If the shortest distance between the lines _1=α i+2 j+2 k+λ ( i-2 j+2 k),λ ∈ R,α >0 and _2=-4 i- k+μ (3 i-2 j-2 k),μ ∈ R is 9, then α is equal to.

We have,

$$\vec{{r}_{1}}=\alpha \hat{i}+2\hat{j}+2\hat{k}+\lambda (\hat{i}-2\hat{j}+2\hat{k})$$r1​​=αi^+2j^​+2k^+λ(i^−2j^​+2k^)

$$\vec{{r}_{2}}=-4\hat{i}-\hat{k}+\mu (3\hat{i}-2\hat{j}-2\hat{k})$$r2​​=−4i^−k^+μ(3i^−2j^​−2k^)

If $\vec{r}=\vec{a}+\lambda \vec{b}$r=a+λb and $\vec{r}=\vec{c}+\lambda \vec{d}$r=c+λd, then shortest distance between two lines is

$$L=\frac{(\vec{a}-\vec{c})\cdot (\vec{b}\times \vec{d})}{|\vec{b}\times \vec{d}|}$$L=∣b×d∣(a−c)⋅(b×d)​

Here,

$$\vec{b}\times \vec{d}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2\end{matrix}|$$b×d=∣i^13​j^​−2−2​k^2−2​∣

$$\Rightarrow \vec{b}\times \vec{d}=8\hat{i}+8\hat{j}+4\hat{k}$$⇒b×d=8i^+8j^​+4k^

$$\Rightarrow \vec{b}\times \vec{d}=4(2\hat{i}+2\hat{j}+\hat{k})$$⇒b×d=4(2i^+2j^​+k^)

$$\Rightarrow |\vec{b}\times \vec{d}|=4\cdot 3$$⇒∣b×d∣=4⋅3

And,

$$\vec{a}-\vec{c}=((\alpha +4)\hat{i}+2\hat{j}+3\hat{k})$$a−c=((α+4)i^+2j^​+3k^)

So, shortest distance is

$$\frac{((\alpha +4)\hat{i}+2\hat{j}+3\hat{k})\cdot 4(2\hat{i}+2\hat{j}+\hat{k})}{4\cdot 3}=9$$4⋅3((α+4)i^+2j^​+3k^)⋅4(2i^+2j^​+k^)​=9

$$\Rightarrow 2(\alpha +4)+4+3=27$$⇒2(α+4)+4+3=27

$$\Rightarrow (\alpha +4)=10$$⇒(α+4)=10

$$\Rightarrow \alpha =6$$⇒α=6