Let a line L be perpendicular to both the lines L1: (x+1)/3 = (y+3)/5 = (z+5)/7 and L2: (x-2)/1 = (y-4)/4 = (z-6)/7. If theta is the acute angle between the lines L and L3: (x-8/7)/2 = (y-4/7)/1 = z/2, then tan theta is equal to:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q16 – JEE-MAIN 2026

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Home Maths 3D GeometryQuestion #17

Let a line L be perpendicular to both the lines

L_{1} : \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}

and

L_{2} : \frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}

. If

θ

is the acute angle between the lines L and

L_{3} : \frac{x - \frac{8}{7}}{2} = \frac{y - \frac{4}{7}}{1} = \frac{z}{2}

, then

\tan θ

is equal to:

Clue 1 of 3: Direction Vector of Perpendicular Line

The direction vector of a line perpendicular to two given lines is found by taking the cross product of their direction vectors.

🧠 Full Solution Path

Step 1: Determine the direction vector of line L✦ Active

The direction vectors of $L_{1}$ and $L_{2}$ are $\vec{d_{1}} = (3, 5, 7)$ and $\vec{d_{2}} = (1, 4, 7)$ respectively. Since line L is perpendicular to both $L_{1}$ and $L_{2}$ , its direction vector $\vec{d_{L}}$ is parallel to $\vec{d_{1}} \times \vec{d_{2}}$ .

\vec{d_{L}} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{matrix} | = \hat{i} (35 - 28) - \hat{j} (21 - 7) + \hat{k} (12 - 5) = 7 \hat{i} - 14 \hat{j} + 7 \hat{k}

We can use a simplified direction vector $\vec{d_{L}^{'}} = (1, - 2, 1)$ .

Step 2: Calculate the cosine of the angle between L and

L_{3}

○ Expand

The direction vector of $L_{3}$ is $\vec{d_{3}} = (2, 1, 2)$ . The dot product $\vec{d_{L}^{'}} \cdot \vec{d_{3}}$ is:

\vec{d_{L}^{'}} \cdot \vec{d_{3}} = (1) (2) + (- 2) (1) + (1) (2) = 2 - 2 + 2 = 2

The magnitudes of the direction vectors are:

| | \vec{d_{L}^{'}} | | = \sqrt{1^{2} + (- 2)^{2} + 1^{2}} = \sqrt{1 + 4 + 1} = \sqrt{6}

| | \vec{d_{3}} | | = \sqrt{2^{2} + 1^{2} + 2^{2}} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

The cosine of the acute angle $θ$ is:

\cos θ = \frac{| \vec{d_{L}^{'}} \cdot \vec{d_{3}} |}{| \vec{d_{L}^{'}} | | \cdot | | \vec{d_{3}} | |} = \frac{| 2 |}{\sqrt{6} \cdot 3} = \frac{2}{3 \sqrt{6}}

💡 Teacher's Secret Hint

Remember to use the absolute value of the dot product for the acute angle.

Step 3: Determine

\tan θ

○ Expand

Using the identity $\sin^{2} θ = 1 - \cos^{2} θ$ :

\sin^{2} θ = 1 - {(\frac{2}{3 \sqrt{6}})}^{2} = 1 - \frac{4}{54} = 1 - \frac{2}{27} = \frac{25}{27}

Since $θ$ is acute, $\sin θ = \sqrt{\frac{25}{27}} = \frac{5}{\sqrt{27}} = \frac{5}{3 \sqrt{3}}$ . Now, calculate $\tan θ$ :

\tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{5}{3 \sqrt{3}}}{\frac{2}{3 \sqrt{6}}} = \frac{5}{3 \sqrt{3}} \times \frac{3 \sqrt{6}}{2} = \frac{5 \sqrt{6}}{2 \sqrt{3}} = \frac{5 \sqrt{2 \times 3}}{2 \sqrt{3}} = \frac{5 \sqrt{2}}{2}

💡 Teacher's Secret Hint

Simplify the radical expressions carefully to match the options.

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Maths Question 16 – JEE-MAIN 2026

Clue 1 of 3: Direction Vector of Perpendicular Line

🧠 Full Solution Path