Vectors & 3D GeometrymediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Foot of Perpendicular from Origin on a Line: 34(a+b+c) = 100 | JEE 2026

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

Question
Let a line LL passing through the point (1,1,1)(1,1,1) be perpendicular to both the vectors 2i^+2j^+k^2\hat i+2\hat j+\hat k and i^+2j^+k^\hat i+2\hat j+\hat k. If P(a,b,c)P(a,b,c) is the foot of perpendicular from the origin on the line LL, then the value of 34(a+b+c)34(a+b+c) is
A5050
B8080
C100100correct
D120120
Solution
Step 1: The direction of LL is the cross product:
∣i^j^k^221121∣=i^(2−2)−j^(2−1)+k^(4−2)=−2i^+3j^−2k^.\begin{vmatrix}\hat i&\hat j&\hat k\\2&2&1\\1&2&1\end{vmatrix}=\hat i(2-2)-\hat j(2-1)+\hat k(4-2)=-2\hat i+3\hat j-2\hat k.
 Using direction (2,−3,2)(2,-3,2) (a scalar multiple), the line is
x−12=y−1−3=z−12=λ.\frac{x-1}{2}=\frac{y-1}{-3}=\frac{z-1}{2}=\lambda.
 Step 2: A general point (the foot) is P=(2λ+1, −3λ+1, 2λ+1)P=(2\lambda+1,\ -3\lambda+1,\ 2\lambda+1). Step 3: OP→\overrightarrow{OP} is perpendicular to the line's direction (2,−3,2)(2,-3,2):
2(2λ+1)−3(−3λ+1)+2(2λ+1)=02(2\lambda+1)-3(-3\lambda+1)+2(2\lambda+1)=0
⇒ 4λ+2+9λ−3+4λ+2=0 ⇒ 17λ+1=0 ⇒ λ=−117.\Rightarrow\ 4\lambda+2+9\lambda-3+4\lambda+2=0\ \Rightarrow\ 17\lambda+1=0\ \Rightarrow\ \lambda=-\frac1{17}.
 Step 4: So
a=−217+1,b=317+1,c=−217+1,a+b+c=3+−2+3−217=3−117=5017.a=\frac{-2}{17}+1,\quad b=\frac{3}{17}+1,\quad c=\frac{-2}{17}+1,\qquad a+b+c=3+\frac{-2+3-2}{17}=3-\frac{1}{17}=\frac{50}{17}.
 Step 5:
34(a+b+c)=34⋅5017=2⋅50=100.34(a+b+c)=34\cdot\frac{50}{17}=2\cdot50=100.
 Correct answer: (3)
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