91Ó°ÊÓ

In the study of geometry, direction vectors are essential when solving problems involving planes. Given a set of points, direction vectors help us understand how these points are positioned relative to one another. To find a direction vector between two points, you subtract the coordinates of one point from the other. For instance, with points \( P(3,4,5) \) and \( Q(1,2,3) \), you obtain the direction vector \( \vec{PQ} = (-2, -2, -2) \). Similarly, for points \( P(3,4,5) \) and \( R(4,7,6) \), the direction vector is \( \vec{PR} = (1, 3, 1) \).
Direction vectors show the relative change in position between points. They are not unique; any scalar multiple of a direction vector is also a valid direction vector.

In the context of planes, a normal vector is a vector that is perpendicular to the plane. This vector is vital because it uniquely determines the plane's orientation. You can find it using the cross product of two direction vectors that lie in the plane. From the previous section, we use \( \vec{PQ} = (-2, -2, -2) \) and \( \vec{PR} = (1, 3, 1) \) to calculate the normal vector.
Using the cross product formula:

• The first component is \((-2\cdot1 - (-2)\cdot3) = 4\).
• The second component is \(-((-2)\cdot1 - (-2)\cdot1) = 0\).
• The third component is \((-2\cdot3 - (-2)\cdot1) = -4\).

Thus, the normal vector is \( \vec{n} = (4, 0, -4) \). This normal vector is essential for writing the plane's equation.

The cross product is a mathematical operation that takes two vectors and produces a third vector perpendicular to both. This is a crucial operation for finding a normal vector to the plane. The formula for the cross product \( \vec{a} \times \vec{b} \) of vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \) is expressed by \[ \vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \] This determinant calculates the components as follows:

• The \( \mathbf{i} \) component as \( a_2b_3 - a_3b_2 \)
• The \( \mathbf{j} \) component as \( a_3b_1 - a_1b_3 \)
• The \( \mathbf{k} \) component as \( a_1b_2 - a_2b_1 \)

When you apply this to our direction vectors, you get \( (4, 0, -4) \) as the normal vector, confirming the earlier result.

An equation of a plane provides a formula for all the points on that plane. Given a normal vector \( \vec{n} = (A, B, C) \) and a point \( (x_0, y_0, z_0) \) on the plane, the general equation of a plane is \[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]For the present case, we utilize the normal vector \( (4, 0, -4) \) and the point \( P(3, 4, 5) \). Substituting these into the formula yields: \[ 4(x - 3) + 0(y - 4) - 4(z - 5) = 0 \]
This simplifies further to \( 4x - 4z = 8 \), which, when divided by 4, becomes \( x - z = 2 \).
This is the final, simplified equation representing the plane through points \( P, Q, \) and \( R \).