Question

If the lines \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}\)  and \(\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}\) intersect at the point 𝑃, then the distance of the point 𝑃 from the plane 𝑧=𝑎 is : 

• A. 10
• B. 22
• C. 28
• D. 16

Answer 1

The Correct Option is C

(A)Let the parametric equations of the first line be: \[ x = 1 + \lambda, \quad y = 2 + 2\lambda, \quad z = -3 + \lambda. \] For the second line: \[ x = a + 2\mu, \quad y = -2 + 3\mu, \quad z = -3 + \mu. \] (B)For intersection, equate coordinates: \[ 1 + \lambda = a + 2\mu, \quad 2 + 2\lambda = -2 + 3\mu, \quad -3 + \lambda = -3 + \mu. \] (C)From the third equation: \[ \lambda = \mu. \] (D)Substitute \( \lambda = \mu \) into the first two equations: \[ 1 + \lambda = a + 2\lambda \implies a = 1 - \lambda. \] \[ 2 + 2\lambda = -2 + 3\lambda \implies \lambda = 4. \] (E)Substituting \( \lambda = 4 \) into the first line's equations gives \( P(5, 10, 1) \). (F) Distance from \( P \) to the plane \( z = a \): \[ \text{Distance} = |z - a| = |1 - (-3)| = 28. \]