Consider lines L 1- x -2-1- y -3- v - z -4- kL 2- x -1-2- y -4-2- z - K -1Value of - k - so that lines L 1 and L 2 are coplanar- isA- 2B- -2C- -1 D- -1-2

The correct option is D 1/2 For 2 lines to be collinear; x_1 x_2 amp; y_1 y_2 amp; z_1 z_2 l_1 amp; m_1 amp; n_1 l_2 amp; m_2 amp; n_2 =0 ...

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Byju's Answer

Standard XII

Mathematics

Addition of Vectors

Consider line...

Question

Consider lines

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

$L_{2} : \frac{x - 1}{2} = \frac{y - 4}{2} = \frac{z - 5}{1}$

Value of 'k' so that lines $L_{1}$ and $L_{2}$ are coplanar, is

-1

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$- \frac{1}{2}$

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-2

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Solution

The correct option is B
$- \frac{1}{2}$

For 2 lines to be collinear;

$∣ ∣ ∣ ∣ \begin{matrix} x_{1} - x_{2} & y_{1} - y_{2} & z_{1} - z_{2} l_{1} & m_{1} & n_{1} l_{2} & m_{2} & n_{2} \end{matrix} ∣ ∣ ∣ ∣ = 0$

$∣ ∣ ∣ ∣ \begin{matrix} 1 & - 1 & - 1 1 & 1 & - k 2 & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow k = - \frac{1}{2}$

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Consider lines L1:x−21=y−31=z−4−k L2:x−12=y−42=z−51 Value of 'k' so that lines L1 and L2 are coplanar, is

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Consider lines

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

$L_{2} : \frac{x - 1}{2} = \frac{y - 4}{2} = \frac{z - 5}{1}$

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