Angle Between A Line And A Plane

To solve this problem, we need to determine the intercept made by the plane P on the y-axis, given that it passes through the intersection of two mutually perpendicular planes and makes a unit length intercept on the positive x-axis.

Step-by-Step Solution:

1. First, identify the equations of the two perpendicular planes:
   • Plane 1:
   • Plane 2:
2. Two planes are perpendicular if their normal vectors are orthogonal. The normal vector of Plane 1 is and for Plane 2 is . Their dot product should be zero:
   • This simplifies to .
   • Rearranging gives: .
3. Solve for from the quadratic equation:
   • Using the quadratic formula: , where , , .
   • .
   • The solutions are and .
   • Since , we select .
4. Substituting into the plane equations, we now have:
   • Plane 1:
   • Plane 2:
5. The equation of Plane , which passes through the intersection of these planes, can be written as:
   • This simplifies to: .
6. Since the plane makes a unit intercept on the x-axis, substitute , , and :
   • Solving:
   • This gives .
7. Substituting in the plane equation, the equation becomes:
   • This simplifies to .
8. Rewriting the equation in x, y, z intercept form:
   • The y-intercept is .

Hence, the y-intercept of the plane is 11.

To solve this problem, we need to understand the scenario given and apply trigonometric principles to find the value of .

In the problem, we are dealing with a vertical tower AB with a height of . A point P on the ground is such that the observer at P can see the height of the tower at an angle of elevation of . Then, the observer moves a distance towards the tower such that the angle of elevation to the top B is .

Given:

• Height of tower
• Initial angle of elevation to height is .
• After moving distance , angle of elevation to top B is .

We start by applying the concept of angle of elevation:

1. From the initial position, with angle :

Where is the horizontal distance from P to the base of the tower at height .

2. From the new position (after moving distance ) with angle to the top B:

Rearrange these equations. First, express in terms of the tangent of angles:

Plugging this into the second equation:

Simplifying gives:

As per double angle identity for tangent:

Using the trigonometric identity and solving gives the expression for :

The correct option from the given list for is:

√7-2, which matches after evaluating the trigonometric expression and eliminating incorrect options based on numerical calculations.

Consider a right circular cone with semi-vertical angle given by tan^-1(3/4). The relationship is tan(θ) = 3/4, where θ is the semi-vertical angle.

The dimensions of the cone can be described in terms of the variables: r (radius of the water surface) and h (depth of the water).

From tan(θ) = r/h and given tan(θ) = 3/4, we derive the relationship:

r = (3/4)h

Water is poured into the tank at 6 m³/h, hence dV/dt = 6 m³/h.

For a cone, the volume is V = (1/3)πr²h. Substituting r = (3/4)h:

V = (1/3)π((3/4)h)²h = (1/3)π(9/16)h³ = (3π/16)h³

Taking the derivative with respect to t:

dV/dt = (9π/16)h²(dh/dt)

Given dV/dt = 6, solve for dh/dt:

6 = (9π/16)h²(dh/dt)

dh/dt = (96/(9πh²))

At h = 4:

dh/dt = (96/(9π(4)²)) = (96/(144π)) = 2/(3π)

The wet curved surface area A of the cone is A = πrl, where l is the slant height, determined by l = √(r² + h²). Given r = (3/4)h:

l = √((3/4)h² + h²) = h√(9/16 + 16/16) = h√(25/16) = (5/4)h

Then A = πr(5/4)h = π(3/4)h(5/4)h = (15π/16)h²

The rate of change of surface area is:

dA/dt = (15π/8)h(dh/dt)

Substitute h = 4 and dh/dt = 2/(3π):

dA/dt = (15π/8)(4)(2/(3π)) = 5/2 m²/h

The rate at which the wet curved surface area increases is 2.5 m²/h.

Confirming this value within the range 5,5:

The computed value 2.5 fits perfectly within this range.

Plane Intersection and Parallel Line

Let the given planes be P₁ : 2x − y + z − 3 = 0 and P₂ : 4x − 3y + 5z + 9 = 0. The equation of the family of planes passing through the line of intersection of P₁ and P₂ is given by:

P₁ + λP₂ = 0 ⇒ (2x − y + z − 3) + λ(4x − 3y + 5z + 9) = 0.

(2 + 4λ)x + (−1 − 3λ)y + (1 + 5λ)z + (−3 + 9λ) = 0.

The given plane is parallel to the line . The direction vector of the line is v_L = (−2, 4, 5). The normal vector of the plane is n_p = (2 + 4λ, −1 − 3λ, 1 + 5λ). Since the plane is parallel to the line, the normal vector of the plane is perpendicular to the direction vector of the line. So, n_p · v_L = 0.

−2(2 + 4λ) + 4(−1 − 3λ) + 5(1 + 5λ) = 0.
−4 − 8λ − 4 − 12λ + 5 + 25λ = 0.
−3 + 5λ = 0 ⇒ λ = .

Substituting λ = in the equation of the plane:

.

.
22x − 14y + 20z + 12 = 0.
11x − 7y + 10z + 6 = 0.

Comparing this with ax + by + cz + 6 = 0, we get a = 11, b = −7, and c = 10.
Therefore, a + b + c = 11 − 7 + 10 = 14.

Step 1: Find the value of and
We are given the equation of the plane and point . The foot of the perpendicular from point A to the plane is denoted by N. The coordinates of N are found by using the formula for the foot of perpendicular from a point to a plane.
From the plane equation , we get the equation of the line joining to : Solving this gives , , and , so the coordinates of N are .
Step 2: Find
The distance is given by: Thus, Step 3: Use the area condition
The area of triangle ABN is given by the formula for the area of a triangle in 3D space: We know that the area is , so we can solve for the unknowns and .
Step 4: Solve for and
Substituting the value of into the area formula and simplifying, we get: Simplifying further: From the area condition, we get a system of equations to solve for and .
Step 5: Final Answer
After solving the system, we find that: Now, calculate : Thus, .

The points and are the intersection points of the given line and the mirror image of the parabola.
From the geometry of the problem, the area of is given by: Thus, . The distance between the points and is computed from the coordinates of the points: Thus, .

We are given the equations of the circle and the hyperbola: and Step 1: Solve the system of equations.
From equation (1), we complete the square: This represents a circle centered at with radius 4. Next, substitute equation (2) into this. Multiply both sides of the equation by 36: Solving this gives us: Substitute this into the circle's equation: Thus, the points of intersection are and .
Step 2: Calculate the centroid. The centroid of is given by the average of the coordinates of , , and . The coordinates of lie on the line . The centroid condition gives the equation: Thus, the centroid lies on the line .