Find the distance between the lines l1 and l2 given by r-2-4k-2-3-6k- and r-3-3-5k-2-3-6k-

Given #160; r⃗ ⃗_⃗ ⃗1⃗ =(î +2ĵ 4k̂ )+α (2î +3ĵ +6k̂ ) r⃗ ⃗_⃗ ⃗2⃗ =(3̂î +3ĵ 5k̂ )+μ (2î +3ĵ +6k̂ ) r⃗ ⃗_⃗ ⃗1⃗ =a⃗ +λb⃗ r⃗ ⃗_⃗ ⃗2⃗ =c⃗ +λd⃗ ...

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Byju's Answer

Standard XII

Physics

Displacement Vector

Find the dist...

Question

Find the distance between the lines $l_{1}$ and $l_{2}$ given by $\to r =^i + 2^j - 4^k + λ (2^i + 3^j + 6^k)$ and $\to r = 3^i + 3^j - 5^k + μ (2^i + 3^j + 6^k)$

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Solution

Given $_{1} = (^i + 2^j - 4^k) + α (2^i + 3^j + 6^k)$
$_{2} = (^3 i + 3^j - 5^k) + μ (2^i + 3^j + 6^k)$
$_{1} = \to a + λ \to b$
$_{2} = \to c + λ \to d$
Distance between them is projection of $\to a - \to c$ along the line perpendicular to both of them
Line DC's perpendicular to both $_{1},_{2}$ is $\frac{\to b \times \to d}{| \to b \times \to d |}$
Distance $= (\to a - \to c) \frac{\to b \times \to d}{| \to b \times \to d |}$
$= \frac{[\to a - \to c \to b \to d]}{| \to b \times \to d |}$
But given lines are parallel lines
Let $P$ be foot of perpendicular from $\to a$ on $_{2}$
$\to a -_{2}$ is perpendicular to $\to b$
$(2 μ + 2, 3 μ + 1, 6 μ - 1)$ is perpendicular to $(2, 3, 6)$
$(2) (2 μ + 2) + (3) (3 μ + 1) + (6) (6 μ - 1) = 0$
$49 μ = - 1$
$μ = \frac{- 1}{49}$
$2 μ + 2 = \frac{96}{49}$
$3 μ + 1 = \frac{46}{49}$
$6 μ - 1 = \frac{- 55}{49}$
Distance $= | \to a -_{2} | = \frac{\sqrt{96^{2} + 46^{2} + 55^{2}}}{49} = 2.45$

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Similar questions

Find the shortest distance between the lines whose vector equations are $r = (^i + 2^j + 3^k) + λ (^i - 3^j + 2^k)$
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\to r = 2^i + 4^j + 5^k + μ (4^i + 6^j + 8^k)

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Find the distance between the lines l1 and l2 given by →r=^i+2^j−4^k+λ(2^i+3^j+6^k) and →r=3^i+3^j−5^k+μ(2^i+3^j+6^k)

Find the distance between the lines $l_{1}$ and $l_{2}$ given by $\to r =^i + 2^j - 4^k + λ (2^i + 3^j + 6^k)$ and $\to r = 3^i + 3^j - 5^k + μ (2^i + 3^j + 6^k)$