Question

If the shortest distance between x - λ = 2y - 1 = -2z and x = y + 2λ = z - λ is √7 / 2√2, then |λ| is _______

Hint:

Shortest distance between skew lines: $SD = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.

Answer 1

Correct Answer: 1

Step 1: $L_1: \frac{x-\lambda}{1} = \frac{y-1/2}{1/2} = \frac{z}{-1/2} \parallel (2, 1, -1)$. Point $A(\lambda, 1/2, 0)$.
Step 2: $L_2: \frac{x}{1} = \frac{y+2\lambda}{1} = \frac{z-\lambda}{1} \parallel (1, 1, 1)$. Point $B(0, -2\lambda, \lambda)$.
Step 3: $\vec{n} = \vec{d_1} \times \vec{d_2} = (2, -3, 1)$. $|\vec{n}| = \sqrt{14}$.
Step 4: $SD = \frac{|(\vec{A}-\vec{B}) \cdot \vec{n}|}{|\vec{n}|} = \frac{|2\lambda - 3(1/2 + 2\lambda) - \lambda|}{\sqrt{14}} = \frac{|-5\lambda - 1.5|}{\sqrt{14}}$.
Step 5: $\frac{|5\lambda + 1.5|}{\sqrt{14}} = \frac{\sqrt{7}}{\sqrt{8}} \Rightarrow |5\lambda + 1.5| = 3.5 \Rightarrow 5\lambda = 2$ or $-5$.
Step 6: $\lambda = -1$ is the integer solution. $|\lambda| = 1$.