Question

The square of the distance of the point of intersection of the line \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+1}{6}\) and the plane \(2x - y + z = 6\) from the point \((-1, -1, 2)\) is _____________
 

Hint:

Always cross-verify your intersection point by checking if it satisfies both the line and the plane equations.

Answer 1

Correct Answer: 61

Step 1: Understanding the Concept:
To find the intersection, we parameterize the line and substitute the coordinates into the plane equation. Once the point of intersection is found, we use the distance formula.
Step 2: Detailed Explanation:
Let the line be \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+1}{6} = \lambda\).
General point on the line: \(x = 2\lambda + 1, y = 3\lambda + 2, z = 6\lambda - 1\).
Substituting this point into the plane \(2x - y + z = 6\):
\[ 2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6 \]
\[ 4\lambda + 2 - 3\lambda - 2 + 6\lambda - 1 = 6 \implies 7\lambda = 7 \implies \lambda = 1 \]
Intersection point \(Q\):
\[ x = 2(1)+1 = 3, y = 3(1)+2 = 5, z = 6(1)-1 = 5 \implies Q = (3, 5, 5) \]
The fixed point is \(P(-1, -1, 2)\).
The distance \(PQ^2\) is:
\[ d^2 = (3 - (-1))^2 + (5 - (-1))^2 + (5 - 2)^2 \]
\[ d^2 = 4^2 + 6^2 + 3^2 = 16 + 36 + 9 = 61 \]
Step 3: Final Answer:
The square of the distance is 61.