Vector Algebra – 3D Lines – JEE Main 29 Jan 2025 Shift 1

Solution:

The direction ratios of $L_1$ are $\vec{d_1} = (1, -1, 2)$.

The direction ratios of $L_2$ are $\vec{d_2} = (-1, 2, 1)$.

Since $L_3$ is perpendicular to both $L_1$ and $L_2$, its direction vector $\vec{d_3}$ is parallel to $\vec{d_1} \times \vec{d_2}$.

$\vec{d_3} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{matrix} \right| = \hat{i}(-1-4) – \hat{j}(1-(-2)) + \hat{k}(2-1) = -5\hat{i} – 3\hat{j} + \hat{k}$.

The line $L_3$ passes through $(\alpha, \beta, \gamma)$ and has direction $(-5, -3, 1)$.

Equation of $L_3$: $\frac{x-\alpha}{-5} = \frac{y-\beta}{-3} = \frac{z-\gamma}{1} = \lambda$.

Any point $A$ on $L_3$ is $(\alpha-5\lambda, \beta-3\lambda, \gamma+\lambda)$.

Since $L_3$ intersects $L_1$, let the point of intersection be $B$.

Any point $B$ on $L_1$ is $(k+1, -k+2, 2k+1)$.

Equating the coordinates of $A$ and $B$:

1) $\alpha – 5\lambda = k+1 \Rightarrow \alpha = 5\lambda + k + 1$

2) $\beta – 3\lambda = -k+2 \Rightarrow \beta = 3\lambda – k + 2$

3) $\gamma + \lambda = 2k+1 \Rightarrow \gamma = -\lambda + 2k + 1$

We need to find $|5\alpha – 11\beta – 8\gamma|$. Substitute the expressions for $\alpha, \beta, \gamma$:

Expression $= |5(5\lambda + k + 1) – 11(3\lambda – k + 2) – 8(-\lambda + 2k + 1)|$

$= |(25\lambda + 5k + 5) – (33\lambda – 11k + 22) – (-8\lambda + 16k + 8)|$

Group terms by $\lambda$ and $k$:

$ = | \lambda(25 – 33 + 8) + k(5 + 11 – 16) + (5 – 22 – 8) |$

$= | \lambda(0) + k(0) – 25 |$

$= |-25| = 25$.

Ans. (3)