Vector Algebra

The vector cross product is anti-commutative by definition. This means that for any two vectors and :
.
The problem states that .
We can substitute the anti-commutative property into the given equation:
.
Rearranging the terms, we get:
.
This implies that the cross product of and must be the zero vector:
.
The magnitude of the cross product is given by , where is the angle between the vectors.
For the cross product to be zero, assuming and are non-zero vectors, we must have .
The angles for which in the range are and .
The problem specifies the range as .
Therefore, the only possible value for is .

The length of the projection of a vector on a vector is Here, Step 1: Evaluate the numerator is a scalar triple product. Using the cyclic property, Since Now, Hence, Step 2: Evaluate the denominator

Step 3: Find Step 4: Compute Answer:

Let's evaluate the expression step by step, from the inside out.
First, calculate and .


.
.
Next, calculate the innermost cross product, let .

If is a vector normal to the plane determined by and then

If is a vector normal to the plane determined by and then
n2 = | =
Vector is parallel to i.e
is parallel to
Given

cosine of acute angle between

Obtuse angle between
and

The correct option is(A): Projection of on and the direction of the projection vector is same as of

The correct answer is 120
x+y=5λ
Cases :

Total number of ways are 120

The given vectors are:

Step 1: Use the Condition for
From , we can write:

This implies that is parallel to , so:

where is a scalar.
Step 2: Use the Dot Product Condition
From , substitute :

Expanding:

Calculate :

Calculate :

Substitute into the equation:

Solve for :

Step 3: Find
Substitute into :

Expand:

Combine terms:

Simplify:
.
Step 4: Find
The magnitude of is:

Simplify:

Simplify further:

Conclusion:
The magnitude of is .

Given the vector equations and properties, solving for b involves using the dot product and cross product information provided. Once is calculated, is found by direct subtraction and scalar multiplication. The solution involves algebraic manipulation of vector components and verifying each option to match the calculated result.

To find the foot of the perpendicular, apply the formula for the point-line distance and the projection of a point onto a line using vector operations. Calculate where is the point on the line closest to and is the direction vector of the line. This provides the specific coordinates of the point , and subsequently, the ratios of these coordinates are compared against the options to identify the incorrect statement.

1. Represent the condition of coplanarity:
   Since the vectors ⃗b − ⃗a, ⃗c − ⃗b, and ⃗d − ⃗c are coplanar, we have:
   [⃗b − ⃗a, ⃗c − ⃗b, ⃗d − ⃗c] = 0
2. Expand the determinant:
   Expanding the determinant using the scalar triple product:
   (⃗b − ⃗a) · ((⃗c − ⃗b) × (⃗d − ⃗c)) = 0
3. Simplify the expression:
   Using the distributive property:
   (⃗b − ⃗a) · (⃗c × ⃗b − ⃗c × ⃗a − ⃗a × ⃗d) = 0
4. Represent as scalar triple products:
   The scalar triple products can be written as:
   [⃗b ⃗c ⃗d] − [⃗b ⃗c ⃗a] − [⃗b ⃗a ⃗d] − [⃗a ⃗c ⃗d] = 0
5. Rearrange to find [⃗a ⃗b ⃗c]:
   Finally, rearranging the terms gives:
   [⃗a ⃗b ⃗c] = [⃗d ⃗c ⃗a] + [⃗b ⃗d ⃗a] + [⃗c ⃗d ⃗b]

Final Answer:

[⃗a ⃗b ⃗c] = [⃗d ⃗c ⃗a] + [⃗b ⃗d ⃗a] + [⃗c ⃗d ⃗b]

The correct answer is 3501

So, the correct answer is 3.

The correct answer is (D) : 5

For four points to be coplanar, the volume of the tetrahedron formed by them must be zero. This is equivalent to the scalar triple product of vectors , , and being zero.

1. Find Vectors:

2. Compute the Scalar Triple Product:

First, find :

Now, compute the dot product with :

Setting the scalar triple product to zero:

Thus, .

We are given:

• ,
• ,
• .

We need to find:

Step 1: Expand

Using the distributive property, expand:

Distribute the terms:

Since , this simplifies to:

Step 2: Compute

Substitute into :

Using the vector triple product identity:

Substitute this back, considering the negative sign:

Given , we have . Substituting this value:

Step 3: Magnitude of

Given , substitute this value:

Compute the magnitude:

Since is a unit vector, . Thus:

Conclusion

The value of is: