Question

The line of the shortest distance between the lines \(\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}\) and \(\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}\) makes an angle of cos⁻¹⁡\(\sqrt{\frac{2}{27}}\) with the plane P: ax – y – z = 0, (a> 0). If the image of the point (1, 1, –5) in the plane P is (α, β, γ), then α + β – γ is equal to ________.

Answer 1

Correct Answer: -2.2 - -2.3

To solve the problem, we begin by analyzing the parameters of the given lines and the plane. 

The first line is given by the parametric equations: 
\(x = 2, y = 1 + t, z = t\), represented in the vector form as \(r_1 = \langle 2, 1, 0 \rangle + t \langle 0, 1, 1 \rangle\).\)
The direction vector is \(\langle 0, 1, 1 \rangle\\)

The second line is given by: 
\(x = 3 + 2s, y = 5 + 2s, z = 1 + s\), represented as \(r_2 = \langle 3, 5, 1 \rangle + s \langle 2, 2, 1 \rangle\).\)
The direction vector here is \(\langle 2, 2, 1 \rangle\\)

The direction vector of the line of shortest distance (LOD) is perpendicular to both direction vectors. We find it by taking the cross product: \(d = \langle 0, 1, 1 \rangle \times \langle 2, 2, 1 \rangle = \langle -1, 2, -2 \rangle.\)

The LOD vector \(d = \langle -1, 2, -2 \rangle\) creates an angle \(\cos^{-1} \left(\sqrt{\frac{2}{27}}\right)\) with plane \(P: ax - y - z = 0\)\). The normal to the plane is \(\langle a, -1, -1 \rangle\\)

The cosine of the angle between \(\langle -1, 2, -2 \rangle\) and \(\langle a, -1, -1 \rangle\) is calculated: 
\(\cos \theta = \frac{|-1 \cdot a + 2 \cdot (-1) + (-2) \cdot (-1)|}{\sqrt{(-1)^2 + 2^2 + (-2)^2} \cdot \sqrt{a^2 + 1^2 + 1^2}}\).
The dot product is \(-a - 2 + 2 = -a\).

Equating the given cosine value: 
\(\frac{|a|}{3\sqrt{a^2 + 2}} = \sqrt{\frac{2}{27}}\)
Squaring both sides and solving for \(a\), 
\(\frac{a^2}{9(a^2 + 2)} = \frac{2}{27}\)
Cross-multiplying yields: 
\(3a^2 = 2a^2 + 4 \Rightarrow a^2 = 4 \Rightarrow a = 2\).

The plane becomes \(2x - y - z = 0\). The image of point \(Q(1, 1, -5)\) is \(Q'( \alpha, \beta, \gamma)\) given by the projection on the plane formula.

x	y	z
\(1 - \lambda \cdot 2\)	\(1 + \lambda\)	\(-5 + \lambda\)

Normal to the plane equation: \(\lambda = \frac{7}{6}\). Calculating coordinates of \(Q'\), 
\(\alpha = 1 - \frac{7}{3}, \beta = 1 + \frac{7}{6}, \gamma = -5 + \frac{7}{6}\).

The values are \(\alpha = -\frac{4}{3}, \beta = \frac{13}{6}, \gamma = -\frac{23}{6}\).
Then \(\alpha + \beta - \gamma = -\frac{4}{3} + \frac{13}{6} + \frac{23}{6} = -2.25\).

The computed value is -2.25, which lies within the range -2.3 to -2.2.