If the planes x + 2y + kz = 0 and 2x + y - 2z | vedclass

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(D) Two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perpendicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0$. Here,the normal

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$2$

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(D) Two planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perpendicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0$. Here,the normal vectors to the planes are $\vec{n_1} = (1, 2, k)$ and $\vec{n_2} = (2, 1, -2)$. Since the planes are at right angles,their normal vectors must be perpendicular,so their dot product is zero: $(1)(2) + (2)(1) + (k)(-2) = 0$. $2 + 2 - 2k = 0$. $4 - 2k = 0$. $2k = 4$. $k = 2$.

If the planes $x + 2y + kz = 0$ and $2x + y - 2z = 0$ are at right angles,then the value of $k$ is

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