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Friday, 9 October

JEE Main 2025 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 7 April 2025, Shift 2, Maths Q16

Question 16 of 75 in this shift, Maths question 16 of 25, Section A.

If the equation of the line passing through the point (0,−12,0)\left(0,-\frac{1}{2},0\right) and perpendicular to the lines r⃗=λ(i^+aj^+bk^)\vec{r}=\lambda(\hat{i}+a\hat{j}+b\hat{k}) and r⃗=(i^−j^−6k^)+μ(−bi^+aj^+5k^)\vec{r}=(\hat{i}-\hat{j}-6\hat{k})+\mu(-b\hat{i}+a\hat{j}+5\hat{k}) is x−1−2=y+4d=z−c−4\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4}, then a+b+c+da+b+c+d is equal to :
  1. (1)10
  2. (2)12
  3. (3)13
  4. (4)14Official answer

Official answer

Option 4

NTA final key.