Magnitude And Directions Of A Vector

Step 1: Recall that two vectors are perpendicular if and only if their dot product equals zero.
Given vectors:

Step 2: Compute the dot product :

Step 3: Set the dot product equal to zero for perpendicular vectors and solve for :

Conclusion: The value of that makes the vectors perpendicular is .

Step 1: Compute the magnitude of the vector . Given , the magnitude is:

Step 2: Find the direction cosines. The direction cosines are the cosines of the angles that the vector makes with the positive , , and axes, respectively. They are given by:

Step 3: Match with the given options. The direction cosines are .

Step 1: Compute the vector . Given: Subtract from : Simplified:

Step 2: Find the magnitude of .

Step 3: Compute the unit vector in the direction of . Divide the vector by its magnitude:

Conclusion: The correct answer is .

Given:

Here, represents the magnitude of vector .

Since is a non-zero vector, its magnitude is also non-zero.

So, we can now divide both sides of the inequality by

Now, we have an absolute value inequality with a constant on the right side.

To solve this, we can take two cases:

Case 1:

is positive

In this case, our inequality becomes:

Now, solve for :

Case 2: 4 - x is negative

In this case, our inequality becomes:

Now, solve for

So, the valid values of x in this case are

Combining both cases, we find that the valid values of are

Step 1: Use the property of perpendicular vectors. Since and are perpendicular:

Step 2: Apply the formula for the magnitude of the difference of vectors:

Given and :

Step 3: Take the square root to find the magnitude:

Conclusion: The value is ( ).

Step 1: Recall the section formula for internal division. For a point R dividing PQ in ratio m:n, its position vector is:

Given ratio 1:2 (m=1, n=2), and position vectors:

Step 2: Apply the section formula:

Step 3: Simplify the expression:

Conclusion: The position vector of point R is .

We need to find the equation of a plane that is perpendicular to both given planes and passes through the point .

Step 1: Find the normal vectors of the given planes

The first plane has normal vector .

The second plane has normal vector .

Step 2: Find the direction vector of the line of intersection (parallel to the desired plane's normal)

The desired plane must be perpendicular to both given planes, so its normal vector must be perpendicular to both and . Thus, we compute the cross product:

We can simplify this to by dividing by 3.

Step 3: Write the equation of the plane

Using the normal vector and the point , the plane equation is:

Final Answer:

(A)

First, substitute into the inequality and simplify: Multiply both sides by (assuming ) to avoid reversing the inequality: Expanding and simplifying yields:
Now, because the inequality assumes is positive (so we do not reverse the inequality sign when multiplying), it implies .
Conclusion: Combining and , we find that must lie in the interval , matching option (A).

We are asked to find the coefficient of in the expansion of .
We can write the expression as: Now, apply the binomial expansion to .
The binomial expansion for is given by:
We need to find the coefficient of .
This corresponds to the term where in the expansion:
Now, calculate and : Thus, the coefficient of is: Thus, the correct answer is option (B), 960.

We are given the relation: This is a set of pairs of integers where .
Step 1: We will find the possible values for for different integer values of .
- For : Therefore, , so .
- For : Therefore, .
- For : Therefore, . - For : This results in no valid solutions for .
Step 2: From the above analysis, we see that the possible values for are .
Therefore, the range of the relation , which consists of the set of all possible values of , is:
Thus, the correct answer is option (E).

For the assignment to be valid, the sum of all probabilities must be equal to 1.
Therefore, we can write: Simplifying: Thus, the value of is .
Thus, the correct answer is option (E), .

We are asked to find the value of .
We can simplify this expression using algebraic identities.
First, recall the identity: Let and .
Thus, the expression becomes: Since (the Pythagorean identity), we are left with:
Next, simplify the expression inside the parentheses. Notice that .
Since , this simplifies to:
Thus, the original expression becomes: Now, we need to compute .
Using the double angle identity, we know that: Since , we have:
Therefore: Substitute this into the equation for :
Thus, the correct answer is option (A), .

Step 1: The given expression is: First, expand the square of the vector: Using the distributive property of the dot product: Since and are unit vectors, we have and .
So the expression becomes: Thus, we get: Now, using the formula for the dot product of two unit vectors: Therefore, the expression becomes: Using the trigonometric identity , we can simplify this to:
Thus, the correct answer is option (B).

We are asked to find the value of . First, recall that the function (inverse sine) is defined on the principal range . This means that for any angle , will return the value of within this range.
Now, consider the angle . This angle is greater than , so we need to adjust it to fall within the principal range.
Since , we have:
Therefore, the result is , as is within the principal range of .
Thus, the correct answer is option (E), .

The function is given by: The maximum value of is , so the maximum value of is .
Thus, the maximum value of occurs when , and is: Thus, the maximum value of is , which corresponds to option (D).

We are given two parallel straight lines:
1. , where
2. , where
We need to find the shortest distance between these two parallel lines.
To solve this, we use the formula for the shortest distance between two parallel lines given by: Here: - (from the first line)
- (from the second line)
- (direction vector of the first line)
- (direction vector of the second line)
Since the lines are parallel, , so the cross product will give a zero vector.
We now use the formula for the shortest distance between two parallel lines: We calculate and , and their dot product: Now, we calculate the magnitude of : Thus, the shortest distance is: Thus, the correct answer is option (C), .

We are given the integral: To simplify this, let's make a substitution. Let: Thus: Also, the limits of integration change accordingly.
When , , and when , .
Now, substitute into the integral:
We can break the integral into three parts:
Now, evaluate each part:
1. The integral of :
2. The integral of :
3. The integral of : The function is odd because is odd and is even.
The integral of an odd function over a symmetric interval (from -1 to 1) is zero:
Now, add the results of the three integrals:
Thus, the value of the integral is 4, which corresponds to option (E).

We are tasked with finding the coefficient of in the expansion of . To do this, we need to consider the expansions of both terms separately.
1. Expansion of :
The binomial expansion of is given by: Thus, the general term in this expansion is .
2. Expansion of :
The expansion of can be found using the multinomial theorem.
The general term in the expansion is: where are non-negative integers such that .
We need the product of the general terms from both expansions that will give .
- The power of from is , and the power of from is .
- Therefore, we need to find and such that .
However, from the nature of the expansions, it is evident that no valid combination of terms will result in a power of , meaning the coefficient of is 0.
Thus, the correct answer is option (C), 0.