To find the shortest distance between the two skew lines L 1 L_1 L 1 L 2 L_2 L 2
d = ∣ ( a ⃗ 2 − a ⃗ 1 ) ⋅ ( b ⃗ 1 × b ⃗ 2 ) ∣ ∣ b ⃗ 1 × b ⃗ 2 ∣ d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} d = ∣ b 1 × b 2 ∣ ∣ ( a 2 − a 1 ) ⋅ ( b 1 × b 2 ) ∣
From the given equations of the lines:
For line L 1 : x + 1 3 = y + 2 1 = z + 1 2 L_1: \frac{x+1}{3} = \frac{y+2}{1} = \frac{z+1}{2} L 1 : 3 x + 1 = 1 y + 2 = 2 z + 1
A point on L 1 L_1 L 1 a ⃗ 1 = − 1 i ^ − 2 j ^ − 1 k ^ \vec{a}_1 = -1\hat{i} - 2\hat{j} - 1\hat{k} a 1 = − 1 i ^ − 2 j ^ − 1 k ^
The direction vector of L 1 L_1 L 1 b ⃗ 1 = 3 i ^ + 1 j ^ + 2 k ^ \vec{b}_1 = 3\hat{i} + 1\hat{j} + 2\hat{k} b 1 = 3 i ^ + 1 j ^ + 2 k ^
For line L 2 : x − 2 1 = y + 2 2 = z − 3 3 L_2: \frac{x-2}{1} = \frac{y+2}{2} = \frac{z-3}{3} L 2 : 1 x − 2 = 2 y + 2 = 3 z − 3
A point on L 2 L_2 L 2 a ⃗ 2 = 2 i ^ − 2 j ^ + 3 k ^ \vec{a}_2 = 2\hat{i} - 2\hat{j} + 3\hat{k} a 2 = 2 i ^ − 2 j ^ + 3 k ^
The direction vector of L 2 L_2 L 2 b ⃗ 2 = 1 i ^ + 2 j ^ + 3 k ^ \vec{b}_2 = 1\hat{i} + 2\hat{j} + 3\hat{k} b 2 = 1 i ^ + 2 j ^ + 3 k ^
Step 1: Calculate a ⃗ 2 − a ⃗ 1 \vec{a}_2 - \vec{a}_1 a 2 − a 1 a ⃗ 2 − a ⃗ 1 = ( 2 − ( − 1 ) ) i ^ + ( − 2 − ( − 2 ) ) j ^ + ( 3 − ( − 1 ) ) k ^ \vec{a}_2 - \vec{a}_1 = (2 - (-1))\hat{i} + (-2 - (-2))\hat{j} + (3 - (-1))\hat{k} a 2 − a 1 = ( 2 − ( − 1 )) i ^ + ( − 2 − ( − 2 )) j ^ + ( 3 − ( − 1 )) k ^ a ⃗ 2 − a ⃗ 1 = 3 i ^ + 0 j ^ + 4 k ^ \vec{a}_2 - \vec{a}_1 = 3\hat{i} + 0\hat{j} + 4\hat{k} a 2 − a 1 = 3 i ^ + 0 j ^ + 4 k ^
Step 2: Calculate the cross product b ⃗ 1 × b ⃗ 2 \vec{b}_1 \times \vec{b}_2 b 1 × b 2
b ⃗ 1 × b ⃗ 2 = ∣ i ^ j ^ k ^ 3 1 2 1 2 3 ∣ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} b 1 × b 2 = i ^ 3 1 j ^ 1 2 k ^ 2 3 b ⃗ 1 × b ⃗ 2 = i ^ ( 1 ⋅ 3 − 2 ⋅ 2 ) − j ^ ( 3 ⋅ 3 − 2 ⋅ 1 ) + k ^ ( 3 ⋅ 2 − 1 ⋅ 1 ) \vec{b}_1 \times \vec{b}_2 = \hat{i}(1\cdot 3 - 2\cdot 2) - \hat{j}(3\cdot 3 - 2\cdot 1) + \hat{k}(3\cdot 2 - 1\cdot 1) b 1 × b 2 = i ^ ( 1 ⋅ 3 − 2 ⋅ 2 ) − j ^ ( 3 ⋅ 3 − 2 ⋅ 1 ) + k ^ ( 3 ⋅ 2 − 1 ⋅ 1 ) b ⃗ 1 × b ⃗ 2 = − i ^ − 7 j ^ + 5 k ^ \vec{b}_1 \times \vec{b}_2 = -\hat{i} - 7\hat{j} + 5\hat{k} b 1 × b 2 = − i ^ − 7 j ^ + 5 k ^
Step 3: Calculate the magnitude ∣ b ⃗ 1 × b ⃗ 2 ∣ |\vec{b}_1 \times \vec{b}_2| ∣ b 1 × b 2 ∣ ∣ b ⃗ 1 × b ⃗ 2 ∣ = ( − 1 ) 2 + ( − 7 ) 2 + 5 2 = 1 + 49 + 25 = 75 = 5 3 |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3} ∣ b 1 × b 2 ∣ = ( − 1 ) 2 + ( − 7 ) 2 + 5 2 = 1 + 49 + 25 = 75 = 5 3
Step 4: Calculate the scalar triple product ( a ⃗ 2 − a ⃗ 1 ) ⋅ ( b ⃗ 1 × b ⃗ 2 ) (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) ( a 2 − a 1 ) ⋅ ( b 1 × b 2 ) ( a ⃗ 2 − a ⃗ 1 ) ⋅ ( b ⃗ 1 × b ⃗ 2 ) = ( 3 ) ( − 1 ) + ( 0 ) ( − 7 ) + ( 4 ) ( 5 ) (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (3)(-1) + (0)(-7) + (4)(5) ( a 2 − a 1 ) ⋅ ( b 1 × b 2 ) = ( 3 ) ( − 1 ) + ( 0 ) ( − 7 ) + ( 4 ) ( 5 ) = − 3 + 0 + 20 = 17 = -3 + 0 + 20 = 17 = − 3 + 0 + 20 = 17
Step 5: Calculate the shortest distance d d d d = ∣ 17 ∣ 5 3 = 17 5 3 d = \frac{|17|}{5\sqrt{3}} = \frac{17}{5\sqrt{3}} d = 5 3 ∣17∣ = 5 3 17
Thus, the shortest distance between L 1 L_1 L 1 L 2 L_2 L 2 17 5 3 \frac{17}{5\sqrt{3}} 5 3 17