The value of $^{'} m^{'}$ for which the straight line $3 x - 2 y + z + 3 = 0 = 4 x - 3 y + 4 z + 1$ is parallel to the plane 2x - y + mz - 2 = 0 is

-2

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Solution

The correct option is A
-2

$\begin{matrix} ∣ ∣ ∣ ∣ \begin{matrix} 3 & - 2 & 1 4 & - 3 & 4 2 & - 1 & m \end{matrix} ∣ ∣ ∣ ∣ & = & 0 \Rightarrow m & = & - 2 \end{matrix}$

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Similar questions

m

3 x - 2 y + z + 3 = 0 = 4 x - 3 y + 4 z + 1

2 x - y + m z - 2 = 0

m

3 x - 2 y + z + 3 = 0 = 4 x - 3 y + 4 z + 1

2 x - y + m x - 2 = 0

2 x - 3 y + 4 = 0, x - 4 y + 4 z + 5

2 x - y + m z - 2 = 0

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

x - y + z + 2 = 0

Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is

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