If the shortest distance between the lines $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ to $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and is $\frac{44}{\sqrt{30}}$ then the largest possible value of $|\lambda|$ is equal ____.

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Three Dimensional Geometry

If the shortest distance between the lines

\frac{x + 2}{- 3} = \frac{y + 5}{2} = \frac{z - 4}{4}

\frac{x - λ}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{1}

and is

\frac{44}{30}

then the largest possible value of

∣ λ ∣

is equal ____.

Answer: 43

Solution

With

a_{1} = (- 2, - 5, 4), d_{1} = (- 3, 2, 4)

and

a_{2} = (λ, 2, 1), d_{2} = (3, - 1, 1)

d_{1} \times d_{2} = (6, 15, - 3)

, magnitude

330

. The shortest distance is

\frac{∣ ( a _{2} - a _{1} ) \cdot ( d _{1} \times d _{2} ) ∣}{∣ d _{1} \times d _{2} ∣} = \frac{∣6 ( λ + 2 ) + 15 ( 7 ) + 9∣}{3 30} = \frac{∣2 λ + 42∣}{30}

Setting this to

\frac{44}{30}

gives

∣2 λ + 42∣ = 44

, so

λ = 1

λ = - 43

. The largest

∣ λ ∣

43

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Integer answer

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