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JEE Main 2023
3D Geometry
3D Geometry
Medium

Question

Let the line of the shortest distance between the lines L1:r→=(i^+2j^+3k^)+λ(i^−j^+k^) and L2:r→=(4i^+5j^+6k^)+μ(i^+j^−k^)\begin{aligned} & \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\\\ & \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k}) \end{aligned} intersect L1\mathrm{L}_1 and L2\mathrm{L}_2 at P\mathrm{P} and Q\mathrm{Q} respectively. If (α,β,γ)(\alpha, \beta, \gamma) is the mid point of the line segment PQ\mathrm{PQ}, then 2(α+β+γ)2(\alpha+\beta+\gamma) is equal to ____________.

Answer: 1

Solution

Key Concepts and Formulas

  1. Skew Lines and Shortest Distance: When two lines in 3D space are skew (non-parallel and non-intersecting), there exists a unique line segment that represents the shortest distance between them. A fundamental property of this line segment is that it is perpendicular to both of the given lines. This means the vector representing the shortest distance is orthogonal to the direction vectors of both skew lines.
  2. Orthogonality Condition: Given two lines L1:r⃗=a⃗1+λb⃗1L_1: \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 and L2:r⃗=a⃗2+μb⃗2L_2: \vec{r} = \vec{a}_2 + \mu \vec{b}_2, let P be a point on L1L_1 and Q be a point on L2L_2. If PQ⃗\vec{PQ} is the vector representing the shortest distance, it must satisfy:
    • PQ⃗⋅b⃗1=0\vec{PQ} \cdot \vec{b}_1 = 0 (Perpendicular to L1L_1)
    • PQ⃗⋅b⃗2=0\vec{PQ} \cdot \vec{b}_2 = 0 (Perpendicular to L2L_2)
  3. Midpoint Formula: The midpoint M of a line segment connecting two points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2) is given by M(x1+x22,y1+y22,z1+z22)M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right).

Step-by-Step Solution

Step 1: Represent the lines and general points. We are given the lines: L1:r⃗=(i^+2j^+3k^)+λ(i^−j^+k^)L_1: \vec{r} = (\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) L2:r⃗=(4i^+5j^+6k^)+μ(i^+j^−k^)L_2: \vec{r} = (4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})

From L1L_1, we identify its fixed point vector a⃗1=i^+2j^+3k^\vec{a}_1 = \hat{i}+2 \hat{j}+3 \hat{k} and its direction vector b⃗1=i^−j^+k^\vec{b}_1 = \hat{i}-\hat{j}+\hat{k}. A general point P on L1L_1 can be expressed in coordinates as P(1+λ,2−λ,3+λ)P(1+\lambda, 2-\lambda, 3+\lambda).

From L2L_2, we identify its fixed point vector a⃗2=4i^+5j^+6k^\vec{a}_2 = 4 \hat{i}+5 \hat{j}+6 \hat{k} and its direction vector b⃗2=i^+j^−k^\vec{b}_2 = \hat{i}+\hat{j}-\hat{k}. A general point Q on L2L_2 can be expressed in coordinates as Q(4+μ,5+μ,6−μ)Q(4+\mu, 5+\mu, 6-\mu).

Step 2: Form the vector PQ⃗\vec{PQ}. The vector PQ⃗\vec{PQ} connects point P on L1L_1 to point Q on L2L_2. We find this vector by subtracting the position vector of P from the position vector of Q: PQ⃗=((4+μ)−(1+λ))i^+((5+μ)−(2−λ))j^+((6−μ)−(3+λ))k^\vec{PQ} = ( (4+\mu) - (1+\lambda) )\hat{i} + ( (5+\mu) - (2-\lambda) )\hat{j} + ( (6-\mu) - (3+\lambda) )\hat{k} Simplifying the components, we get: PQ⃗=(3+μ−λ)i^+(3+μ+λ)j^+(3−μ−λ)k^\vec{PQ} = (3+\mu-\lambda)\hat{i} + (3+\mu+\lambda)\hat{j} + (3-\mu-\lambda)\hat{k}

Step 3: Apply the orthogonality conditions. For PQ⃗\vec{PQ} to be the line segment of shortest distance, it must be perpendicular to both direction vectors b⃗1\vec{b}_1 and b⃗2\vec{b}_2.

Condition 1: PQ⃗⋅b⃗1=0\vec{PQ} \cdot \vec{b}_1 = 0 The direction vector b⃗1=(1,−1,1)\vec{b}_1 = (1, -1, 1). (3+μ−λ)(1)+(3+μ+λ)(−1)+(3−μ−λ)(1)=0(3+\mu-\lambda)(1) + (3+\mu+\lambda)(-1) + (3-\mu-\lambda)(1) = 0 3+μ−λ−3−μ−λ+3−μ−λ=03+\mu-\lambda - 3-\mu-\lambda + 3-\mu-\lambda = 0 3−3λ−μ=0⇒μ=3−3λ(Equation 1)3 - 3\lambda - \mu = 0 \quad \Rightarrow \quad \mu = 3 - 3\lambda \quad \text{(Equation 1)}

