Two lines whose equation are L 1- x -3-2- y -2-3- z -1- and L 2- x -2-3- y -3-2- z -2-3 lie in the same plane- If L 1 intersects a plane x - y - z -15 at P- then distance of P from 3-4-3 isA- 3B- -5C- -15D- -10

The correct option is A 3Lines are coplanar if 1 amp;1 amp;1 2 amp; 3 amp; λ 3 amp;2 amp;3 =0 ⇒ 20 5λ =0⇒λ =4 Any point on L_1 is (3 ...

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Geometrical Representation of a Complex Number

Two lines who...

Question

Two lines whose equation are $L_{1} : \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{λ}$ and $L_{2} : \frac{x - 2}{3} = \frac{y - 3}{2} = \frac{z - 2}{3}$ lie in the same plane. If $L_{1}$ intersects a plane $x + y + z = 15$ at P, then distance of P from (3, 4, 3) is

3

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\sqrt{5}

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\sqrt{10}

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\sqrt{15}

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L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}

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

L_{1}

L_{2}

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}

L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

(1, 1, 1)

(- 1, - 2, - 1)

L_{1}

L_{2}

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}, L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

L_{1}

L_{2}

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

L_{2} : \frac{x}{1} = \frac{y}{- 1} = \frac{z - 5}{3}

L_{1}

L_{2}

L_{1}

L_{2}

L_{1}

L_{2}

\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}

\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}

L_{1}

L_{2}

L_{1}

L_{2}

L_{1}

L_{2}

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