Vectors & 3D GeometrymediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Vectors & 3D Geometry: Distance Point Image Line Sum Possible Values (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
If the distance of the point (a,2,5)(a,2,5) from the image of (1,2,7)(1,2,7) in the line
x1=y−11=z−22\dfrac{x}{1}=\dfrac{y-1}{1}=\dfrac{z-2}{2}
is 4, then the sum of all possible values of aa is:
A11
B9
C6correct
D4
Solution
Step 1: Find the foot of perpendicular from P(1,2,7)P(1,2,7) to the line A general point on the line is M=(t, t+1, 2t+2)M=(t,\ t+1,\ 2t+2). The direction vector is (1,1,2)(1,1,2).
PM→⋅(1,1,2)=0  ⟹  (t−1)+(t−1)+2(2t−5)=0  ⟹  6t=12  ⟹  t=2\overrightarrow{PM}\cdot(1,1,2) = 0 \implies (t-1)+(t-1)+2(2t-5)=0 \implies 6t=12 \implies t=2
 So M=(2,3,6)M=(2,3,6). Step 2: Compute the image QQ of PP in the line
Q=2M−P=(2⋅2−1, 2⋅3−2, 2⋅6−7)=(3,4,5)Q = 2M-P = (2\cdot2-1,\ 2\cdot3-2,\ 2\cdot6-7) = (3,4,5)
 Step 3: Set up the distance equation
(a−3)2+(2−4)2+(5−5)2=4  ⟹  (a−3)2+4=16  ⟹  (a−3)2=12\sqrt{(a-3)^2+(2-4)^2+(5-5)^2} = 4 \implies (a-3)^2+4 = 16 \implies (a-3)^2 = 12
 This gives a2−6a+9=12a^2-6a+9=12, i.e., a2−6a−3=0a^2-6a-3=0. Step 4: Apply Vieta's formula By Vieta's formulas, the sum of the two roots of a2−6a−3=0a^2-6a-3=0 is 66. Answer: (3)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Vectors & 3D Geometry · medium
A line with direction ratios 1,−1,21,-1,2 intersects the lines x2=y3=z+13\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z+1}{3} and x+1−1=y−21=z4\dfrac{x+1}{-1}=\dfrac{y-2}{1}=\dfrac{z}{4} at the points PP and QQ respectively. If the length of the line segment PQPQ is α\alpha, then 225α2225\alpha^2 is equal to
Vectors & 3D Geometry · medium
Let OO be the origin with OP→=a⃗\overrightarrow{OP}=\vec{a} and OQ→=b⃗\overrightarrow{OQ}=\vec{b}. If RR is on OP→\overrightarrow{OP} such that OP→=5OR→\overrightarrow{OP}=5\overrightarrow{OR}, and MM is such that OQ→=5RM→\overrightarrow{OQ}=5\overrightarrow{RM}, then PM→\overrightarrow{PM} equals:
Vectors & 3D Geometry · medium
Let △PQR\triangle PQR be such that PP and QQ lie on x+38=y−42=z+12\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2} and are at distance 6 from R(1,2,3)R(1,2,3). If (α,β,γ)(\alpha,\beta,\gamma) is the centroid of △PQR\triangle PQR, then α+β+γ\alpha+\beta+\gamma equals:
Vectors & 3D Geometry · medium
Let the vectors a⃗=−i^+j^+3k^\vec a=-\hat i+\hat j+3\hat k and b⃗=i^+3j^+k^\vec b=\hat i+3\hat j+\hat k. For some λ,μ∈R\lambda,\mu\in\mathbb{R}, let c⃗=λa⃗+μb⃗\vec c=\lambda\vec a+\mu\vec b. If c⃗⋅(3i^−6j^+2k^)=10\vec c\cdot(3\hat i-6\hat j+2\hat k)=10 and c⃗⋅(i^+j^+k^)=−2\vec c\cdot(\hat i+\hat j+\hat k)=-2, then ∣c⃗∣2|\vec c|^2 is equal to

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath