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MCQMediumJEE 2025Equation of Line in 3D

JEE Mathematics 2025 Question with Solution

If the equation of the line passing through the point (0,−12,0)\left( 0, -\frac{1}{2}, 0 \right) and perpendicular to the lines r1=λ(i^+aj^+bk^)andr2=(i^−j^−6k^)+μ(−bi^+aj^+5k^)\mathbf{r_1} = \lambda ( \hat{i} + a \hat{j} + b \hat{k}) \quad \text{and} \quad \mathbf{r_2} = ( \hat{i} - \hat{j} - 6 \hat{k} ) + \mu( -b \hat{i} + a \hat{j} + 5 \hat{k}), is x−1−2=y+4d=z−c−4\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}, then a+b+c+da + b + c + d is equal to:

  • A

    1010

  • B

    1414

  • C

    1313

  • D

    1212

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The required line passes through (0,−12,0)\left( 0, -\frac{1}{2}, 0 \right) and is perpendicular to the lines with direction vectors d1⃗=(1,a,b)\vec{d_1} = (1,a,b) and d2⃗=(−b,a,5)\vec{d_2} = (-b,a,5).

Find: a+b+c+da+b+c+d.

Since the required line is perpendicular to both given lines, its direction ratios are parallel to the cross product of their direction ratios.

d1⃗×d2⃗=∣i^j^k^1ab−ba5∣\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & a & b \\ -b & a & 5 \end{vmatrix}
=i^(5a−ab)−j^(5+b2)+k^(a+ab)= \hat{i}(5a-ab) - \hat{j}(5+b^2) + \hat{k}(a+ab)

From x−1−2=y+4d=z−c−4\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}, the direction ratios of the required line are (−2,d,−4)(-2,d,-4), so

(5a−ab,−(5+b2),a+ab)∝(−2,d,−4)(5a-ab, -(5+b^2), a+ab) \propto (-2,d,-4)

Also, the line passes through (0,−12,0)\left(0,-\frac{1}{2},0\right). Substituting this point into

x−1−2=y+4d=z−c−4\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}

gives

0−1−2=−12+4d=0−c−4\frac{0 - 1}{-2} = \frac{-\frac{1}{2} + 4}{d} = \frac{0 - c}{-4}
12=72d=c4\frac{1}{2} = \frac{\frac{7}{2}}{d} = \frac{c}{4}

Hence,

d=7,c=2d = 7, \qquad c = 2

Now compare direction ratios. From the solution working, solving the proportionality conditions gives

a=2,b=3a = 2, \qquad b = 3

Therefore,

a+b+c+d=2+3+2+7=14a+b+c+d = 2+3+2+7 = 14

So the correct option is B.

Note: The solution states option D, but its own final computed value is 1414, which matches option B in the given options.

Using point on the required line first

Given: The line is x−1−2=y+4d=z−c−4\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4} and it passes through (0,−12,0)\left( 0, -\frac{1}{2}, 0 \right).

Find: a+b+c+da+b+c+d.

First use the point condition to determine cc and dd.

−1−2=72d=−c−4\frac{-1}{-2} = \frac{\frac{7}{2}}{d} = \frac{-c}{-4}
12=72d=c4\frac{1}{2} = \frac{\frac{7}{2}}{d} = \frac{c}{4}

So,

d=7,c=2d = 7, \qquad c = 2

The required line is perpendicular to both given lines, so its direction ratios (−2,7,−4)(-2,7,-4) must be perpendicular to (1,a,b)(1,a,b) and (−b,a,5)(-b,a,5).

Hence,

(−2,7,−4)⋅(1,a,b)=0(-2,7,-4) \cdot (1,a,b) = 0 −2+7a−4b=0-2 + 7a - 4b = 0

And,

(−2,7,−4)⋅(−b,a,5)=0(-2,7,-4) \cdot (-b,a,5) = 0 2b+7a−20=02b + 7a - 20 = 0

Solve the system:

7a−4b=27a - 4b = 2 7a+2b=207a + 2b = 20

Subtracting,

6b=18⇒b=36b = 18 \Rightarrow b = 3

Then,

7a+6=20⇒7a=14⇒a=27a + 6 = 20 \Rightarrow 7a = 14 \Rightarrow a = 2

Therefore,

a+b+c+d=2+3+2+7=14a+b+c+d = 2+3+2+7 = 14

So the correct option is B.

Common mistakes

  • Using the direction vectors of the two given lines incorrectly. The required line is perpendicular to both, so its direction ratios must be orthogonal to both given direction vectors or parallel to their cross product. Do not equate them directly.

  • Forgetting to use the given point (0,−12,0)\left(0,-\frac{1}{2},0\right) in the symmetric equation of the required line. This point is necessary to find cc and dd before evaluating the final sum.

  • Misreading the options because the solution labels the correct option inconsistently. The computed value in the working is 1414, so the answer must be the option containing 1414, which is B in the provided options.

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