Let a plane P contain the points hat{i}, hat{ | vedclass

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(A) The plane $P$ passes through the points $A_1(1, 0, 0)$, $A_2(0, 1, 0)$, and $A_3(1, 1, 1)$.Two vectors lying in the plane are $\vec{v_1} = A_2 - A

Vedclass · Accepted Answer

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

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(A) The plane $P$ passes through the points $A_1(1, 0, 0)$, $A_2(0, 1, 0)$, and $A_3(1, 1, 1)$.Two vectors lying in the plane are $\vec{v_1} = A_2 - A_1 = (-1, 1, 0)$ and $\vec{v_2} = A_3 - A_1 = (0, 1, 1)$.The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{v_1} 	imes \vec{v_2}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & 0 \ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1-0) - \hat{j}(-1-0) + \hat{k}(-1-0) = \hat{i} + \hat{j} - \hat{k}$.The direction ratios of the normal are $(1, 1, -1)$.The line passes through $A(3, 0, -5)$ and is parallel to the normal vector $\vec{n} = (1, 1, -1)$.The equation of the line is $\frac{x-3}{1} = \frac{y-0}{1} = \frac{z-(-5)}{-1}$, which simplifies to $\frac{x-3}{1} = \frac{y}{1} = \frac{z+5}{-1}$.

Let a plane $P$ contain the points $\hat{i}, \hat{j}$ and $\hat{i}+\hat{j}+\hat{k}$. Let $L$ be the line through the point $A(3, 0, -5)$ and parallel to the vector $\hat{i}-\hat{j}+\hat{k}$. The equation of the normal to the plane $P$ passing through point $A$ is:

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