Question

The plane \(2x−y+z=4\) intersects the line segment joining the points \(A(a,−2,4)\) and \(B(2,b,−3)\) at the point C in the ratio \(2:1\) and the distance of the point C from the origin is \(\sqrt5\). If \(ab<0 \) and \(P\) is the point \((a−b,b,2b−a)\). Then \(CP^2 \) is equal to

• A. \(\frac{16}{3}\)
• B. \(\frac{17}{3}\)
• C. \(\frac{73}{3}\)
• D. \(\frac{97}{3}\)

Answer 1

The Correct Option is B

(A)The coordinates of \( C \) using the section formula: \[ C = \left(\frac{a + 4}{3}, \frac{2b - 2}{3}, \frac{2 - 2}{3}\right). \] (B)Substituting \( C \) into the plane equation \( 2x - y + z = 4 \): \[ 2\left(\frac{a + 4}{3}\right) - \left(\frac{2b - 2}{3}\right) + \left(\frac{2}{3}\right) = 4. \] Simplify: \[ \frac{2a + 8 + 2b - 2 + 2}{3} = 4 \implies 2a + 2b = 4 \implies a + b = 2. \quad (1) \] (C)The distance of \( C \) from the origin: \[ \left(\frac{a + 4}{3}\right)^2 + \left(\frac{2b - 2}{3}\right)^2 + \left(\frac{2}{3}\right)^2 = 5. \] Solve using \( a + b = 2 \) and simplify: \[ (b + 6)^2 + (2b - 2)^2 = 41 \implies 5b^2 + 4b - 1 = 0. \] (D)Roots are \( b = -1 \) or \( b = \frac{1}{5} \). Using \( ab < 0 \), \( (a, b) = (1, -1) \). (E)Coordinates of \( C \): \[ C = \left(\frac{5}{3}, \frac{-4}{3}, \frac{2}{3}\right). \] Coordinates of \( P \): \[ P = (2, -1, -3). \] (F) Compute \( CP^2 \): \[ CP^2 = \left(\frac{5}{3} - 2\right)^2 + \left(\frac{-4}{3} + 1\right)^2 + \left(\frac{2}{3} + 3\right)^2. \] Simplify: \[ CP^2 = \frac{17}{3}. \]