L 1 : x + 1 1 = y 1 / 2 = z − 1 / 12 , L 2 : x 1 = y + 2 1 = z − 1 1 6 \mathrm{L}_1: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}_2: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}} L 1 : 1 x + 1 = 1 / 2 y = − 1 / 12 z , L 2 : 1 x = 1 y + 2 = 6 1 z − 1
d 1 = \mathrm{d}_1= d 1 = L 1 & L 2 \mathrm{L}_1 ~\& \mathrm{~L}_2 L 1 & L 2
= ∣ ( a ⃗ 2 − a ⃗ 1 ) ⋅ ( b ⃗ 1 × b ⃗ 2 ) ∣ ( b ⃗ 1 × b ⃗ 2 ) ∣ ∣ d 1 = 2 L 3 : x − 1 2 = y + 8 − 7 = z − 4 5 , L 4 : x − 1 2 = y − 2 1 = z − 6 − 3 d 2 = s h o r t e s t d i s t a n c e b e t w e e n L 3 & L 4 d 2 = 12 3 H e n c e = 32 3 d 1 d 2 = 32 3 × 2 12 3 = 16 \begin{aligned} & =\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}\right| \\ & d_1=2 \\ & \mathrm{~L}_3: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+8}{-7}=\frac{\mathrm{z}-4}{5}, \\ & \mathrm{~L}_4: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-6}{-3} \\ & \mathrm{~d}_2=\text { shortest distance between } \mathrm{L}_3 ~\& \mathrm{~L}_4 \\ & \mathrm{~d}_2=\frac{12}{\sqrt{3}} \\ & { Hence } =\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16 \end{aligned} = ( b 1 × b 2 ) ( a 2 − a 1 ) ⋅ ( b 1 × b 2 ) d 1 = 2 L 3 : 2 x − 1 = − 7 y + 8 = 5 z − 4 , L 4 : 2 x − 1 = 1 y − 2 = − 3 z − 6 d 2 = s h or t es t d i s t an ce b e tw ee n L 3 & L 4 d 2 = 3 12 He n ce = d 2 32 3 d 1 = 3 12 32 3 × 2 = 16