Condition 2: PQ⃗⋅b⃗2=0\vec{PQ} \cdot \vec{b}_2 = 0 The direction vector b⃗2=(1,1,−1)\vec{b}_2 = (1, 1, -1). (3+μ−λ)(1)+(3+μ+λ)(1)+(3−μ−λ)(−1)=0(3+\mu-\lambda)(1) + (3+\mu+\lambda)(1) + (3-\mu-\lambda)(-1) = 0 3+μ−λ+3+μ+λ−(3−μ−λ)=03+\mu-\lambda + 3+\mu+\lambda - (3-\mu-\lambda) = 0 3+μ−λ+3+μ+λ−3+μ+λ=03+\mu-\lambda + 3+\mu+\lambda - 3+\mu+\lambda = 0 3+3μ+λ=0(Equation 2)3 + 3\mu + \lambda = 0 \quad \text{(Equation 2)}

Step 4: Solve the system of linear equations for λ\lambda and μ\mu. We have the system of equations:

  1. μ=3−3λ\mu = 3 - 3\lambda
  2. λ+3μ=−3\lambda + 3\mu = -3

Substitute Equation 1 into Equation 2: λ+3(3−3λ)=−3\lambda + 3(3 - 3\lambda) = -3 λ+9−9λ=−3\lambda + 9 - 9\lambda = -3 −8λ=−12-8\lambda = -12 λ=−12−8=32\lambda = \frac{-12}{-8} = \frac{3}{2}

Now, substitute the value of λ\lambda back into Equation 1 to find μ\mu: μ=3−3(32)=3−92=6−92=−32\mu = 3 - 3\left(\frac{3}{2}\right) = 3 - \frac{9}{2} = \frac{6-9}{2} = -\frac{3}{2} So, the parameters for the points P and Q are λ=32\lambda = \frac{3}{2} and μ=−32\mu = -\frac{3}{2}.

Step 5: Determine the coordinates of points P and Q. Using λ=32\lambda = \frac{3}{2} for point P: P(1+32,2−32,3+32)=P(52,12,92)P\left(1+\frac{3}{2}, 2-\frac{3}{2}, 3+\frac{3}{2}\right) = P\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right) Using μ=−32\mu = -\frac{3}{2} for point Q: Q(4+(−32),5+(−32),6−(−32))=Q(8−32,10−32,12+32)=Q(52,72,152)Q\left(4+\left(-\frac{3}{2}\right), 5+\left(-\frac{3}{2}\right), 6-\left(-\frac{3}{2}\right)\right) = Q\left(\frac{8-3}{2}, \frac{10-3}{2}, \frac{12+3}{2}\right) = Q\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right)

Step 6: Calculate the midpoint (α,β,γ)(\alpha, \beta, \gamma) of the line segment PQ. Using the midpoint formula for P(52,12,92)P\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right) and Q(52,72,152)Q\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right): α=52+522=1022=52\alpha = \frac{\frac{5}{2} + \frac{5}{2}}{2} = \frac{\frac{10}{2}}{2} = \frac{5}{2} β=12+722=822=42=2\beta = \frac{\frac{1}{2} + \frac{7}{2}}{2} = \frac{\frac{8}{2}}{2} = \frac{4}{2} = 2 γ=92+1522=2422=122=6\gamma = \frac{\frac{9}{2} + \frac{15}{2}}{2} = \frac{\frac{24}{2}}{2} = \frac{12}{2} = 6 So, the midpoint is (α,β,γ)=(52,2,6)(\alpha, \beta, \gamma) = \left(\frac{5}{2}, 2, 6\right).

Step 7: Calculate the required expression 2(α+β+γ)2(\alpha+\beta+\gamma). First, find the sum α+β+γ\alpha+\beta+\gamma: α+β+γ=52+2+6=52+42+122=5+4+122=212\alpha+\beta+\gamma = \frac{5}{2} + 2 + 6 = \frac{5}{2} + \frac{4}{2} + \frac{12}{2} = \frac{5+4+12}{2} = \frac{21}{2} Now, multiply by 2: 2(α+β+γ)=2×212=212(\alpha+\beta+\gamma) = 2 \times \frac{21}{2} = 21

Common Mistakes & Tips

  • Sign Errors: Be meticulous with signs, especially when performing dot products and solving algebraic equations. A small error can propagate and lead to an incorrect final result.
  • Verifying Orthogonality: A good check is to compute the cross product b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2. The vector PQ⃗\vec{PQ} should be parallel to this cross product. For this problem, b⃗1×b⃗2=(i^−j^+k^)×(i^+j^−k^)=2j^+2k^\vec{b}_1 \times \vec{b}_2 = (\hat{i}-\hat{j}+\hat{k}) \times (\hat{i}+\hat{j}-\hat{k}) = 2\hat{j}+2\hat{k}. Our calculated PQ⃗=3j^+3k^\vec{PQ} = 3\hat{j}+3\hat{k}, which is indeed parallel to 2j^+2k^2\hat{j}+2\hat{k}, confirming our values of λ\lambda and μ\mu.
  • Arithmetic with Fractions: Take extra care when adding and dividing fractions to avoid calculation errors, particularly when finding the midpoint coordinates.

Summary

The problem requires finding the coordinates of the midpoint of the line segment that represents the shortest distance between two given skew lines. We achieved this by first parameterizing general points P and Q on each line. Then, we formed the vector PQ⃗\vec{PQ} and used the property that this shortest distance vector must be perpendicular to the direction vectors of both lines. This led to a system of two linear equations in λ\lambda and μ\mu, which were solved to find their unique values. Substituting these values back into the parameterized points P and Q gave their exact coordinates. Finally, the midpoint formula was applied to P and Q to find (α,β,γ)(\alpha, \beta, \gamma), and the expression 2(α+β+γ)2(\alpha+\beta+\gamma) was calculated. The final result of this process is 21.

Final Answer

The final answer is 21\boxed{21}.

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