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JEE Mathematics 2025 Question with Solution

Let A be the point of intersection of the lines L1:x−71=y−50=z−3−1andL2:x−13=y+34=z+75L_1 : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{-1} \quad \text{and} \quad L_2 : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5} Let B and C be the points on the lines L1L_1 and L2L_2, respectively, such that AB=AC=15AB = AC = \sqrt{15}. Then the square of the area of the triangle ABC is:

  • A

    5454

  • B

    6363

  • C

    5757

  • D

    6060

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The lines are

L1:x−71=y−50=z−3−1L_1: \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{-1}

and

L2:x−13=y+34=z+75L_2: \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5}

Also, AB=AC=15AB = AC = \sqrt{15}.

Find: The square of the area of triangle ABC.

For L1L_1, since y−50\frac{y-5}{0} implies y=5y=5, its parametric form is

x=7+t,y=5,z=3−tx = 7 + t, \quad y = 5, \quad z = 3 - t

For L2L_2, the parametric form is

x=1+3s,y=−3+4s,z=−7+5sx = 1 + 3s, \quad y = -3 + 4s, \quad z = -7 + 5s

To find the point of intersection AA, use the yy-coordinate:

−3+4s=5-3 + 4s = 5

So,

4s=84s = 8 s=2s = 2

Substituting s=2s=2 in L2L_2,

x=1+3(2)=7,y=−3+4(2)=5,z=−7+5(2)=3x = 1 + 3(2) = 7, \quad y = -3 + 4(2) = 5, \quad z = -7 + 5(2) = 3

Hence,

A=(7,5,3)A = (7,5,3)

The direction vectors of the two lines are

d1⃗=(1,0,−1),d2⃗=(3,4,5)\vec{d_1} = (1,0,-1), \quad \vec{d_2} = (3,4,5)

Their magnitudes are

∣d1⃗∣=12+02+(−1)2=2|\vec{d_1}| = \sqrt{1^2+0^2+(-1)^2} = \sqrt{2} ∣d2⃗∣=32+42+52=50|\vec{d_2}| = \sqrt{3^2+4^2+5^2} = \sqrt{50}

The cosine of the angle θ\theta between them is

cos⁡θ=∣d1⃗⋅d2⃗∣d1⃗∣ ∣d2⃗∣∣=∣3+0−52×50∣=15\cos\theta = \left|\frac{\vec{d_1}\cdot\vec{d_2}}{|\vec{d_1}|\,|\vec{d_2}|}\right| = \left|\frac{3+0-5}{\sqrt{2}\times\sqrt{50}}\right| = \frac{1}{5}

Now,

sin⁡2θ=1−cos⁡2θ=1−125=2425\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{1}{25} = \frac{24}{25}

Therefore,

sin⁡θ=265\sin\theta = \frac{2\sqrt{6}}{5}

The area of triangle ABCABC is

Area=12×AB×AC×sin⁡θ\text{Area} = \frac{1}{2} \times AB \times AC \times \sin\theta

Substituting AB=AC=15AB = AC = \sqrt{15},

Area=12×15×15×265\text{Area} = \frac{1}{2} \times \sqrt{15} \times \sqrt{15} \times \frac{2\sqrt{6}}{5} Area=36\text{Area} = 3\sqrt{6}

So,

(Area)2=(36)2=54(\text{Area})^2 = (3\sqrt{6})^2 = 54

Therefore, the square of the area of the triangle is 5454. The correct option is A.

Use equal sides and angle between the lines

Given: AB=AC=15AB = AC = \sqrt{15} and points BB and CC lie on lines L1L_1 and L2L_2 through the common point AA.

Find: The square of the area of triangle ABC.

Since AA is the intersection point, vectors AB⃗\vec{AB} and AC⃗\vec{AC} are along the direction vectors of L1L_1 and L2L_2. So the included angle of triangle ABCABC is the angle between

(1,0,−1)and(3,4,5)(1,0,-1) \quad \text{and} \quad (3,4,5)

Compute

cos⁡θ=∣3+0−52×50∣=15\cos\theta = \left|\frac{3+0-5}{\sqrt{2}\times\sqrt{50}}\right| = \frac{1}{5}

Hence,

sin⁡θ=265\sin\theta = \frac{2\sqrt{6}}{5}

Now directly use

Area=12absin⁡θ\text{Area} = \frac{1}{2}ab\sin\theta

with a=b=15a=b=\sqrt{15}:

Area=12×15×15×265=36\text{Area} = \frac{1}{2}\times\sqrt{15}\times\sqrt{15}\times\frac{2\sqrt{6}}{5} = 3\sqrt{6}

Therefore,

(Area)2=54(\text{Area})^2 = 54

Thus, the correct option is A.

Common mistakes

  • Treating the condition as AB−AC=15AB - AC = \sqrt{15} is incorrect here because the solution working uses AB=AC=15AB = AC = \sqrt{15}. Use the relation supported by the solution before applying the area formula.

  • Using the coordinates of arbitrary points on the two lines without noticing that AA is the common intersection point is wrong. First find AA, then recognize that AB⃗\vec{AB} and AC⃗\vec{AC} lie along the direction vectors of the two lines.

  • Computing the angle incorrectly by using point coordinates instead of direction vectors leads to the wrong result. Use the direction vectors (1,0,−1)(1,0,-1) and (3,4,5)(3,4,5) for the angle between the lines.

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