If the shortest distance between the lines r=- i+3 k+( i-a j) and r=- j+2 k+( i- j+ k) is …

Question

If the shortest distance between the lines $\vec r=-\hat i+3\hat k+\lambda(\hat i-a\hat j)$ and $\vec r=-\hat j+2\hat k+\mu(\hat i-\hat j+\hat k)$ is $\sqrt{\dfrac23}$, the integral value of $a$ is:

Answer: 2

Accepted Answer

Direction vectors $\vec d_1=(1,-a,0)$, $\vec d_2=(1,-1,1)$; connecting vector $(1,-1,-1)$.

$$\vec d_1\times\vec d_2 = (-a,\ -1,\ a-1)$$
$$\text{Distance} = \frac{|(1,-1,-1)\cdot(-a,-1,a-1)|}{|\vec d_1\times\vec d_2|} = \frac{|-a+1-a+1|}{\sqrt{a^2+1+(a-1)^2}} = \frac{|2-2a|}{\sqrt{a^2+1+(a-1)^2}}$$

Setting this equal to $\sqrt{2/3}$ and squaring:
$$\frac{4(1-a)^2}{2a^2-2a+2} = \frac23 \implies 12(1-a)^2 = 2(2a^2-2a+2) \implies 4a^2-10a+4=0 \implies 2a^2-5a+2=0$$
$$a = 2 \text{ or } \frac12$$

The integral value is $a=2$.

Solutions · 1