Determine the vector equation of the plane passing through the intersection of the planes \(\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6\) and \(\vec{r}. (2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})=-5\) , and the point (1, 1, 1)?
Vector Equation of a Plane
To determine the vector equation of a plane that passes through the intersection of two given planes and a specific point, we will follow a systematic approach. The general equation of a plane passing through the intersection of two planes \(P_1: \vec{r} \cdot \vec{n_1} = d_1\) and \(P_2: \vec{r} \cdot \vec{n_2} = d_2\) is given by:
\[ (\vec{r} \cdot \vec{n_1} - d_1) + \lambda (\vec{r} \cdot \vec{n_2} - d_2) = 0 \] This can be rearranged as:
\[ \vec{r} \cdot (\vec{n_1} + \lambda \vec{n_2}) = d_1 + \lambda d_2 \]
Identifying Given Plane Equations
We are given two plane equations:
- First plane (\(P_1\)): \( \vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6 \)
- Second plane (\(P_2\)): \( \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})=-5 \)
From these equations, we can identify the normal vectors (\(\vec{n_1}\), \(\vec{n_2}\)) and constants (\(d_1\), \(d_2\)):
- For \(P_1\): \( \vec{n_1} = \hat{\imath}+\hat{\jmath}+\hat{k} \) and \( d_1 = 6 \)
- For \(P_2\): \( \vec{n_2} = 2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k} \) and \( d_2 = -5 \)
Forming the Combined Plane Equation
Now, substitute these values into the general equation for a plane passing through the intersection:
\[ \vec{r} \cdot ((\hat{\imath}+\hat{\jmath}+\hat{k}) + \lambda (2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})) = 6 + \lambda (-5) \]
Combine the terms on the left side:
\[ \vec{r} \cdot ((1+2\lambda)\hat{\imath} + (1+3\lambda)\hat{\jmath} + (1+4\lambda)\hat{k}) = 6 - 5\lambda \quad \textbf{(Equation 1)} \]
Using the Given Point to Find \(\lambda\)
The problem states that the desired plane also passes through the point (1, 1, 1). This means the position vector \(\vec{r} = \hat{\imath}+\hat{\jmath}+\hat{k}\) must satisfy Equation 1. Substitute \(\vec{r} = \hat{\imath}+\hat{\jmath}+\hat{k}\) into Equation 1:
\[ (\hat{\imath}+\hat{\jmath}+\hat{k}) \cdot ((1+2\lambda)\hat{\imath} + (1+3\lambda)\hat{\jmath} + (1+4\lambda)\hat{k}) = 6 - 5\lambda \]
Perform the dot product:
\[ 1(1+2\lambda) + 1(1+3\lambda) + 1(1+4\lambda) = 6 - 5\lambda \]
Simplify and solve for \(\lambda\):
\[ 1+2\lambda + 1+3\lambda + 1+4\lambda = 6 - 5\lambda \] \[ 3 + 9\lambda = 6 - 5\lambda \] \[ 9\lambda + 5\lambda = 6 - 3 \] \[ 14\lambda = 3 \] \[ \lambda = \frac{3}{14} \]
Substituting \(\lambda\) Back into the Equation
Now, substitute the value of \( \lambda = \frac{3}{14} \) back into Equation 1:
\[ \vec{r} \cdot \left( \left(1+2\left(\frac{3}{14}\right)\right)\hat{\imath} + \left(1+3\left(\frac{3}{14}\right)\right)\hat{\jmath} + \left(1+4\left(\frac{3}{14}\right)\right)\hat{k} \right) = 6 - 5\left(\frac{3}{14}\right) \]
Calculate the coefficients for \(\hat{\imath}\), \(\hat{\jmath}\), and \(\hat{k}\):
- Coefficient of \(\hat{\imath}\): \( 1+2\left(\frac{3}{14}\right) = 1+\frac{6}{14} = 1+\frac{3}{7} = \frac{7+3}{7} = \frac{10}{7} \)
- Coefficient of \(\hat{\jmath}\): \( 1+3\left(\frac{3}{14}\right) = 1+\frac{9}{14} = \frac{14+9}{14} = \frac{23}{14} \)
- Coefficient of \(\hat{k}\): \( 1+4\left(\frac{3}{14}\right) = 1+\frac{12}{14} = 1+\frac{6}{7} = \frac{7+6}{7} = \frac{13}{7} \)
Calculate the right-hand side:
\[ 6 - 5\left(\frac{3}{14}\right) = 6 - \frac{15}{14} = \frac{84-15}{14} = \frac{69}{14} \]
So, the equation becomes:
\[ \vec{r} \cdot \left( \frac{10}{7}\hat{\imath} + \frac{23}{14}\hat{\jmath} + \frac{13}{7}\hat{k} \right) = \frac{69}{14} \]
Final Vector Equation of the Plane
To clear the denominators and obtain integer coefficients, multiply the entire equation by the least common multiple of the denominators (7, 14, 7), which is 14:
\[ 14 \times \left[ \vec{r} \cdot \left( \frac{10}{7}\hat{\imath} + \frac{23}{14}\hat{\jmath} + \frac{13}{7}\hat{k} \right) \right] = 14 \times \frac{69}{14} \]
\[ \vec{r} \cdot \left( \left(14 \times \frac{10}{7}\right)\hat{\imath} + \left(14 \times \frac{23}{14}\right)\hat{\jmath} + \left(14 \times \frac{13}{7}\right)\hat{k} \right) = 69 \]
\[ \vec{r} \cdot (20\hat{\imath} + 23\hat{\jmath} + 26\hat{k}) = 69 \]
This is the vector equation of the plane passing through the intersection of the given planes and the point (1, 1, 1).
Comparing with Options
Let's compare our derived equation with the given options:
| Option | Vector Equation |
|---|---|
| 1 | \( \vec{r} \cdot(10 \hat{\imath}+13 \hat{\jmath}+23 \hat{k})=69 \) |
| 2 | \( \vec{r} \cdot(20 \hat{\imath}+23 \hat{\jmath}+26 \hat{k})=69 \) |
| 3 | \( \vec{r} \cdot(30 \hat{\imath}+23 \hat{\jmath}+13 \hat{k})=69 \) |
| 4 | \( \vec{r} \cdot(10 \hat{\imath}+23 \hat{\jmath}+13 \hat{k})=69 \) |
Our calculated equation, \( \vec{r} \cdot (20\hat{\imath} + 23\hat{\jmath} + 26\hat{k}) = 69 \), matches option 2.
Important Questions from Equation of a Plane
If the foot of the perpendicular drawn from (-2, 1, 0) on a plane is (1, -2, 1), then the equation of the plane is
The equation of the plane which contain the points (0, 6, 0) and (-2, -3, 4) and which is parallel to the ray with direction ratios (2, 3, -2) is:
The image of the point (–3, 8, 4) in the plane 6x – 3y – 2z + 1 = 0, is -
The equation of the plane through the point (1, 2, –3) and normal to the straight line joining the points (–1, 3, 4) and (5, 2, –1) is-
The equation of the plane passing through the intersection of the planes 2x + y + 2z = 9, 4x – 5y – 4z = 1 and the point (3, 2, 1) is
