JEE Main 2026 — Three Dimensional Geometry Question with Solution

The direction vector of line $L_1$ is $\overrightarrow{d_1}=3\hat{i}+5\hat{j}+7\hat{k}$.

The direction vector of line $L_2$ is $\overrightarrow{d_2}=\hat{i}+4\hat{j}+7\hat{k}$.

Since line $L$ is perpendicular to both $L_1$ and $L_2$, its direction vector $\vec{d}$ is given by the cross product of $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$:

$\vec{d}=\overrightarrow{d_1}\times \overrightarrow{d_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 5& 7\\ 1& 4& 7\end{array}\right|$

$\vec{d}=\hat{i}(35-28)-\hat{j}(21-7)+\hat{k}(12-5)=7\hat{i}-14\hat{j}+7\hat{k}$

The direction ratios of $L$ are proportional to $1,-2,1$.

The direction vector of line $L_3$ is $\overrightarrow{d_3}=2\hat{i}+\hat{j}+2\hat{k}$.

The acute angle $\theta$ between $L$ and $L_3$ is given by:

$\cos \theta =\frac{|\vec{d}\cdot \overrightarrow{d_3}|}{|\vec{d}||\overrightarrow{d_3}|}$

$\cos \theta =\frac{|(1)(2)+(-2)(1)+(1)(2)|}{\sqrt{1^2+(-2{)}^2+1^2}\sqrt{2^2+1^2+2^2}}$

$\cos \theta =\frac{|2-2+2|}{\sqrt{6}\sqrt{9}}=\frac{2}{3\sqrt{6}}$

From $\cos \theta =\frac{2}{3\sqrt{6}}$, considering a right-angled triangle with adjacent side $2$ and hypotenuse $3\sqrt{6}$:

The opposite side is $\sqrt{(3\sqrt{6}{)}^2-2^2}=\sqrt{54-4}=\sqrt{50}=5\sqrt{2}$.

Therefore, $\tan \theta =\frac{5\sqrt{2}}{2}$.

Answer: $\frac{5}{2}\sqrt{2}$