If the image of the point P(1,0,3) in the line joining the points A(4,7,1) and B(3,5,3) is Q(α, β, γ) , then α+β+γ is equal to

Correct answer: B. (46)/(3) aligned & P(1,0,3) & A(4,7,1), B(3,5,3) & Line AB x-3 1 = y-5 2 = z-3 -2 =λ aligned Let foot of perpendicular of P on AB be aligned & R (λ+3,2 λ+5,-2 λ+3) & (λ+3-1)(1)+(2 λ+5-0)(2)+(-2 λ+3-3) & (-2)=0 & λ+2+4 λ+10+4 λ=0 & λ=-(4)/(3) & R ((5)/(3), (7)/(3), (17)/(3)) & Q ((10)/(3)-1, (14)/(3)-0, (34)/(3)-3) ((7)/(3), (14)/(3), (25)/(3)) & α+β+γ=(7+14+25)/(3)=(46)/(3) aligned

JEE Main 20252 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Three Dimensional Geometry

If the image of the point P(1,0,3) in the line joining the points...

If the image of the point

\mathrm{P}(1,0,3)

in the line joining the points

\mathrm{A}(4,7,1)

and

\mathrm{B}(3,5,3)

\mathrm{Q}(\alpha, \beta, \gamma)

, then

\alpha+\beta+\gamma

is equal to

Options

A $\frac{47}{3}$
B $\frac{46}{3}$
C $18$
D $13$

Correct answer

B. $\frac{46}{3}$

Step-by-step solution

\begin{aligned}& \mathrm{P}(1,0,3) \\ & \mathrm{A}(4,7,1), \mathrm{B}(3,5,3) \\ & \text{ Line } \mathrm{AB} \Rightarrow \frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-5}{2}=\frac{\mathrm{z}-3}{-2}=\lambda\end{aligned}

Let foot of perpendicular of P on AB be

\begin{aligned} & \mathrm{R} \equiv(\lambda+3,2 \lambda+5,-2 \lambda+3) \\ & \Rightarrow(\lambda+3-1)(1)+(2 \lambda+5-0)(2)+(-2 \lambda+3-3) \\ & (-2)=0 \\ & \Rightarrow \lambda+2+4 \lambda+10+4 \lambda=0 \\ & \Rightarrow \lambda=-\frac{4}{3} \\ & \Rightarrow \mathrm{R} \equiv\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right) \\ & \mathrm{Q} \equiv\left(\frac{10}{3}-1, \frac{14}{3}-0, \frac{34}{3}-3\right) \equiv\left(\frac{7}{3}, \frac{14}{3}, \frac{25}{3}\right) \\ & \Rightarrow \alpha+\beta+\gamma=\frac{7+14+25}{3}=\frac{46}{3}\end{aligned}

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