(a) To find the vector equation of the line L, we first need to find the direction vector of the line. This is given by the difference between the coordinates of the two points the line passes through.

So, the direction vector, d = (4-1, 6-2, -2-3) = (3, 4, -5)

The vector equation of a line is given by r = a + td, where a is a point on the line, t is a scalar and d is the direction vector.

So, the vector equation of the line L is r = (1, 2, 3) + t(3, 4, -5)

(b) To find the point of intersection between the line L and the plane Π, we substitute the parametric equations of the line into the equation of the plane.

So, we have x + y - z = 24 becomes (1+3t) + (2+4t) - (3-5t) = 24

Solving this equation gives t = 2

Substituting t = 2 back into the parametric equations of the line gives the point of intersection as (7, 10, -7)

(c) The cosine of the angle between the line L and the normal to the plane Π is given by the dot product of the direction vector of the line and the normal vector of the plane, divided by the product of their magnitudes.

The normal vector of the plane is (1, 1, -1)

So, cos(θ) = ((3, 4, -5) . (1, 1, -1)) / (||(3, 4, -5)|| ||(1, 1, -1)||)

Calculating this gives cos(θ) = -2/√50

(d) The Cartesian equation of a plane is given by ax + by + cz = d, where (a, b, c) is a normal vector to the plane.

Since the plane is perpendicular to Π, its normal vector is parallel to the line L, so we can use the direction vector of L, (3, 4, -5), as the normal vector.

Substituting the coordinates of a point on the line, say (1, 2, 3), into the equation gives 3*1 + 4*2 - 5*3 = d

Solving this gives d = -1

So, the Cartesian equation of the plane is 3x + 4y - 5z = -1

Question

(a) To find the vector equation of the line L, we first need to find the direction vector of the line. This is given by the difference between the coordinates of the two points the line passes through.

So, the direction vector, d = (4-1, 6-2, -2-3) = (3, 4, -5)

The vector equation of a line is given by r = a + td, where a is a point on the line, t is a scalar and d is the direction vector.

So, the vector equation of the line L is r = (1, 2, 3) + t(3, 4, -5)

(b) To find the point of intersection between the line L and the plane Π, we substitute the parametric equations of the line into the equation of the plane.

So, we have x + y - z = 24 becomes (1+3t) + (2+4t) - (3-5t) = 24

Solving this equation gives t = 2

Substituting t = 2 back into the parametric equations of the line gives the point of intersection as (7, 10, -7)

(c) The cosine of the angle between the line L and the normal to the plane Π is given by the dot product of the direction vector of the line and the normal vector of the plane, divided by the product of their magnitudes.

The normal vector of the plane is (1, 1, -1)

So, cos(θ) = ((3, 4, -5) . (1, 1, -1)) / (||(3, 4, -5)|| ||(1, 1, -1)||)

Calculating this gives cos(θ) = -2/√50

(d) The Cartesian equation of a plane is given by ax + by + cz = d, where (a, b, c) is a normal vector to the plane.

Since the plane is perpendicular to Π, its normal vector is parallel to the line L, so we can use the direction vector of L, (3, 4, -5), as the normal vector.

Substituting the coordinates of a point on the line, say (1, 2, 3), into the equation gives 3*1 + 4*2 - 5*3 = d

Solving this gives d = -1

So, the Cartesian equation of the plane is 3x + 4y - 5z = -1

Knowee AI · Accepted Answer