Product Of Two Vectors

Concept: Vector Algebra - Vector Triple Product and Dot Product.

Step 1: Apply the Vector Triple Product expansion. The given equation is . We use the standard vector triple product expansion identity . Applying this to our equation gives: .

Step 2: Simplify the equation using given magnitudes. We are given the magnitudes , , and . We know that . Substituting this into our expanded equation yields: .

Step 3: Isolate vector and square both sides. To utilize the magnitude of , let's isolate it: . Now, take the square of the magnitude of both sides: .

Step 4: Expand the squared expression and solve for the dot product. Expanding the left side gives: . This simplifies to: . Now, substitute the known squared magnitudes ( ): . This simplifies to , which means , and therefore . This gives .

Step 5: Calculate using the dot product formula. By definition, . Substituting the knowns: . Solving for gives . Since is the reciprocal of , we have . Squaring this value gives . $ $

Concept: Vector Algebra - Dot and Cross Products.

Step 1: Deduce dot products from the cross product condition. We are given . By definition, the cross product is a vector perpendicular to both and . Therefore, vector is perpendicular to both and . This implies their dot products are zero: and .

Step 2: Expand the given magnitude equation. We are given . Squaring both sides gives . The expansion of this square is: .

Step 3: Substitute known magnitude values. We are given the magnitudes , , and . Squaring these yields , , and . The sum of these squares is .

Step 4: Simplify the equation using the dot products. Substitute the sum of squares (which is 1) and the zero dot products (from Step 1) into the expanded equation: .

This simplifies to , which means , leading to .

Step 5: Determine the angle between the vectors. The dot product is defined as , where is the angle between and . Since and neither magnitude is zero, it must be that . Therefore, the angle is . $ $

Concept: Vector Algebra - Dot and Cross Products.

Step 1: Use the dot product to find the magnitude of vector . We are given . Using the formula for the dot product, . Substituting the known values ( and ), we get .

Step 2: Solve for . Since , the equation becomes . Multiplying both sides by gives us .

Step 3: Set up the magnitude of the target cross product. We need to evaluate . Let . The magnitude of a cross product is , where is the angle between and .

Step 4: Apply the perpendicularity condition. The problem states that is perpendicular to . Therefore, the angle , and . The expression simplifies to .

Step 5: Calculate the final value. Expand as . Substitute all known magnitudes: . This simplifies to . $ $

Concept: Vector Algebra - Magnitudes, Dot Product, and Cross Product of Vectors.

Step 1: Isolate the term not involved in the cross product. We are given the vector equation . Since we need to find the value of , which involves vectors and , let's isolate vector on one side: .

Step 2: Square both sides to utilize the dot product. Squaring both sides of the equation gives . Expanding the left side using the property gives: .

Step 3: Substitute unit vector magnitudes and solve for the dot product. Since , , and are unit vectors, their magnitudes are 1 ( ). Substituting these values: . This simplifies to , which means . Using the dot product definition , we get , leading to .

Step 4: Calculate from . We know the trigonometric identity . Therefore, . Substituting yields . (We take the positive root because the angle between two vectors is defined in , where sine is always positive).

Step 5: Calculate the magnitude of the cross product. The magnitude of the cross product is given by the formula . Substituting the known values: . $ $

Concept: Three vectors are coplanar if their scalar triple product is zero. Equivalently, the determinant formed by their components is zero.

Step 1: Write vectors in component form.

Step 2: Form the determinant condition.

Step 3: Expand the determinant. Since vectors are coplanar:

Concept: Three vectors are coplanar if their scalar triple product is zero. Equivalently, the determinant formed by their components must be zero.

Step 1: Write the vectors in component form.

Step 2: Form the determinant for coplanarity.

Step 3: Expand the determinant. Thus,

Concept: The cross product of two vectors and is given by the determinant The magnitude of the cross product is where are the components of the cross product.

Step 1: Write the vectors in component form.

Step 2: Compute the cross product using determinant form.

Step 3: Find the magnitude of the resulting vector.

Concept: The dot product of two vectors is given by where and are the magnitudes of the vectors and is the angle between them.

Step 1: Use the given condition. Substitute into the formula:

Step 2: Since the vectors are non-zero, their magnitudes are not zero. Therefore,

Step 3: Find the angle. Thus, the vectors are perpendicular.

Concept: A vector perpendicular to both vectors and is obtained using the cross product: The unit vector in that direction is:
Step 1: {Find the cross product .}

Step 2: {Find the magnitude of the vector.}

Step 3: {Find the unit vector.}
Thus, the unit vector perpendicular to both vectors is in the direction of .