The equation of a plane which passes through | vedclass

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Question

(A) The normal vector $\vec{n}$ to the plane is the vector joining the points $(3, 4, -1)$ and $(2, -1, 5)$.$\vec{n} = (2 - 3)\hat{i} + (-1 - 4)\hat{j

Vedclass · Accepted Answer

$x + 5y - 6z + 19 = 0$

Vedclass · Answer

(A) The normal vector $\vec{n}$ to the plane is the vector joining the points $(3, 4, -1)$ and $(2, -1, 5)$.$\vec{n} = (2 - 3)\hat{i} + (-1 - 4)\hat{j} + (5 - (-1))\hat{k} = -1\hat{i} - 5\hat{j} + 6\hat{k}$.Since the plane is normal to this line,the normal vector to the plane is $\vec{n} = -\hat{i} - 5\hat{j} + 6\hat{k}$ or $\vec{n} = \hat{i} + 5\hat{j} - 6\hat{k}$.The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $a\hat{i} + b\hat{j} + c\hat{k}$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.Substituting $(x_0, y_0, z_0) = (2, -3, 1)$ and $(a, b, c) = (1, 5, -6)$:$1(x - 2) + 5(y - (-3)) - 6(z - 1) = 0$$x - 2 + 5y + 15 - 6z + 6 = 0$$x + 5y - 6z + 19 = 0$.

The equation of a plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by

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