Finding the Equation of a Line Parallel to Two Planes

The question asks for the equation of a line L that passes through a given point P(5, -6, 7) and is parallel to two given planes: \(x + y + z = 1\) and \(2x – y - 2z = 3\).

The equation of a line in three-dimensional space can be written in symmetric form as:

where \((x_0, y_0, z_0)\) is a point on the line and \(\langle a, b, c \rangle\) is the direction vector of the line.

We are given that the line L passes through the point P(5, -6, 7). So, \((x_0, y_0, z_0) = (5, -6, 7)\). This means the numerators in the equation will be \((x - 5)\), \((y - (-6)) = (y + 6)\), and \((z - 7)\). Let's look at the options based on the point:

• Option 1: Numerators are \((x - 5)\), \((y + 6)\), \((z - 7)\). This matches the point P.
• Option 2: Numerators are \((x + 5)\), \((y - 6)\), \((z + 7)\). This does not match the point P. Option 2 is incorrect.
• Option 3: Numerators are \((x - 5)\), \((y + 6)\), \((z - 7)\). This matches the point P.
• Option 4: Numerators are \((x - 5)\), \((y + 6)\), \((z - 7)\). This matches the point P.

Now, we need to find the direction vector \(\vec{v} = \langle a, b, c \rangle\) of the line L.

A line is parallel to a plane if its direction vector is perpendicular to the normal vector of the plane.

The normal vector of the first plane \(x + y + z = 1\) is \(\vec{n_1} = \langle 1, 1, 1 \rangle\).

The normal vector of the second plane \(2x – y - 2z = 3\) is \(\vec{n_2} = \langle 2, -1, -2 \rangle\).

Since line L is parallel to both planes, its direction vector \(\vec{v}\) must be perpendicular to both \(\vec{n_1}\) and \(\vec{n_2}\). A vector that is perpendicular to two vectors is parallel to their cross product.

Therefore, the direction vector \(\vec{v}\) is parallel to \(\vec{n_1} \times \vec{n_2}\). Let's calculate the cross product:

So, a possible direction vector for line L is \(\vec{v} = \langle -1, 4, -3 \rangle\). Any scalar multiple of this vector is also a valid direction vector.

Using the point (5, -6, 7) and the direction vector \(\langle -1, 4, -3 \rangle\), the equation of the line is:

Let's compare this equation with the remaining options (Options 1, 3, and 4):

• Option 1: \(\frac{{x - 5}}{{ - 1}} = \frac{{y + 6}}{4} = \frac{{z - 7}}{{ - 3}}\). The direction vector is \(\langle -1, 4, -3 \rangle\). This matches our calculated direction vector.
• Option 3: \(\frac{{x - 5}}{{ - 1}} = \frac{{y + 6}}{{ - 4}} = \frac{{z - 7}}{3}\). The direction vector is \(\langle -1, -4, 3 \rangle\). This vector is not a scalar multiple of \(\langle -1, 4, -3 \rangle\) as the components are not in the same ratio (\(-1/-1 = 1\), \(-4/4 = -1\), \(3/-3 = -1\)).
• Option 4: \(\frac{{x - 5}}{{ - 1}} = \frac{{y + 6}}{{ - 4}} = \frac{{z - 7}}{{ - 3}}\). The direction vector is \(\langle -1, -4, -3 \rangle\). This vector is not a scalar multiple of \(\langle -1, 4, -3 \rangle\) as the components are not in the same ratio (\(-1/-1 = 1\), \(-4/4 = -1\), \(-3/-3 = 1\)).

Therefore, the equation of the line L is \(\frac{{x - 5}}{{ - 1}} = \frac{{y + 6}}{4} = \frac{{z - 7}}{{ - 3}}\).

Revision Table: Key Concepts for Line Equations

Additional Information on 3D Lines and Planes

In three-dimensional geometry, understanding the relationship between lines and planes is crucial. A line's direction is defined by its direction vector, while a plane's orientation is defined by its normal vector.

• When a line is parallel to a plane, it means the line never intersects the plane. The direction vector of the line and the normal vector of the plane form a 90-degree angle, making their dot product zero.
• When a line is parallel to two non-parallel planes, it implies that the line is parallel to the line of intersection of these two planes. The direction vector of the intersection line is always parallel to the cross product of the normal vectors of the two planes. Thus, the direction vector of a line parallel to both planes is also parallel to this cross product.
• The symmetric form of the line equation is derived from the vector form \(\vec{r} = \vec{r_0} + t\vec{v}\), where \(\vec{r} = \langle x, y, z \rangle\), \(\vec{r_0} = \langle x_0, y_0, z_0 \rangle\), \(\vec{v} = \langle a, b, c \rangle\), and \(t\) is a scalar parameter. Equating components and solving for \(t\) gives: \[ x = x_0 + ta \implies t = \frac{x - x_0}{a} \] \[ y = y_0 + tb \implies t = \frac{y - y_0}{b} \] \[ z = z_0 + tc \implies t = \frac{z - z_0}{c} \] Setting these equal gives the symmetric form: \(\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}\). This form is useful when the direction numbers \(a, b, c\) are non-zero. If one of the direction numbers is zero, the equation is adjusted accordingly (e.g., if \(a=0\), then \(x=x_0\) and \(\frac{y - y_0}{b} = \frac{z - z_0}{c}\)).

Calculating the cross product correctly is key to finding the direction vector in problems where a line's direction is constrained by its relationship to planes.