If the shortest distance between the lines $\vec{r}_{1}=\alpha \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k}), \lambda \in R$, $\alpha>0$ and $\vec{r}_{2}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k}), \mu \in R$ is 9, then $\alpha$ is equal to ______.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Three Dimensional Geometry

If the shortest distance between the lines

r_{1} = α \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k}), λ \in R

α > 0

and

r_{2} = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k}), μ \in R

is 9, then

α

is equal to ______.

Answer: 6

Solution

The shortest distance between two skew lines

r_{1} = a_{1} + λ b_{1}

and

r_{2} = a_{2} + μ b_{2}

is:

d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}

Here

a_{1} = (α, 2, 2)

b_{1} = (1, - 2, 2)

a_{2} = (- 4, 0, - 1)

b_{2} = (3, - 2, - 2)

b_{1} \times b_{2} = \hat{i} 13 \hat{j} - 2 - 2 \hat{k} 2 - 2 = (8, 8, 4)

∣ b_{1} \times b_{2} ∣ = 64 + 64 + 16 = 12

a_{2} - a_{1} = (- 4 - α, - 2, - 3)

(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = 8 (- 4 - α) + 8 (- 2) + 4 (- 3) = - 32 - 8 α - 16 - 12 = - 60 - 8 α

Given

d = 9

\frac{∣ - 60 - 8 α ∣}{12} = 9

⟹

∣ - 60 - 8 α ∣ = 108

Since

α > 0

- 60 - 8 α < 0

, so

60 + 8 α = 108

⟹

8 α = 48

⟹

α = 6

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer

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