Three Dimensional Geometry Planes

Given planes are  $ x+y+z-1=0.....(1)$ and $x+2y-4z-5=0.........(2)$
Therefore equation of planes bisecting these planes are
$\dfrac{x+y+z-1}{\sqrt{3}}=\pm\dfrac{x+2y-4z-5}{\sqrt{21}}$

In Euclidean geometry, a point has no dimensions, a line has one dimension (length), and a plane has two dimensions (length and width).

The equation of a plane that passes through a point (x0, y0, z0) and has normal vector n = (a, b, c) is given by the formula a(x - x0) + b(y - y0) + c(z - z0) = 0. Substituting the given values, we get 2(x - 1) - 1(y - 2) + 3(z - 3) = 0, which simplifies to 2x - y + 3z = 8.

The Miller index for a plane is determined by the reciprocals of the intercepts of the plane with the crystallographic axes. In this case, the Miller index is (234).

To find the equation of the plane, we can use the point-normal form: (\vec{r} \cdot \vec{n} = d), where (\vec{r}) is a vector from the origin to any point on the plane, (\vec{n}) is the normal vector, and (d) is a constant. Plugging in the given values, we get: ((x - 1) \cdot 2 + (y - 2) \cdot (-1) + (z - 3) \cdot 3 = d) which simplifies to: (2x - y + 3z = d). Since the plane passes through the point ((1, 2, 3)), we can plug in these values to find the value of (d): (2(1) - (2) + 3(3) = d) which gives (d = 10). Therefore, the equation of the plane is: (2x - y + 3z = 10).

To find the equation of the plane, we can use the vector equation of a plane: r = r0 + s * v1 + t * v2, where r0 is a point on the plane, v1 and v2 are vectors parallel to the plane, and s and t are scalar parameters. We can choose r0 to be the point (1, 2, 3), and v1 and v2 to be the vectors (2, 3, 4) - (1, 2, 3) = (1, 1, 1) and (3, 4, 5) - (1, 2, 3) = (2, 2, 2), respectively. Substituting these values into the vector equation, we get: r = (1, 2, 3) + s * (1, 1, 1) + t * (2, 2, 2). To convert this into an equation of the plane, we can set s = x - 1 and t = y - 2, which gives: r = (1, 2, 3) + (x - 1) * (1, 1, 1) + (y - 2) * (2, 2, 2). Expanding this equation, we get: r = (1 + x - 1, 2 + y - 2, 3 + x - 1 + 2y - 4) = (x, y, x + 2y - 3). Therefore, the equation of the plane is x + 2y + 3z = 14.

The distance from a point (x0, y0, z0) to a plane Ax + By + Cz + D = 0 is given by the formula: distance = |Ax0 + By0 + Cz0 + D| / √(A^2 + B^2 + C^2). Substituting the values of the point and the plane, we get: distance = |2 * 2 + 3 * 3 - 4 * 4 + 10| / √(2^2 + 3^2 + (-1)^2) = |4 + 9 - 16 + 10| / √(4 + 9 + 1) = |7| / √14 = 7 / √14 ≈ 2. Therefore, the distance from the point (2, 3, 4) to the plane 2x + 3y - z = 10 is approximately 2 units.

To find the line of intersection of two planes, we can solve the system of equations formed by the two plane equations. Substituting the second equation into the first equation, we get: x + y + (2x - y + z) = 6, which simplifies to 3x + z = 4. This equation represents a plane parallel to the y-axis. To find the line of intersection, we can choose a point on this plane, such as (1, 0, 1), and find the direction vector of the line. The direction vector can be found by taking the cross product of the normal vectors of the two planes. The normal vector of the first plane is (1, 1, 1), and the normal vector of the second plane is (2, -1, 1). Taking the cross product of these vectors, we get: (1, 1, 1) x (2, -1, 1) = (-2, -3, 3). Therefore, the direction vector of the line of intersection is (-2, -3, 3). Using the point (1, 0, 1) and the direction vector (-2, -3, 3), we can write the parametric equations of the line of intersection as: x = 1 - 2t, y = 0 - 3t = -3t, z = 1 + 3t.

The angle between two planes can be found using the formula: angle = cos^-1((n1 · n2) / (|n1| * |n2|)), where n1 and n2 are the normal vectors of the planes. The normal vector of the first plane is (2, 1, -1), and the normal vector of the second plane is (1, -1, 2). Substituting these values into the formula, we get: angle = cos^-1(((2, 1, -1) · (1, -1, 2)) / (|(2, 1, -1)| * |(1, -1, 2)|)) = cos^-1((2 - 1 + 2) / (√(2^2 + 1^2 + (-1)^2) * √(1^2 + (-1)^2 + 2^2))) = cos^-1(3 / √6 * √6) = cos^-1(3 / 6) = cos^-1(1/2) ≈ 60°. Therefore, the angle between the planes 2x + y - z = 3 and x - y + 2z = 5 is approximately 60°.

To find the equation of the tangent plane, we need to find the partial derivatives of the function (f(x, y) = x^2 + y^2) and evaluate them at the given point. The partial derivatives are (f_x(x, y) = 2x) and (f_y(x, y) = 2y). At the point ((1, 2, 5)), the partial derivatives are (f_x(1, 2) = 2) and (f_y(1, 2) = 4). Using the point-normal form, the equation of the tangent plane is (z - 5 = 2(x - 1) + 4(y - 2)), which simplifies to (z = 5 + 2x + 4y).

The intersection of two planes is a line.

The equation of a plane in three-dimensional space is given by $Ax + By + Cz + D = 0$, where A, B, C, and D are constants.

The equation of a line that is parallel to the plane $Ax + By + Cz + D = 0$ and passes through the point $(x_0, y_0, z_0)$ is given by $x = x_0 + At$, $y = y_0 + Bt$, $z = z_0 + Ct$, where A, B, and C are the coefficients of the plane equation.