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Maths Question 14 – JEE-MAIN 2025

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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 :

The image of a point in a line is found by first determining the foot of the perpendicular from the point to the line.

Step 1: Find the equation of the line and a general point on it.✦ Active

The direction vector of the line joining A(4,7,1) and B(3,5,3) is d→=B−A=(3−4,5−7,3−1)=(−1,−2,2). The parametric equation of the line is L:(x,y,z)=(4−λ,7−2λ,1+2λ). Let M(4−λ,7−2λ,1+2λ) be a general point on the line.

Step 2: Determine the foot of the perpendicular from P to the line.○ Expand

The vector PM→ from P(1,0,3) to M is (4−λ−1,7−2λ−0,1+2λ−3)=(3−λ,7−2λ,2λ−2). Since PM→ is perpendicular to the line, their dot product is zero: PM→⋅d→=0.

(−1)(3−λ)+(−2)(7−2λ)+(2)(2λ−2)=0 −3+λ−14+4λ+4λ−4=0 9λ−21=0⇒λ=219=73

Substituting λ=73 into M's coordinates gives the foot of the perpendicular M(4−73,7−2(73),1+2(73))=M(53,73,173).

Step 3: Calculate the image Q and the required sum.○ Expand

The foot of the perpendicular M is the midpoint of P(1,0,3) and its image Q(α,β,γ). Using the midpoint formula:

1+α2=53⇒3+3α=10⇒α=73 0+β2=73⇒3β=14⇒β=143 3+γ2=173⇒9+3γ=34⇒γ=253

The image is Q(73,143,253). Therefore, the sum α+β+γ is:

α+β+γ=73+143+253=7+14+253=463
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