Three Dimensional Geometry

Given: The plane equation is:

Step 1: Find the direction ratios

The line perpendicular to the plane has its direction ratios given by the coefficients of the plane equation:

Step 2: Use the symmetric form equation

The equation of the required line passing through (7, 5, 3) and having direction ratios (3, 2, 1) is:

Final Answer:

Step 1: Parametric Equations of the Line
The given line equation is: This can be rewritten as:

Step 2: Use the Given Point P(x,1,4)
Since the point lies on the line, we substitute into the equation:

Step 3: Solve for

Step 4: Find
Substituting into the equation for :

Final Answer: .

Step 1: Use the direction cosine formula
The cosine of the angle made by a vector with the positive x-axis is given by: where is the x-component of .

Step 2: Compute the magnitude of
Given: The magnitude is:

Step 3: Compute

Final Answer: is .

We are given two lines and . The parametric equation of a line joining two points and is given by: where is a parameter. For the line joining and , the parametric equations are: Thus, the parametric form of is . For the line joining and , the parametric equations are: Since these lines are collinear and intersect at a fixed point, the correct point would be .

The correct option is (B) :

The distance from a point to the plane is given by the formula: Here, the point is , and the equation of the plane is . Thus, , , , and . Substitute the values into the distance formula: Simplifying the numerator: Now, simplify the denominator: Thus, the distance is: To rationalize the denominator: Therefore, the distance from the point to the plane is .

The correct option is (D) :

We are given three points on the plane: To find the normal vector to the plane, we need two vectors on the plane. These can be obtained by subtracting the coordinates of the points: Next, we find the cross product to get a vector perpendicular to both and . The cross product is computed as follows: Expanding this determinant: ### Step 2: Normalize the normal vector To find the unit normal vector, we divide by its magnitude. The magnitude of is: Thus, the unit normal vector is:

The correct option is (E) :

The equation of a line in space passing through a point and parallel to a vector is given by: We are given the point and the line joining and .
Step 1: Find the direction vector of the line joining and The direction vector is:
Step 2: Use the formula for the equation of the line The equation of the line through and parallel to the direction vector is: Simplifying this, we get:

The correct option is (B) :

The equation of a plane is generally given by the formula: where are the direction ratios of the normal to the plane. In this case, the plane is perpendicular to the x-axis. A plane perpendicular to the x-axis has a normal vector along the x-axis, so the direction ratios are . Thus, the equation of the plane is: This plane passes through the point , and since it is parallel to the yz-plane, the equation of the plane is simply .

The correct option is (A) :

Step 1: Find two vectors in the plane

Given points: , , .

Vector .

Vector .

Step 2: Find the normal vector to the plane

Compute the cross product :

Step 3: Write the plane equation

Using point and normal vector :

Simplify:

Step 4: Find x-intercept

For x-intercept, set :

We are given two lines with direction cosines proportional to:

Line 1:

Line 2:

To find the angle between these lines, we use the dot product formula for direction ratios:

First, compute the dot product of the direction ratios:

Next, compute the magnitudes of the direction ratios:

For Line 1:

For Line 2:

Now, substitute into the dot product formula:

Thus, the angle is:

Given that and are the angles made by a line with the positive directions of the and axes, respectively, we know from the property of direction cosines that:

We are asked to find the value of . Using the double-angle identity for cosine, , we can rewrite the expression as:

Simplify the expression:

Substitute the direction cosines property:

Thus, the value of is .

and are the point of intersection of the lines and
So, the point f intersection can be found as ,
Any point on can be written as
Any point on can be written as
To find the intersection of and , the coordinates of the point written in the two different ways should be equal.
Hence,



The three equations are consistent and on solving we get: and .
Hence, desired point of intersection is .
So, the correct option is (B) :