Vector Algebra

The given equation of the line is:

The direction vector of the line is:

The given plane equation is:

The normal vector of the plane is:

The angle between the line and the plane is given by:

Computing the dot product:

Computing the magnitudes:

Computing :

Final Answer:

Step 1: Use the formula for the magnitude of the cross product
The magnitude of the cross product of two vectors and is given by: Given:

Step 2: Compute the sine value

Step 3: Calculate the magnitude

Final Answer: .

Step 1: Use the condition for perpendicular vectors
Two vectors and are perpendicular if their dot product is zero: Given vectors:

Step 2: Compute the dot product
Using the formula: Expanding:

Step 3: Solve the quadratic equation
The quadratic equation is: Using the quadratic formula: Here, , , and :

Final Answer: is .

Step 1: Use the cross product formula
The cross product of two vectors and is given by:

Step 2: Compute the determinant
Expanding along the first row: Computing the determinants:

Step 3: Substitute values and compare
Given that: Comparing the -components:

Final Answer: is 1.

Step 1: Use the dot product formula
The dot product of two vectors and is given by: Given: The magnitude of is: The given angle is .

Step 2: Compute the dot product
Using the formula: Since , we substitute:

Final Answer: is .

Step 1: Understand the section formula
The given condition is: This means that the point divides the line segment in the ratio: We use the section formula: where and , with the ratio .

Step 2: Compute the coordinates of
Using the section formula:

Final Answer: The coordinates of are (4,3,3).

The magnitude of the cross product of two vectors is given by: where is the angle between the two vectors. Substituting the given values: The possible values of for are:

So, the correct option is (B) :

Let and . The dot product formula is given by: where is the angle between the vectors. We are given that , so . ### Step 1: Calculate the dot product ### Step 2: Calculate the magnitudes of and ### Step 3: Use the dot product formula Substitute the values into the dot product formula: Multiply both sides by : ### Step 4: Square both sides Simplify the equation: Thus:

The correct option is (C) :

We are given the vectors We need to find the value of . First, compute the cross product . Using the determinant form for the cross product: Expanding the determinant: Calculating the 2x2 determinants: Now, compute the dot product : Using the distributive property of the dot product: Thus, the value of is .

The correct option is (E) :

We are asked to find the value of . We can expand the dot product as follows: We can simplify this expression using the properties of dot products: Note that and . Therefore, we get: Substitute the given values and : Thus, the value of is .

The correct option is (E) :

We are given the magnitudes of the vectors and , and the magnitude of their difference. We can use the following identity for the magnitude of the difference between two vectors: Substituting the given values: We have: Simplifying: Subtract 24 from both sides:

The correct option is (E) : 0

We are given the following vector equations: First, express in terms of and the given equation: Simplify: Use the dot product equation We are given . Compute the dot product: Simplify: Now, substitute : Solve for :

The correct option is (C) :

Given:

Then, first find magnitude of

Then the unit vector in the direction of a is

as per the question the vector having a magnitude of 5

The direction cosines satisfy the equation: Substituting given values: Calculating each term: Converting to denominator 105: Since , we take the negative value:
Final Answer: