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The angle between the lines

l1:x+11=2−y2=z−11l_1: \frac{x + 1}{1} = \frac{2 - y}{2} = \frac{z - 1}{1} and l2:x−14=2y−46=z−12l_2: \frac{x - 1}{4} = \frac{2y - 4}{6} = \frac{z - 1}{2} is

Solution

✅ Correct Option: 4

The angle between two lines in 3D requires finding the direction vectors of each line, then using the dot product formula.

Given: l1:x+11=2−y2=z−11l_1: \frac{x + 1}{1} = \frac{2 - y}{2} = \frac{z - 1}{1}

The middle term has "2 - y" instead of "y - 2". Rewriting:

x+11=−(y−2)2=z−11\frac{x + 1}{1} = \frac{-(y - 2)}{2} = \frac{z - 1}{1}

x+11=y−2−2=z−11\frac{x + 1}{1} = \frac{y - 2}{-2} = \frac{z - 1}{1}

Direction vector of l1l_1: b1⃗=(1,−2,1)\vec{b_1} = (1, -2, 1)


Given: l2:x−14=2y−46=z−12l_2: \frac{x - 1}{4} = \frac{2y - 4}{6} = \frac{z - 1}{2}

The middle term simplifies: 2y−4=2(y−2)2y - 4 = 2(y - 2)

x−14=2(y−2)6=z−12\frac{x - 1}{4} = \frac{2(y - 2)}{6} = \frac{z - 1}{2}

x−14=y−23=z−12\frac{x - 1}{4} = \frac{y - 2}{3} = \frac{z - 1}{2}

Direction vector of l2l_2: b2⃗=(4,3,2)\vec{b_2} = (4, 3, 2)


b1⃗⋅b2⃗=(1)(4)+(−2)(3)+(1)(2)\vec{b_1} \cdot \vec{b_2} = (1)(4) + (-2)(3) + (1)(2)

=4−6+2= 4 - 6 + 2

=0= 0


When the dot product of two direction vectors is zero, the lines are perpendicular to each other.

The angle between perpendicular lines is π2\frac{\pi}{2}.

Therefore, the angle between the lines is π2\frac{\pi}{2}.

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