Three Dimensional Geometry

Let, is the given triangle
Slope of
Slope of
The equation of line passing through A and perpendicular to BO is



and equation of line passing through 0 and perpendicular to AB is

The intersection point of Eqs. (i) and (ii) (0, 0), which is the required orthocentre.

The correct answer is Option C)

O(0, 0, 0), A(a, 0, 0), B(0, a, 0), R(0, 0, a), D(a, a, 0), K(a, 0, a), L(0, a, a,), P(a, a, a) direction-cosines of diagonal Op is proportional to (a, a, a)

so, Op's direct cosines are:

The angle between two diagonals of a cube is cos^–11/3.

Discover More Topics From This Chapter: Three-Dimensional Geometry

We need to find the sine of the angle between the straight line and the plane .

First, rewrite the equation of the line in the standard form: . So, the direction vector of the line is .

The normal vector to the plane is .

Let be the angle between the line and the plane. The angle between the direction vector of the line and the normal vector of the plane is the complement of , i.e., .

So,

Therefore, the sine of the angle between the line and the plane is .

Thus, the correct option is (D) .

Let the point be and the line be .

A general point on the line is . The vector is:

$ \vec{d} = (2, 3, 6) \vec{PQ} \cdot \vec{d} = 0 t = -\frac{1}{49} Q \vec{PQ} d = |\vec{PQ}| \sqrt{14357} \sqrt{293 \cdot 49} = 7\sqrt{293} $

If a line makes an angle of with each of the x and y-axis, then we need to find the acute angle made by the line with the z-axis.

Let be the angles the line makes with the x, y, and z-axis respectively. Then .

We are given that and . Therefore:

Since we want the acute angle, we take the positive value:

Therefore, the acute angle made by the line with the z-axis is .

Thus, the correct option is (A) .

We need to determine which octant the point (1, -3, 4) lies in.

The signs of the coordinates in each octant are as follows:

1. (+, +, +)
2. (-, +, +)
3. (-, -, +)
4. (+, -, +)
5. (+, +, -)
6. (-, +, -)
7. (-, -, -)
8. (+, -, -)

The point (1, -3, 4) has signs (+, -, +). Comparing this to the list of octants, we see it corresponds to the 4th octant.

Therefore, the correct option is (C) Fourth.

If the side of a cube is increased by 5%, then we need to find the percentage increase in the surface area of the cube.

Let the original side of the cube be 'a'. Then the original surface area is .

If the side is increased by 5%, the new side is .

The new surface area is .

The increase in surface area is .

The percentage increase in surface area is .

Since 10.25% is closest to 10%, the correct option is (A) 10%.

We can find the area of the quadrilateral ABCD by dividing it into two triangles, for example, triangle ABC and triangle ADC.
The total area will be the sum of the areas of these two triangles.
The area of a triangle with vertices P, Q, R in 3D space is given by half the magnitude of the cross product of two vectors forming adjacent sides, e.g., Area = .

First, calculate the area of triangle ABC. The vertices are A(0, 4, 1), B(2, 3, -1), C(4, 5, 0). Find the vectors and :

Calculate the cross product :

Calculate the magnitude of the cross product: The area of triangle ABC is:

Next, calculate the area of triangle ADC. The vertices are A(0, 4, 1), D(2, 6, 2), C(4, 5, 0). Find the vectors and :

Calculate the cross product :

Calculate the magnitude of the cross product: The area of triangle ADC is:

The total area of the quadrilateral ABCD is the sum of the areas of the two triangles:

Alternatively, if the quadrilateral is planar (which can be confirmed because the normal vectors and are parallel), the area can be calculated as half the magnitude of the cross product of its diagonals. Diagonal = (4, 1, -1) Diagonal = D - B = (2 - 2, 6 - 3, 2 - (-1)) = (0, 3, 3) Both methods yield the same result.

The area of the quadrilateral ABCD is (A): 9 sq. units.

Let the direction cosines of the first line be . Let the direction cosines of the second line be .

Let be the angle between the two lines. The cosine of the angle between two lines with direction cosines and is given by the formula:

Substitute the given direction cosines into the formula:

Simplify the expression:

We need to find the angle such that and . We know that . Since is negative, lies in the second quadrant. Therefore, .

The angle between two lines is conventionally taken as the acute angle (or ). If the angle obtained from the dot product formula is obtuse ( ), the acute angle between the lines is . In this case, is obtuse. The acute angle between the lines is .

The angle between the lines is (C): .

The equation of the line joining two points and is given by the parametric form: Substituting the given points and into the formula, we get: This simplifies to: Now, multiplying through by the denominators to match the options:

So, the correct answer is (B) : .

To find the equation of the line passing through the points and , we use the vector form of a line in three-dimensional space. The equation for a line passing through the point with direction ratios is given by:

First, let's determine the direction ratios by subtracting the coordinates of the given points:

• Direction ratio for :
• Direction ratio for :
• Direction ratio for :

Thus, the direction ratios are . However, these ratios can be simplified by dividing each by 2:

Resulting simplified direction ratios are .

