Find the equation of the line perpendicular t | vedclass

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(B) The given equation of the plane is $2x + 4y - 5z = 10$. The normal vector to the plane is $\bar{n} = (2, 4, -5)$. $A$ line perpendicular to the pl

Vedclass · Accepted Answer

$\bar{r} = (2k, 4k, -5k), k \in R$

Vedclass · Answer

(B) The given equation of the plane is $2x + 4y - 5z = 10$. The normal vector to the plane is $\bar{n} = (2, 4, -5)$. $A$ line perpendicular to the plane will have the same direction as the normal vector of the plane. Since the line passes through the origin $(0, 0, 0)$,the vector equation of the line is given by $\bar{r} = \bar{a} + k\bar{b}$,where $\bar{a} = (0, 0, 0)$ and $\bar{b} = (2, 4, -5)$. Therefore,the equation of the line is $\bar{r} = (0, 0, 0) + k(2, 4, -5)$,which simplifies to $\bar{r} = (2k, 4k, -5k)$ for $k \in R$.

Find the equation of the line perpendicular to the plane $2x + 4y - 5z = 10$ passing through the origin.

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