Let the position vectors of three vertices of a triangle be 4p+q−3r, −5p+q+2r, and 2p−q+2r. If the position vectors of the orthocenter and the circumcenter of the triangle are 2p+q+r and αp+βq+γr respectively, then α+2β+5γ is equal to:
Get the complete, step-by-step math solution for: "Let the position vectors of three vertices of a triangle be 4 {p} + {q} - 3 {r}, -5 {p} + {q} + 2 {r}, and 2 {p} - {q} + 2 {r}. If the position vector...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Define position vectors of vertices, orthocenter, and circumcenter
We are given the position vectors of the three vertices of the triangle, denoted as A, B, and C. We are also given the position vector of the orthocenter, H, and the circumcenter, O. We need to find the values of α, β, and γ from the circumcenter's position vector.
Step 2: Apply Euler's Line Theorem
For any triangle, the orthocenter (H), centroid (G), and circumcenter (O) are collinear, and the centroid divides the segment HO in the ratio 2:1. This relationship is known as Euler's Line Theorem, which states that H+2O=3G. Since the centroid G=3A+B+C, we can substitute this into the Euler's Line equation to get H+2O=A+B+C.
Step 3: Substitute known vectors into Euler's Line equation
Now, we substitute the given position vectors for A, B, C, H, and O into the Euler's Line equation. This will allow us to form an equation in terms of p, q, and r.
Step 4: Simplify and equate coefficients
We simplify both sides of the equation by collecting the coefficients of p, q, and r. On the right side, (4−5+2)p=p, (1+1−1)q=q, and (−3+2+2)r=r. Since p, q, and r are non-coplanar vectors, their coefficients on both sides of the equation must be equal.
Step 5: Solve for α, β, and γ
Equating the coefficients of p, q, and r from the simplified equation, we solve for α, β, and γ. In this case, all three coefficients turn out to be 41.
Step 6: Calculate α+2β+5γ
Finally, we substitute the calculated values of α, β, and γ into the expression α+2β+5γ to find the final answer.