Imagine you're designing a solar panel installation on a roof. The solar panel needs to be placed on a plane that is oriented such that it is at a distance of 5 units from the roof's center (considered as the origin in your design), with the normal to this plane being in the direction of the vector $2\hat{i} - 3\hat{j} + 6\hat{k}$.

Question: What is the vector equation of the plane on which the solar panel should be installed?

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To find the vector equation of the plane, we use the standard formula for a plane's equation:

$\vec{r} \cdot \hat{n} = d$

Where:

• $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is the position vector of a point on the plane,
• $\hat{n} = 2\hat{i} - 3\hat{j} + 6\hat{k}$ is the normal vector to the plane,
• $d = 5$ is the perpendicular distance of the plane from the origin.

The equation of the plane becomes:

$(2\hat{i} - 3\hat{j} + 6\hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 5$

Expanding this:

$2x - 3y + 6z = 5$

Thus, the vector equation of the plane is:

$2x - 3y + 6z = 5$

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Would you like more details on the method or further clarifications?

Here are 5 related questions to extend your understanding:

1. How would you determine if a given point lies on the plane described by the equation $2x - 3y + 6z = 5$?
2. What is the angle between two planes if their normal vectors are known?
3. How would the equation change if the distance from the origin was doubled?
4. Can you find a parametric equation for this plane?
5. How would you calculate the distance of a point from the plane if the point is not on it?

Tip: The dot product between two vectors is a useful tool for finding angles, projections, and determining perpendicularity between lines or planes.