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Inverse of a Function

Let L 1, and ...

Question

Let $L_{1}$ , and $L_{2}$ denotes the lines
$\to r =^i + λ (-^i + 2^j + 2^k), λ \in R$ and
$\to r = μ (2^i -^j + 2^k), μ \in R$ respectively.
If $L_{3}$ is a line which is perpendicular to both $L_{1}$ and $L_{2}$ and cuts both of them, then which of the following option describe(s) $L_{3}$ ?

\to r = \frac{2}{9} (4^i +^j +^k) + t (2^i + 2^j -^k), t \in R

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\to r = \frac{2}{9} (2^i -^j + 2^k) + t (2^i + 2^j -^k), t \in R

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\to r = \frac{1}{3} (2^i +^k) + t (2^i + 2^j -^k), t \in R

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\to r = t (2^i + 2^j -^k), t \in R

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Solution

The correct options are
A $\to r = \frac{2}{9} (4^i +^j +^k) + t (2^i + 2^j -^k), t \in R$
B $\to r = \frac{2}{9} (2^i -^j + 2^k) + t (2^i + 2^j -^k), t \in R$
C $\to r = \frac{1}{3} (2^i +^k) + t (2^i + 2^j -^k), t \in R$

Equation of line $L_{1} : \to r =^i + λ (-^i + 2^j + 2^k), λ \in R$
Equation of line $L_{2} : \to r = μ (2^i -^j + 2^k), μ \in R$
Therefore, Point on lines $L_{1}$ and $L_{2}$ are $A (1 - λ, 2 λ, 2 λ)$ and $B (2 μ, - μ, 2 μ)$
$∴ - - \to A B = (2 μ + λ - 1)^i + (- μ - 2 λ)^j + (2 μ - 2 λ)^k$
$L_{3}$ is perpendicular to both $L_{1}$ and $L_{2}$
$∴ L_{3} ∥ (L_{1} \times L_{2})$
$L_{3} = m ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k - 1 & 2 & 2 2 & - 1 & 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3 m (2^i + 2^j -^k) = t (2^i + 2^j -^k m, t \in R)$
Direction ratios of AB will be $(2 t, 2 t, - t)$ which is propotional to $(2, 2, - 1)$
Hence, $\frac{2 μ + λ - 1}{2} = \frac{- μ - 2 λ}{2} = \frac{2 μ - 2 λ}{- 1} = c (L e t)$
$2 μ + λ - 1 = 2 c . . . (i)$
$- μ - 2 λ = 2 c . . . (i i)$
$2 μ - 2 λ = - c . . . (i i i)$
On solving equations we get,
$λ = \frac{1}{9}, μ = \frac{2}{9}$
$∴ A = (\frac{8}{9}, \frac{2}{9}, \frac{2}{9})$ , $B = (\frac{4}{9}, - \frac{2}{9}, \frac{4}{9})$
Mid Point of $- - \to A B = (\frac{2}{3}, 0, \frac{1}{3})$
$∴$ Equation of line $L_{3}$ can be
$\to r = \frac{8}{9}^i + \frac{2}{9}^j + \frac{2}{9}^k + t (2^i + 2^j -^k) t \in R$
$\to r = \frac{2}{9} (4^i +^j +^k) + t (2^i + 2^j -^k) t \in R$ or
$\to r = \frac{4}{9}^i - \frac{2}{9}^j + \frac{2}{9}^k + t (2^i + 2^j -^k) t \in R$
$\to r = \frac{2}{9} (2^i -^j +^k) + t (2^i + 2^j -^k) t \in R$ or
$\to r = \frac{2}{3}^i + \frac{1}{3}^k + t (2^i + 2^j -^k) t \in R$
$\to r = \frac{1}{3} (2^i +^k) + t (2^i + 2^j -^k) t \in R$

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Similar questions

L_{1}

L_{2}

\to r =^i + λ (-^i + 2^j + 2^k), λ \in R

\to r = μ (2^i -^j + 2^k), μ \in R

L_{3}

L_{1}

L_{2}

L_{3}

L_{1} : ¯ r = 3^i -^j + 2^k + λ (2^i + 4^j -^k);

L_{2} : ¯ r =^i +^j - 3^k + μ (4^i + 2^j + 4^k);

L_{3} : ¯ r = 3^i - 2^j - 2^k + t (2^i + 4^j - 2^k)

Find the shortest distance between the lines
$r = (^i + 2^j +^k) + λ (^i -^j +^k) a n d r = (2^i -^j -^k) + μ (2^i +^j + 2^k)$

\to r =^i + 2^j +^k + λ (^i -^j +^k)

\to r = 2^i -^j -^k + μ (2^i +^j + 2^k)

l_{1}

l_{2}

\to r = 2^i +^j +^k + λ (^i - 2^j +^k)

\to r =^i - 3^j -^k + λ (5^i + 2^j -^k)

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