Question

Let O be the origin and OA = a1i + a2j + a3k. Find the equation of the plane that contains vectors I + j + k and 2i + j + k

Answer 1

Answer on Question #63635 – Math – Analytic Geometry

Question

Let $O$ be the origin and $OA = a1i + a2j + a3k$. Find the equation of the plane that contains vectors $l + j + k$ and $2i + j + k$

Solution

The equation of a plane is

$$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$

where $\begin{pmatrix} a \\ b \end{pmatrix}$ is the normal vector to the plane, $P(x_0, y_0, z_0)$ is the point on the plane. Since the plane contains the vector $\vec{i} + \vec{j} + \vec{k}$, then it passes through the origin and $(x_0, y_0, z_0) = (0, 0, 0)$.

We can use the cross product of two vectors $\vec{i} + \vec{j} + \vec{k}$ and $2\vec{i} + \vec{j} + \vec{k}$ as the normal vector to the plane since both of them are in the plane (any vector that is orthogonal to both of these will also be orthogonal to the plane):

$$\begin{array}{l} \vec{n} = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & 1 \\ 2 & 1 & 1 \end{array} \right| = \left| \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right| \vec{i} - \left| \begin{array}{cc} 1 & 1 \\ 2 & 1 \end{array} \right| \vec{j} + \left| \begin{array}{cc} 1 & 1 \\ 2 & 1 \end{array} \right| \vec{k} = (1 \cdot 1 - 1 \cdot 1) \vec{i} - \\ - (1 \cdot 1 - 2 \cdot 1) \vec{j} + (1 \cdot 1 - 2 \cdot 1) \vec{k} = (1 - 1) \vec{i} - (1 - 2) \vec{j} + (1 - 2) \vec{k} = 0 \vec{i} + \vec{j} - \vec{k}. \end{array}$$

The equation of the plane is then

$$\begin{array}{l} 0 \cdot (x - 0) + 1 \cdot (y - 0) - 1 \cdot (z - 0) = 0; \\ y - z = 0. \end{array}$$

Answer: the equation of the plane is $y - z = 0$.

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