JEE Challenger
More from Three Dimensional Geometry
MathHardAdv 2020 • Paper 1

Angle Bisector Line of Two Intersecting Lines in 3D Space

Consider two straight lines L1L_1 and L2L_2 defined by: L1:x−11=y−1=z−13andL2:x−1−3=y−1=z−11L_1 : \frac{x-1}{1} = \frac{y}{-1} = \frac{z-1}{3} \quad \text{and} \quad L_2 : \frac{x-1}{-3} = \frac{y}{-1} = \frac{z-1}{1} Suppose a straight line L:x−αl=y−1m=z−γ−2L : \frac{x-\alpha}{l} = \frac{y-1}{m} = \frac{z-\gamma}{-2} is contained in the plane containing L1L_1 and L2L_2, and passes through their point of intersection. If the line LL bisects the acute angle between L1L_1 and L2L_2, which of the following options is/are TRUE?

Options

A

α−γ=3\alpha - \gamma = 3

Correct
B

l+m=2l + m = 2

Correct
C

α−γ=1\alpha - \gamma = 1

D

l+m=0l + m = 0

Topics & Concepts

Three Dimensional GeometryLine in 3DPlane in 3D

Step-by-Step Solution

To determine which of the given options are true, we proceed step-by-step:

Step 1: Find the Point of Intersection of L1L_1 and L2L_2

The equations of the lines are: L1:x−11=y−1=z−13L_1 : \frac{x-1}{1} = \frac{y}{-1} = \frac{z-1}{3} L2:x−1−3=y−1=z−11L_2 : \frac{x-1}{-3} = \frac{y}{-1} = \frac{z-1}{1}

Setting y=0y = 0 in both line equations:

  • For L1L_1: x−11=0  ⟹  x=1\frac{x-1}{1} = 0 \implies x = 1 and z−13=0  ⟹  z=1\frac{z-1}{3} = 0 \implies z = 1.
  • For L2L_2: x−1−3=0  ⟹  x=1\frac{x-1}{-3} = 0 \implies x = 1 and z−11=0  ⟹  z=1\frac{z-1}{1} = 0 \implies z = 1.

Thus, both lines intersect at the point P(1,0,1)P(1, 0, 1). Since line LL passes through the point of intersection of L1L_1 and L2L_2, the point P(1,0,1)P(1, 0, 1) must also lie on line LL.

Step 2: Determine the Direction Vector of the Acute Angle Bisector

The direction vectors of L1L_1 and L2L_2 are: v⃗1=i^−j^+3k^\vec{v}_1 = \hat{i} - \hat{j} + 3\hat{k} v⃗2=−3i^−j^+k^\vec{v}_2 = -3\hat{i} - \hat{j} + \hat{k}

Their magnitudes are: ∣v⃗1∣=12+(−1)2+32=11|\vec{v}_1| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{11} ∣v⃗2∣=(−3)2+(−1)2+12=11|\vec{v}_2| = \sqrt{(-3)^2 + (-1)^2 + 1^2} = \sqrt{11}

Now, compute the dot product v⃗1⋅v⃗2\vec{v}_1 \cdot \vec{v}_2: v⃗1⋅v⃗2=(1)(−3)+(−1)(−1)+(3)(1)=−3+1+3=1\vec{v}_1 \cdot \vec{v}_2 = (1)(-3) + (-1)(-1) + (3)(1) = -3 + 1 + 3 = 1

Since v⃗1⋅v⃗2>0\vec{v}_1 \cdot \vec{v}_2 > 0, the angle between v⃗1\vec{v}_1 and v⃗2\vec{v}_2 is acute. Because ∣v⃗1∣=∣v⃗2∣|\vec{v}_1| = |\vec{v}_2|, the vector along the acute angle bisector is given by: v⃗b=v⃗1+v⃗2=(1−3)i^+(−1−1)j^+(3+1)k^=−2i^−2j^+4k^\vec{v}_b = \vec{v}_1 + \vec{v}_2 = (1 - 3)\hat{i} + (-1 - 1)\hat{j} + (3 + 1)\hat{k} = -2\hat{i} - 2\hat{j} + 4\hat{k}

Dividing by −2-2, a simplified direction vector for the acute angle bisector is: v⃗b′=i^+j^−2k^\vec{v}_b' = \hat{i} + \hat{j} - 2\hat{k}

Step 3: Find ll and mm

The given line LL is: L:x−αl=y−1m=z−γ−2L : \frac{x-\alpha}{l} = \frac{y-1}{m} = \frac{z-\gamma}{-2}

Comparing its direction ratios (l,m,−2)(l, m, -2) with the direction vector of the acute angle bisector (1,1,−2)(1, 1, -2), we get: l=1andm=1l = 1 \quad \text{and} \quad m = 1

Thus: l+m=1+1=2l + m = 1 + 1 = 2

This shows that Option B is TRUE and Option D is FALSE.

Step 4: Find α\alpha and γ\gamma

Since P(1,0,1)P(1, 0, 1) lies on line LL: 1−αl=0−1m=1−γ−2\frac{1-\alpha}{l} = \frac{0-1}{m} = \frac{1-\gamma}{-2}

Substitute l=1l = 1 and m=1m = 1: 1−α1=−1  ⟹  1−α=−1  ⟹  α=2\frac{1-\alpha}{1} = -1 \implies 1 - \alpha = -1 \implies \alpha = 2 1−γ−2=−1  ⟹  1−γ=2  ⟹  γ=−1\frac{1-\gamma}{-2} = -1 \implies 1 - \gamma = 2 \implies \gamma = -1

Now, evaluate α−γ\alpha - \gamma: α−γ=2−(−1)=3\alpha - \gamma = 2 - (-1) = 3

This shows that Option A is TRUE and Option C is FALSE.

Conclusion

The correct options are (A) and (B).

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