JEE Main 2025 (Online) 7th April Evening ShiftQuestion 1
If the equation of the line passing through the point $\left(0,-\frac{1}{2},0\right)$ and perpendicular to the lines $\vec{r}=\lambda\left(\hat{i}+a\hat{j}+b\hat{k}\right)$ and $\vec{r}=\left(\hat{i}-\hat{j}-6\hat{k}\right)+\mu\left(-b\hat{i}+a\hat{j}+5\hat{k}\right)$ is $\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4}$, then $a+b+c+d$ is equal to :
3D Geometry
Solution
Direction of required line $\parallel(1,a,b)\times(-b,a,5)=\big(a(5-b),\,-(5+b^2),\,a(1+b)\big)\parallel(-2,d,-4)$. Line passes through $(0,-\frac12,0)$ and has form $\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4}$: $x=0$ gives parameter $\frac12$, so $y=-\frac12\Rightarrow d=7$, and $\frac{-c}{-4}=\frac12\Rightarrow c=2$. Ratio of $x,z$ components: $\frac{5-b}{1+b}=\frac12\Rightarrow b=3$. Then $(2a,-14,4a)=\lambda(-2,7,-4)$: $\lambda=-2$, $a=2$. $a+b+c+d=2+3+2+7=14$. Option (D).