If the distance of the point (4,6,8) from the plane \( \vec{r} \cdot (6\hat{i} - 12\hat{j} + 4\hat{k}) = a \) is 1, then \( a \) is:
-2 or -30
Understanding the Distance from a Point to a Plane
The question asks us to find the value(s) of a variable 'a' in the equation of a plane, given the distance from a specific point to that plane. We are given the point (4,6,8) and the plane's equation in vector form \( \vec{r} \cdot (6\hat{i} - 12\hat{j} + 4\hat{k}) = a \). The distance is given as 1.
Formula for Distance
The distance \( D \) from a point \( P(x_0, y_0, z_0) \) to a plane with the Cartesian equation \( Ax + By + Cz + D' = 0 \) is given by the formula:
\( D = \frac{|Ax_0 + By_0 + Cz_0 + D'|}{\sqrt{A^2 + B^2 + C^2}} \)
Alternatively, if the plane equation is \( Ax + By + Cz = D'' \), the formula is:
\( D = \frac{|Ax_0 + By_0 + Cz_0 - D''|}{\sqrt{A^2 + B^2 + C^2}} \)
Converting the Plane Equation
The given plane equation is in vector form:
\( \vec{r} \cdot (6\hat{i} - 12\hat{j} + 4\hat{k}) = a \)
Here, \( \vec{r} \) is the position vector of any point on the plane, which can be written as \( x\hat{i} + y\hat{j} + z\hat{k} \). The vector \( \vec{n} = 6\hat{i} - 12\hat{j} + 4\hat{k} \) is the normal vector to the plane.
Substituting \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \) into the equation:
\( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (6\hat{i} - 12\hat{j} + 4\hat{k}) = a \)
Taking the dot product gives the Cartesian equation of the plane:
\( 6x - 12y + 4z = a \)
Comparing this with the form \( Ax + By + Cz = D'' \), we have \( A = 6 \), \( B = -12 \), \( C = 4 \), and \( D'' = a \).
Applying the Distance Formula
The given point is \( (x_0, y_0, z_0) = (4, 6, 8) \). The distance \( D \) is given as 1.
Using the distance formula:
\( 1 = \frac{|(6)(4) + (-12)(6) + (4)(8) - a|}{\sqrt{6^2 + (-12)^2 + 4^2}} \)
Now, let's calculate the values inside the absolute value and the square root:
- Numerator: \( (6)(4) + (-12)(6) + (4)(8) - a = 24 - 72 + 32 - a \)
- Numerator: \( 56 - 72 - a = -16 - a \)
- Denominator: \( \sqrt{6^2 + (-12)^2 + 4^2} = \sqrt{36 + 144 + 16} = \sqrt{196} \)
- Denominator: \( \sqrt{196} = 14 \)
Substitute these values back into the distance equation:
\( 1 = \frac{|-16 - a|}{14} \)
Multiply both sides by 14:
\( 14 = |-16 - a| \)
Solving for 'a'
The absolute value equation \( |-16 - a| = 14 \) means that the expression inside the absolute value can be either 14 or -14.
Case 1: \( -16 - a = 14 \)
\( -a = 14 + 16 \)
\( -a = 30 \)
\( a = -30 \)
Case 2: \( -16 - a = -14 \)
\( -a = -14 + 16 \)
\( -a = 2 \)
\( a = -2 \)
Thus, the possible values for \( a \) are -2 or -30.
| Step | Description | Calculation |
|---|---|---|
| 1 | Identify point and plane equation parameters. | Point: \( (x_0, y_0, z_0) = (4, 6, 8) \) Plane: \( 6x - 12y + 4z = a \) \( A=6, B=-12, C=4, D''=a \) Distance \( D=1 \) |
| 2 | Write down distance formula. | \( D = \frac{|Ax_0 + By_0 + Cz_0 - D''|}{\sqrt{A^2 + B^2 + C^2}} \) |
| 3 | Substitute values into formula. | \( 1 = \frac{|(6)(4) + (-12)(6) + (4)(8) - a|}{\sqrt{6^2 + (-12)^2 + 4^2}} \) |
| 4 | Simplify numerator and denominator. | Numerator: \( |24 - 72 + 32 - a| = |-16 - a| \) Denominator: \( \sqrt{36 + 144 + 16} = \sqrt{196} = 14 \) |
| 5 | Set up the absolute value equation. | \( 1 = \frac{|-16 - a|}{14} \implies |-16 - a| = 14 \) |
| 6 | Solve for 'a' (Case 1). | \( -16 - a = 14 \implies a = -30 \) |
| 7 | Solve for 'a' (Case 2). | \( -16 - a = -14 \implies a = -2 \) |
| 8 | Final possible values for 'a'. | -2 or -30 |
Revision Table: Point Plane Distance
| Concept | Description | Formula |
|---|---|---|
| Vector Equation of Plane | \( \vec{r} \cdot \vec{n} = d \) where \( \vec{n} \) is the normal vector and \( d \) is a constant. | \( \vec{r} \cdot \vec{n} = d \) |
| Cartesian Equation of Plane | \( Ax + By + Cz = D'' \) where \( (A, B, C) \) are components of the normal vector. | \( Ax + By + Cz = D'' \) |
| Normal Vector | A vector perpendicular to the plane. If the equation is \( Ax + By + Cz = D'' \), the normal vector is \( A\hat{i} + B\hat{j} + C\hat{k} \). | \( \vec{n} = A\hat{i} + B\hat{j} + C\hat{k} \) |
| Distance from Point \( (x_0, y_0, z_0) \) to Plane \( Ax + By + Cz = D'' \) | The shortest distance from the point to any point on the plane. | \( D = \frac{|Ax_0 + By_0 + Cz_0 - D''|}{\sqrt{A^2 + B^2 + C^2}} \) |
Additional Information: Planes and Distances
Understanding the equation of a plane and the distance formulas is crucial in 3D geometry. The normal vector \( \vec{n} \) dictates the orientation of the plane. The constant 'a' in the vector equation \( \vec{r} \cdot \vec{n} = a \) relates to the distance of the plane from the origin, measured along the normal vector.
The distance formula used here is a direct application of projecting the vector connecting the point to any point on the plane onto the normal vector, and taking the magnitude. The absolute value ensures the distance is always non-negative. There are often two possible planes satisfying a given distance from a point if the normal vector is fixed, corresponding to the point being on either side of the plane. In this problem, fixing the normal vector \( (6, -12, 4) \) leads to two possible values for 'a', representing two parallel planes, each at a distance of 1 unit from the point (4,6,8).
This concept is widely used in various fields, including physics (calculating distances in electric fields), computer graphics (rendering and collision detection), and engineering.
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