If-the-line-x-1-2-y-1-3-z-1-4-x-3-1-y-k-2-z-1-intersect-the-value-of-k-is-

Question Number 132327 by liberty last updated on 13/Feb/21

If the line {\begin{cases} \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} \\ \frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1} \end{cases}

intersect . the value of k is

Answered by Ar Brandon last updated on 13/Feb/21

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

L_{2} : 3 i + α j + μ (i + 2 j + k) = (3 + μ) i + (α + 2 μ) j + k

{\begin{cases} 1 + 2 λ = 3 + μ \\ 3 λ - 1 = α + 2 μ \\ 4 λ + 1 = 1 \end{cases} \Rightarrow {\begin{cases} λ = 0 \\ μ = - 2 \\ α = 3 \end{cases}

k = α = 3 and i - j + k is point of intersection

Answered by EDWIN88 last updated on 15/Feb/21

let \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = λ \to {\begin{cases} x = 2 λ + 1 \\ y = 3 λ - 1 \\ z = 4 λ + 1 \end{cases}

let \frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1} = ϑ \to {\begin{cases} x = ϑ + 3 \\ y = 2 ϑ + k \\ z = ϑ \end{cases}

the line are intersect the we get

{\begin{cases} 2 λ + 1 = ϑ + 3 \\ 3 λ - 1 = 2 ϑ + k \\ 4 λ + 1 = ϑ \end{cases}; \to {\begin{cases} ϑ = - 5 \\ 2 λ = 3; λ = - \frac{3}{2} \end{cases}

then k = 3 λ - 2 ϑ - 1 = - \frac{9}{2} + 10 - 1 = - \frac{9}{2} + 9 = \frac{9}{2}

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