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Mathematics Question on 3D Geometry

Let (α,β,γ)(\alpha, \beta, \gamma) be the point (8,5,7)(8, 5, 7) in the line x−12=y+13=z−25\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}. Then α+β+γ\alpha + \beta + \gamma is equal to

A

16

B

18

C

14

D

20

Answer

14

Explanation

Solution

Let P(8,5,7)P(8, 5, 7), a⃗=⟨2,3,5⟩\vec{a} = \langle 2, 3, 5 \rangle, F(2λ+1,3λ−1,5λ+2)F(2\lambda + 1, 3\lambda - 1, 5\lambda + 2).

The vector PF→\overrightarrow{PF} is:

PF→=⟨2λ−7,3λ−6,5λ−5⟩.\overrightarrow{PF} = \langle 2\lambda - 7, 3\lambda - 6, 5\lambda - 5 \rangle.

Since PF→⋅a⃗=0\overrightarrow{PF} \cdot \vec{a} = 0, we have:

(2λ−7)(2)+(3λ−6)(3)+(5λ−5)(5)=0.(2\lambda - 7)(2) + (3\lambda - 6)(3) + (5\lambda - 5)(5) = 0.

Simplify:

4λ−14+9λ−18+25λ−25=0,4\lambda - 14 + 9\lambda - 18 + 25\lambda - 25 = 0, 38λ−57=0  ⟹  λ=32.38\lambda - 57 = 0 \implies \lambda = \frac{3}{2}.

The coordinates of FF are:

F=(2λ+1,3λ−1,5λ+2)=(4,72,192).F = (2\lambda + 1, 3\lambda - 1, 5\lambda + 2) = \left( 4, \frac{7}{2}, \frac{19}{2} \right).

The coordinates of Q(α,β,γ)Q(\alpha, \beta, \gamma) are:

α+82=2λ+1,β+52=3λ−1,γ+72=5λ+2.\frac{\alpha + 8}{2} = 2\lambda + 1, \quad \frac{\beta + 5}{2} = 3\lambda - 1, \quad \frac{\gamma + 7}{2} = 5\lambda + 2.

Solving each equation:

α=4λ−6,β=6λ−7,γ=10λ−3.\alpha = 4\lambda - 6, \quad \beta = 6\lambda - 7, \quad \gamma = 10\lambda - 3.

Substitute λ=32\lambda = \frac{3}{2}:

α=4(32)−6=6−6=0,\alpha = 4\left(\frac{3}{2}\right) - 6 = 6 - 6 = 0, β=6(32)−7=9−7=2,\beta = 6\left(\frac{3}{2}\right) - 7 = 9 - 7 = 2, γ=10(32)−3=15−3=12.\gamma = 10\left(\frac{3}{2}\right) - 3 = 15 - 3 = 12.

Finally, compute α+β+γ\alpha + \beta + \gamma:

α+β+γ=0+2+12=14.\alpha + \beta + \gamma = 0 + 2 + 12 = 14.

Final Answer: 14