To find the equation of the plane when you know a point and the foot of the perpendicular, use the point-normal form of the plane equation. The direction of the perpendicular vector is normal to the plane, and you can use it as the normal vector for the plane equation.
The correct answer is: (B): \( 2x - y + 2z + 1 = 0 \)
We are given that \( (2, 3, -1) \) is the foot of the perpendicular from the point \( (4, 2, 1) \) to a plane, and we are tasked with finding the equation of the plane.
Step 1: Find the direction vector of the perpendicular
The direction of the perpendicular from the point \( (4, 2, 1) \) to the plane is given by the vector connecting the point \( (4, 2, 1) \) to the foot of the perpendicular \( (2, 3, -1) \). The direction vector \( \vec{v} \) is:
\( \vec{v} = (2 - 4, 3 - 2, -1 - 1) = (-2, 1, -2) \)
Step 2: Use the point-normal form of the plane equation
The equation of a plane can be written as:
\( \vec{n} \cdot (\vec{r} - \vec{P}) = 0 \)
Here, \( \vec{n} \) is the normal vector to the plane, \( \vec{r} = (x, y, z) \) is a point on the plane, and \( \vec{P} = (2, 3, -1) \) is the foot of the perpendicular.
Since \( \vec{v} = (-2, 1, -2) \) is the direction of the perpendicular, it is also normal to the plane. Thus, the normal vector to the plane is \( \vec{n} = (-2, 1, -2) \).
Step 3: Set up the equation of the plane
Now, use the point-normal form of the plane equation with \( \vec{n} = (-2, 1, -2) \) and \( \vec{P} = (2, 3, -1) \):
-2(x - 2) + 1(y - 3) - 2(z + 1) = 0
Step 4: Simplify the equation
Now, expand and simplify the equation:
-2x + 4 + y - 3 - 2z - 2 = 0
-2x + y - 2z - 1 = 0
Multiplying through by -1 gives:
2x - y + 2z + 1 = 0
Conclusion:
The equation of the plane is \( 2x - y + 2z + 1 = 0 \), so the correct answer is (B): \( 2x - y + 2z + 1 = 0 \).
Show that the following lines intersect. Also, find their point of intersection:
Line 1: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]
Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
The vector equations of two lines are given as:
Line 1: \[ \vec{r}_1 = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(4\hat{i} + 6\hat{j} + 12\hat{k}) \]
Line 2: \[ \vec{r}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(6\hat{i} + 9\hat{j} + 18\hat{k}) \]
Determine whether the lines are parallel, intersecting, skew, or coincident. If they are not coincident, find the shortest distance between them.
Determine the vector equation of the line that passes through the point \( (1, 2, -3) \) and is perpendicular to both of the following lines:
\[ \frac{x - 8}{3} = \frac{y + 16}{7} = \frac{z - 10}{-16} \quad \text{and} \quad \frac{x - 15}{3} = \frac{y - 29}{-8} = \frac{z - 5}{-5} \]
You are given a dipole of charge \( +q \) and \( -q \) separated by a distance \( 2l \). A sphere 'A' of radius \( R \) passes through the centre of the dipole as shown below and another sphere 'B' of radius \( 2R \) passes through the charge \( +q \). Then the electric flux through the sphere A is