choose \(\vec{d}_A = (x,y,z)\) such that \(-4x + 2y - z = 0\). - Choose \(x=1, y=2\), then \(-4(1) + 2(2) - z = 0 \Rightarrow -4 + 4 - z = 0 \Rightarrow z=0\). - So \(\vec{d}_A = (1,2,0)\). - Equation of altitude from \(A\): $$\vec{r}_A = (2,1,0) + t(1,2,0)$$ - Similarly, for altitude from \(B\), find \(\vec{d}_B\) perpendicular to \(\overrightarrow{AC} = (-1,5,2)\). - Find \(\vec{d}_B = (x,y,z)\) such that \(-1x + 5y + 2z = 0\). - Choose \(x=0, y=2\), then \(-1(0) + 5(2) + 2z = 0 \Rightarrow 10 + 2z = 0 \Rightarrow z = -5\). - So \(\vec{d}_B = (0,2,-5)\). - Equation of altitude from \(B\): $$\vec{r}_B = (5,4,3) + s(0,2,-5)$$ 6. **Step 4: Find intersection of altitudes from A and B** - Set \(\vec{r}_A = \vec{r}_B\): \[ 2 + t = 5 + 0s \\ 1 + 2t = 4 + 2s \\ 0 + 0t = 3 - 5s \] - From first equation: \(t = 3\). - From third equation: \(0 = 3 - 5s \Rightarrow s = \frac{3}{5} = 0.6\). - From second equation: \(1 + 2(3) = 4 + 2(0.6) \Rightarrow 1 + 6 = 4 + 1.2 \Rightarrow 7 = 5.2\) which is false. 7. **Step 5: Recalculate direction vector for altitude from A** - Try another vector perpendicular to \(\overrightarrow{BC} = (-4,2,-1)\). - Let \(x=1, z=2\), then \(-4(1) + 2y - 2 = 0 \Rightarrow -4 + 2y - 2 = 0 \Rightarrow 2y = 6 \Rightarrow y=3\). - So \(\vec{d}_A = (1,3,2)\). - New altitude from \(A\): $$\vec{r}_A = (2,1,0) + t(1,3,2)$$ 8. **Step 6: Solve system again with new \(\vec{d}_A\)** - Equate \(\vec{r}_A = \vec{r}_B\): \[ 2 + t = 5 + 0s \\ 1 + 3t = 4 + 2s \\ 0 + 2t = 3 - 5s \] - From first: \(t = 3\). - Substitute \(t=3\) into second and third: \[ 1 + 3(3) = 4 + 2s \Rightarrow 10 = 4 + 2s \Rightarrow 2s = 6 \Rightarrow s=3 \] \[ 0 + 2(3) = 3 - 5s \Rightarrow 6 = 3 - 5(3) \Rightarrow 6 = 3 - 15 \Rightarrow 6 = -12\) false again. 9. **Step 7: Use altitude from C instead of B** - Vector \(\overrightarrow{AB} = B - A = (5-2,4-1,3-0) = (3,3,3)\). - Altitude from \(C\) is perpendicular to \(\overrightarrow{AB}\). - Find \(\vec{d}_C = (x,y,z)\) such that \(3x + 3y + 3z = 0 \Rightarrow x + y + z = 0\). - Choose \(x=1, y=1\), then \(z = -2\). - So \(\vec{d}_C = (1,1,-2)\). - Equation of altitude from \(C\): $$\vec{r}_C = (1,6,2) + u(1,1,-2)$$ 10. **Step 8: Find intersection of altitudes from A and C** - Equate \(\vec{r}_A = \vec{r}_C\): \[ 2 + t = 1 + u \\ 1 + 3t = 6 + u \\ 0 + 2t = 2 - 2u \] - From first: \(u = 1 + u = 2 + t \Rightarrow u = 1 + t\). - Substitute into second: \[1 + 3t = 6 + (1 + t) \Rightarrow 1 + 3t = 7 + t \Rightarrow 2t = 6 \Rightarrow t = 3\). - Then \(u = 1 + 3 = 4\). - Check third: \[2t = 2 - 2u \Rightarrow 2(3) = 2 - 2(4) \Rightarrow 6 = 2 - 8 \Rightarrow 6 = -6\) false. 11. **Step 9: Use vector cross product to find orthocenter** - The orthocenter \(H\) satisfies: \[ (\overrightarrow{H} - A) \cdot \overrightarrow{BC} = 0 \\ (\overrightarrow{H} - B) \cdot \overrightarrow{AC} = 0 \] - Let \(H = (x,y,z)\). - From \(A\): \[(x-2)(-4) + (y-1)(2) + (z-0)(-1) = 0 \Rightarrow -4x + 8 + 2y - 2 - z = 0 \Rightarrow -4x + 2y - z + 6 = 0\] - From \(B\): \[(x-5)(-1) + (y-4)(5) + (z-3)(2) = 0 \Rightarrow -x + 5 + 5y - 20 + 2z - 6 = 0 \Rightarrow -x + 5y + 2z - 21 = 0\] - From \(C\): \[(x-1)(3) + (y-6)(3) + (z-2)(3) = 0 \Rightarrow 3x - 3 + 3y - 18 + 3z - 6 = 0 \Rightarrow 3x + 3y + 3z - 27 = 0\] 12. **Step 10: Solve system of equations:** \[ -4x + 2y - z = -6 \\ -x + 5y + 2z = 21 \\ 3x + 3y + 3z = 27 \] - Multiply second by 3: \[-3x + 15y + 6z = 63\] - Add to third: \[(3x + 3y + 3z) + (-3x + 15y + 6z) = 27 + 63 \Rightarrow 18y + 9z = 90\] - Divide by 9: \[2y + z = 10 \Rightarrow z = 10 - 2y\] - Substitute \(z\) into first: \[-4x + 2y - (10 - 2y) = -6 \Rightarrow -4x + 2y - 10 + 2y = -6 \Rightarrow -4x + 4y = 4 \Rightarrow -x + y = 1\] - Substitute \(z\) into second: \[-x + 5y + 2(10 - 2y) = 21 \Rightarrow -x + 5y + 20 - 4y = 21 \Rightarrow -x + y = 1\] - Both give \(-x + y = 1\). - From \(-x + y = 1\), we get \(y = x + 1\). - Substitute \(y\) and \(z\) into third: \[3x + 3(x+1) + 3(10 - 2(x+1)) = 27 \Rightarrow 3x + 3x + 3 + 30 - 6x - 6 = 27 \Rightarrow 6x + 33 - 6x - 6 = 27 \Rightarrow 27 = 27\] (true) - Choose \(x = 0\) for simplicity: \[y = 0 + 1 = 1\] \[z = 10 - 2(1) = 8\] 13. **Final answer:** The orthocenter is \(\boxed{(0,1,8)}\).