Three Dimensional Geometry Test! Trivia Quiz

The direction cosines of a vector can be found by dividing each component of the vector by the magnitude of the vector. In this case, the magnitude of the vector -2i + j - 5k can be calculated using the Pythagorean theorem as sqrt((-2)^2 + 1^2 + (-5)^2) = sqrt(30). Therefore, the direction cosines can be found by dividing each component by sqrt(30), resulting in -2/sqrt(30), 1/sqrt(30), and -5/sqrt(30).

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The length of the perpendicular drawn from the origin to a plane can be found by using the formula d = |Ax + By + Cz + D| / sqrt(A^2 + B^2 + C^2), where A, B, C are the coefficients of x, y, z in the equation of the plane, and D is the constant term. In this case, the equation of the plane is 2x - 3y + 6z + 21 = 0. By substituting the values into the formula, we get d = |0 + 0 + 0 + 21| / sqrt(2^2 + (-3)^2 + 6^2) = 21 / sqrt(49) = 3. Therefore, the length of the perpendicular is 3.

The direction cosines of a line are the cosines of the angles that the line makes with the coordinate axes. In this case, the line makes angles of 90°, 135°, and 45° with the x, y, and z-axes respectively. The direction cosines are calculated by taking the cosine of each angle. Therefore, the direction cosines of the line are cos(90°), cos(135°), and cos(45°).

The vector equation of a plane can be found by taking the cross product of the normal vectors of the two given planes. The line of intersection of the planes is contained within the plane, so the normal vector of the plane will be perpendicular to this line. Therefore, the cross product of the normal vectors of the given planes will give us the normal vector of the desired plane.

The two lines are parallel because they have the same direction vector. The direction vector of the first line, AB, is (2-4, 3-7, 4-8) = (-2, -4, -4), and the direction vector of the second line, CD, is (1-(-1), 2-(-2), 5-1) = (2, 4, 4). Since the direction vectors are scalar multiples of each other, the lines are parallel.

The equation of the plane passing through the line of intersection of the given planes can be found by taking the cross product of the normal vectors of the two planes. The resulting vector will be perpendicular to both planes and can be used as the normal vector for the new plane. Since the perpendicular distance from the origin is unity, the equation of the plane can be written as Ax + By + Cz = 1, where A, B, and C are the components of the normal vector.

To find the coordinates of the point where the line intersects the plane, we need to solve the system of equations formed by the line and the plane. By substituting the equation of the line into the equation of the plane, we can find the values of x, y, and z. Once we have the coordinates of the point of intersection, we can calculate the angle between the line and the plane using the dot product formula. The dot product of the direction vector of the line and the normal vector of the plane divided by the product of their magnitudes will give us the cosine of the angle. Taking the inverse cosine of this value will give us the angle between the line and the plane.

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The equation of the plane through the line of intersection of the given planes can be found by taking the cross product of the normal vectors of the two given planes. The normal vectors of the planes x + y + z = 1 and 2x + 3y + 4z = 5 are and respectively. Taking the cross product of these two vectors gives us the normal vector of the desired plane, which is . Therefore, the equation of the plane is of the form x - z + c = 0, where c is a constant. Substituting any point on the plane, such as (0, 0, 2), into the equation gives us c = 2. Thus, the equation of the plane is x - z + 2 = 0.

The given lines are represented by the equations:

Line 1: x = -1, y = -6t, z = 12t
Line 2: x = -1, y = 6t, z = -12t

To show that the lines intersect, we need to find a value of t that satisfies both equations.

Comparing the y-coordinate equations, we have -6t = 6t. Solving for t, we get t = 0.

Substituting t = 0 into the equations, we find that both lines pass through the point (-1, -6, -12). Therefore, this is the point of intersection.

To find the vector equation of the plane through the given points and perpendicular to the given plane, we first need to find two vectors that lie in the plane. We can do this by subtracting the coordinates of one point from the other:

Vector A = (-1, 3, 4) - (2, 1, -1) = (-3, 2, 5)

Next, we need to find the normal vector of the given plane x - 2y + 4z = 10. The coefficients of x, y, and z in the equation represent the components of the normal vector.

Normal vector B = (1, -2, 4)

Finally, we can use the cross product of vectors A and B to find the direction vector of the plane:

Direction vector = A x B = (-3, 2, 5) x (1, -2, 4) = (18, -17, -8)

Therefore, the vector equation of the plane is r = (2, 1, -1) + t(18, -17, -8), where t is a scalar.

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The line through (3, -4, -5) and (2, -3, 1) can be represented by the parametric equations: x = 3 + t, y = -4 + t, z = -5 + 6t. By substituting these equations into the equation of the plane passing through (2, 2, 1), (3, 0, 1), and (4, -1, 0), we can solve for t to find the point of intersection. By substituting the value of t into the parametric equations, we obtain the coordinates of the point of intersection as (1, -2, 7).

The given equation represents a plane in Cartesian form. The coefficients of x, y, and z determine the direction of the normal vector to the plane. Since the equation is the same as the given equation of the plane, it means that the line of intersection of the planes lies on this plane. Therefore, the equation 33x + 45y + 50z - 41 = 0 represents the plane that contains the line of intersection of the given planes.