Now, substitute these values into the line equation form using the point :

To compare with the given options, we observe that by multiplying each term by -1, we get the correct formation:

This matches the given correct option: .

Therefore, the correct option is:

To find the angle between two lines whose direction cosines are given, we can use the formula for the cosine of the angle between two vectors. The direction cosines for the lines are given as and .

The cosine of the angle between two lines is given by the dot product of their direction cosines:

Substitute the given direction cosines:

Perform the dot product calculation:

Adding these results gives:

This simplifies to:

Knowing that the cosine of the angle is , we can relate it to the angle as:

However, as we correctly relate the provided solutions, the angle of is indeed because in geometrical context, angles tend to be acute or obtuse

Thus, the correct answer is:

To find the equation of the plane that meets the coordinate axes at points , , and , with the centroid of triangle being at the point , we use the following approach:

• The general form of a plane that intersects the , , and axes at points , , and respectively is: .
• The coordinates of point are , of point are , and of point are .
• The centroid of triangle is: .
• We are given that the centroid is at the point . So, we have:
• Substituting these values into the plane equation , we get: .

Therefore, the equation of the plane is , which matches the second option given.

The correct answer is: (C):

We are tasked with finding the equation of the plane that passes through the points , , and .

Step 1: Find two vectors on the plane

Let the points be , , and . To find the equation of the plane, we first need two vectors that lie on the plane. We can find these vectors by subtracting the coordinates of one point from the others:

Step 2: Find the normal vector to the plane

The normal vector to the plane is given by the cross product of and :

Now, compute the cross product using the determinant formula:

Step 3: Calculate the cross product

Expanding the determinant, we get:

Now compute each 2x2 determinant:

Step 4: Use the point-normal form of the plane equation

The equation of the plane can be written as:

Here, is a general point on the plane, and is a point on the plane. Substituting and , we get:

Step 5: Simplify the equation

Expanding and simplifying:

Conclusion:
The equation of the plane is , so the correct answer is (C): .

The correct answer is: (B):

We are given that is the foot of the perpendicular from the point 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 to the plane is given by the vector connecting the point to the foot of the perpendicular . The direction vector is:

Step 2: Use the point-normal form of the plane equation

The equation of a plane can be written as:

Here, is the normal vector to the plane, is a point on the plane, and is the foot of the perpendicular.

Since is the direction of the perpendicular, it is also normal to the plane. Thus, the normal vector to the plane is .

Step 3: Set up the equation of the plane

Now, use the point-normal form of the plane equation with and :

-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 , so the correct answer is (B): .

The correct answer is: (A): (2, 6, -4)

We are tasked with finding the point of intersection of the line with the plane .

Step 1: Parametrize the line equation
The given line equation is symmetric, so we introduce a parameter and express , , and in terms of :

Now solve for , , and in terms of :

Step 2: Substitute into the plane equation
The equation of the plane is . Substitute , , and into this equation:

Step 3: Simplify the equation
Expand each term:

Now, combine like terms:

Step 4: Solve for
Now, solve for :

Step 5: Find the coordinates of the point of intersection
Substitute back into the parametric equations for , , and :

Conclusion:
The point of intersection is , so the correct answer is (A): (2, 6, -4).

The correct answer is: (C):

We are given the point and the line equation:

Step 1: Parametrize the line equation

We start by parametrizing the given line equation. Let be the parameter. The parametric form of the line is:

Step 2: Find the direction vector of the line

The direction vector of the line is the vector of coefficients of , which is:

Step 3: Find the vector from the point to the line

The point on the line corresponding to the parameter is . The vector from this point to is:

Simplifying the vector components:

Step 4: Use the formula for the perpendicular distance

The perpendicular distance from a point to a line is given by the formula:

Step 5: Compute the cross product

We first compute the cross product . We have:

Using the determinant formula for the cross product, we calculate:

After evaluating the determinant, we get:

Step 6: Find the magnitude of the cross product

Now, we find the magnitude of :

Step 7: Solve for the perpendicular distance

Finally, we divide the magnitude of the cross product by the magnitude of , which is:

After simplifying the expression, we find that the perpendicular distance is:

Conclusion:
The length of the perpendicular drawn from the point to the line is , so the correct answer is (C): .

The correct answer is: (C):

We are given that the line makes an angle of with both the X-axis and the Y-axis, and we are asked to find the acute angle it makes with the Z-axis.

Step 1: Understand the direction cosines

The direction cosines of a line are the cosines of the angles the line makes with the coordinate axes. If the line makes angles , , and with the X, Y, and Z axes respectively, then the direction cosines are given by:

Step 2: Use the known angles

We are told that the line makes an angle of with both the X-axis and the Y-axis. Therefore, the direction cosines for the X and Y axes are:

Step 3: Use the property of direction cosines

The sum of the squares of the direction cosines of a line is always equal to 1. Therefore, we have:

Substitute into the equation:

Simplifying:

Step 4: Solve for

Now, take the square root of both sides:

Therefore, the acute angle is:

Conclusion:
The acute angle made by the Z-axis is , so the correct answer is (C): .