What is the equation of the plane passing thr | vedclass

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(A) The given line is $\vec{r} = (2, 3, 4) + k(3, 4, 5)$.The direction vector of this line is $\vec{v} = (3, 4, 5)$.Since the plane is perpendicular t

Vedclass · Accepted Answer

$3x + 4y + 5z = 7$

Vedclass · Answer

(A) The given line is $\vec{r} = (2, 3, 4) + k(3, 4, 5)$.The direction vector of this line is $\vec{v} = (3, 4, 5)$.Since the plane is perpendicular to the line,the direction vector of the line acts as the normal vector $\vec{n}$ to the plane.Thus,$\vec{n} = (3, 4, 5)$.The plane passes through the point $(x_0, y_0, z_0) = (1, 1, 0)$.The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is given by $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.Substituting the values: $3(x - 1) + 4(y - 1) + 5(z - 0) = 0$.Expanding this: $3x - 3 + 4y - 4 + 5z = 0$.Simplifying: $3x + 4y + 5z = 7$.

What is the equation of the plane passing through the point $(1, 1, 0)$ and perpendicular to the line $\vec{r} = (2, 3, 4) + k(3, 4, 5)$,where $k \in R$?

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