Question

Let the image of the point \(\left(\frac{5}{3},\frac{5}{3},\frac{8}{3}\right)\) in the plane x - 2y + z - 2 = 0 be P. If the distance of the point Q(6,-2,α), α > 0, from P is 13, then α is equal to___.

Answer 1

Correct Answer: 15

The image of the point \( \left( \frac{5}{3}, \frac{5}{3}, \frac{8}{3} \right) \) in the plane \( x - 2y + z - 2 = 0 \) is given as P. We are tasked to find \( \alpha \) given that the distance from point Q(6, -2, a) to point P is 13.

1. Finding the Image Point P:

Given Point: \( A = \left( \frac{5}{3}, \frac{5}{3}, \frac{8}{3} \right) \)

Plane Equation: \( x - 2y + z - 2 = 0 \)

Let the image point P be \( (x, y, z) \).

1. Midpoint: The midpoint of AP lies on the plane.

Midpoint \( M = \left( \frac{x + \frac{5}{3}}{2}, \frac{y + \frac{5}{3}}{2}, \frac{z + \frac{8}{3}}{2} \right) \)

Substitute M into the plane equation:

\( \frac{x + \frac{5}{3}}{2} - 2\left( \frac{y + \frac{5}{3}}{2} \right) + \frac{z + \frac{8}{3}}{2} - 2 = 0 \)

\( x + \frac{5}{3} - 2y - \frac{10}{3} + z + \frac{8}{3} - 4 = 0 \)

\( x - 2y + z - \frac{9}{3} - 4 = 0 \)

\( x - 2y + z - 7 = 0 \)

2. Line AP is Perpendicular to the Plane:

The direction vector of AP is \( \vec{AP} = \left( x - \frac{5}{3}, y - \frac{5}{3}, z - \frac{8}{3} \right) \)

The normal vector of the plane is \( \vec{n} = (1, -2, 1) \)

Since AP is perpendicular to the plane, \( \vec{AP} \) is parallel to \( \vec{n} \). Therefore, \( \vec{AP} = \lambda \vec{n} \) for some scalar \( \lambda \).

\( x - \frac{5}{3} = \lambda, \quad y - \frac{5}{3} = -2\lambda, \quad z - \frac{8}{3} = \lambda \)

\( x = \lambda + \frac{5}{3}, \quad y = -2\lambda + \frac{5}{3}, \quad z = \lambda + \frac{8}{3} \)

3. Solve for x, y, z:

Substitute these into the midpoint equation:

\( (\lambda + \frac{5}{3}) - 2(-2\lambda + \frac{5}{3}) + (\lambda + \frac{8}{3}) - 7 = 0 \)

\( \lambda + \frac{5}{3} + 4\lambda - \frac{10}{3} + \lambda + \frac{8}{3} - 7 = 0 \)

\( 6\lambda + \frac{3}{3} - 7 = 0 \)

\( 6\lambda + 1 - 7 = 0 \)

\( 6\lambda = 6 \)

\( \lambda = 1 \)

Then:

\( x = 1 + \frac{5}{3} = \frac{8}{3}, \quad y = -2 + \frac{5}{3} = -\frac{1}{3}, \quad z = 1 + \frac{8}{3} = \frac{11}{3} \)

So, \( P = \left( \frac{8}{3}, -\frac{1}{3}, \frac{11}{3} \right) \)

2. Calculating alpha:

P = (8/3, -1/3, 11/3)

Q = (6, -2, a)

Distance PQ = 13

\(13^2 = (6-8/3)^2 + (-2+1/3)^2 + (a-11/3)^2\)

\(169 = (10/3)^2 + (-5/3)^2 + (a-11/3)^2\)

\(169 = 100/9 + 25/9 + (a-11/3)^2\)

\(169 = 125/9 + (a-11/3)^2\)

\((a-11/3)^2 = 169 - 125/9 = 1521/9 - 125/9 = 1396/9\)

\(a-11/3 = \pm \sqrt{1396}/3\)

\(a = (11 \pm \sqrt{1396})/3\)

 

Correct final answer:  \(a = (11 \pm \sqrt{1396})/3\)