Vectors

We have,
and are mutually perpendicular vectors.

and
Now,

Step 1: Use the condition for coplanarity.
For three vectors to be coplanar, their scalar triple product must be zero:
Step 2: Compute the cross product.
First, compute the cross product using the determinant formula: Simplifying this gives:
Step 3: Compute the dot product.
Next, compute the dot product : Simplifying this gives:
Step 4: Solve for .
Simplifying further:
Step 5: Conclusion.
Thus, the value of is , making option (C) the correct answer.

Step 1: Recall the vector property of special points.
The incentre of a triangle satisfies the relation where are the lengths of the opposite sides.

Step 2: Compare with the given condition.
The given expression exactly matches the standard vector condition of the incentre.

Step 3: Conclusion.
Hence, the point is the incentre of the triangle.

Step 1: Condition for collinearity.
Two vectors are collinear if one is a scalar multiple of the other. Hence, for the given vectors and , we set up the following equation:
Step 2: Solve for and .
Solving this equation, we find and , corresponding to option (A).

Step 1: Use the condition for coplanar vectors.
For three vectors to be coplanar, the scalar triple product must be zero. Use this condition to form an equation relating , , and .
Step 2: Solve the equation.
After solving the equation, we find that must be in geometric progression (G.P.).
Step 3: Conclusion.
The correct answer is (C) are in G.P..

Step 1: Use determinant properties.
Using linearity of determinants, we expand: Step 2: Expand column-wise.
Step 3: Use determinant properties.
Determinants with two identical vectors are zero. Hence, Step 4: Form the ratio.
Step 5: Conclusion.
The required value is .

Step 1: Use the condition for collinearity.
If two vectors are collinear, then their corresponding components are proportional.
Step 2: Form ratios.
Step 3: Simplify the ratios.
Step 4: Solve for .
Step 5: Conclusion.
The value of is .

Step 1: Use the formula for volume of a parallelepiped.
The volume is given by Equivalently, the volume is the absolute value of the scalar triple product.

Step 2: Find .

Step 3: Find and .

Step 4: Compute the scalar triple product.

Step 1: Find vectors and .

Step 2: Compute the dot product.

Step 3: Conclude the angle.
Since the dot product is zero, the vectors are perpendicular. Hence,

Step 1: Use the given vector relation.

Step 2: Interpret the equation.
This equation represents a point dividing both segments and internally in the same ratio.

Step 3: Find the position vector of the intersection point.

Step 1: Condition for to lie in the plane of and .
For to lie in the plane of and , it must be a linear combination of and . Thus, we express as:
Step 2: Solve for .
Substituting the components of , , and into the equation and solving for , we find:
Step 3: Conclusion.
Thus, , corresponding to option (C).

Step 1: Formula for median vector.
The vector of the median through vertex is given by

Step 2: Add the given vectors.


Step 3: Find the median vector.


Step 4: Find the length of the median.


Step 5: Conclusion.
The length of the median through is

Step 1: Use the condition for coplanarity.
Three vectors are coplanar if the scalar triple product is zero:

Step 2: Form the determinant.


Step 3: Evaluate the determinant.


Step 4: Final conclusion.

Step 1: Understanding the volume of the parallelepiped.
The volume of a parallelepiped with coterminous edges is given by the scalar triple product: The volume of the new parallelepiped formed by the vectors can be found by using the properties of the scalar triple product.
Step 2: Applying the identity.
Using the distributive property of the scalar triple product, we calculate the volume, which turns out to be 36 units . Thus, the correct answer is option (A).

Step 1: Compute .
Thus,
Step 2: Compute the cross product .
We use the determinant formula for the cross product: After calculating the determinant, we get:
Step 3: Compute .
Next, calculate the cross product of with . The result simplifies, and after performing the operations, we get:
Step 4: Take the dot product.
Finally, take the dot product of with the resulting vector . After simplifying, the final value is . Therefore, the correct answer is option (B).

Step 1: Understanding the midpoint condition.
We are given that the point is the midpoint of a line segment intercepted between the coordinate axes. Let the intercepts on the x-axis and y-axis be and , respectively. Since the midpoint is , we use the midpoint formula to find the equation of the line.

Step 2: Using the midpoint formula.
The midpoint of and is: From this, we find:
Step 3: Finding the equation of the line.
The equation of the line passing through and can be written in slope-intercept form. The slope is: Thus, the equation of the line is: Simplifying this equation, we get:
Step 4: Conclusion.
Thus, the equation of the line is , which makes option (B) the correct answer.

Step 1: Find the cross products.
To solve for and , compute the cross products:
Step 2: Take the dot product.
Now, take the dot product of the two resulting vectors:
Step 3: Conclusion.
Thus, the value is -74, corresponding to option (A).

Step 1: Understanding the concept of direction cosines.
The direction cosines of a line are the cosines of the angles that the line makes with the coordinate axes. To bisect the angle between the positive directions of the Y and Z axes, the direction cosines must be equal for the Y and Z axes and zero for the X axis, since the line lies in the Y-Z plane. Thus, the direction cosines are .

Step 2: Conclusion.
Thus, the correct answer is , corresponding to option (C).

To find the equation of a vector that lies in the plane of vectors a and b and has a projection of 1/√3 on vector c, we can use vector projection and vector addition.

Let's denote the vector we want to find as .
The projection of on vector is given by:

Step 1: Set up the projection equation
Since we want the projection to be , we can write the equation:

Step 2: Express as a linear combination of and
Since lies in the plane of vectors and , we can express as a linear combination of and : where and are scalar coefficients.

Step 3: Substitute into the projection equation
Now, substitute into the projection equation: Expanding the dot product:

Step 4: Use the cross product to find a relation between , , and
Since lies in the plane of and , we can also express as: where denotes the cross product. This means that is perpendicular to and lies in the plane of and . Therefore, we can take the cross product of and to obtain . Hence: Solving for , we get:

Step 5: Substitute this expression for into the projection equation
Now, substitute this expression for into the projection equation: Expanding the double cross product:

Step 6: Simplify the equation
Simplifying the expression:

Step 7: Solve for the scalar coefficients and
This equation provides a relationship between the scalar coefficients and . We can now choose any value for that satisfies the equation and then solve for the corresponding value of . The vector will then be given by:

We are tasked with evaluating: Step 1: Simplify using the identity for the sum of arctangents. The addition formula for arctangents is: Apply this identity to pairs of terms: Simplify the expression: Similarly: Simplify: Step 2: Combine the results. Now, the expression becomes: Step 3: Simplify further. Since , we have: Since , the final result is:

We are given the equation: Step 1: Square both sides to simplify the equation.
First, square both sides: Expanding the left-hand side: Simplify the expression: Rearrange the equation: Step 2: Eliminate fractions by multiplying through by .
Multiply both sides by to clear the denominators: Step 3: Differentiate both sides with respect to .
Now, differentiate both sides with respect to : Using the chain rule, we get: Simplifying the equation: Rearranging terms to isolate : Factoring out : Solving for : Step 4: Simplify the expression for .
Factor both the numerator and the denominator: Cancel the common factor : Final Answer:

We are given the following vectors:
Step 1: Calculate the dot product The dot product is calculated as follows:
Step 2: Calculate the magnitude of Next, we calculate the magnitude of vector :
Step 3: Use the given equation to find the angle We are given that: Substituting the values we calculated: Thus, the angle between and , , satisfies: Since we know and , we need to find the magnitude of : Substituting into the cosine equation: This yields . By calculation, we find that the angle is approximately .
Step 4: Calculate the projection of on The projection of vector onto is given by: From our previous calculations: Thus, the projection of onto is .

Step 1: Equation from Point
Passes through .
Step 2: Area Integral
.
.
.
Step 3: Solve
From .
.
.
.
Final Answer: (D)

We are given three vectors , , and with magnitudes 2, 3, and 4, respectively. We are also given the following conditions: 1) is perpendicular to , 2) is perpendicular to , 3) is perpendicular to . We need to find the magnitude of the vector sum .
Step 1: Use the perpendicularity conditions For each pair of vectors, their dot product must be zero because they are perpendicular. - - -
Step 2: Simplify the equations From the first condition: From the second condition: From the third condition: Thus, we have the system of three equations: 1) 2) 3) These equations imply that , , and are mutually perpendicular to each other.
Step 3: Calculate the magnitude of Since the vectors are perpendicular to each other, the magnitude of their sum is given by the Pythagorean theorem: Substitute the magnitudes of , , and : Thus, the magnitude of is .