On occasion,the notation = [A,θ] will be a shorthand notation for .
-If vector Is added to vector ,the result is a third vector that is perpendicular toAnd has a magnitude equal to 3 What is the ratio of the magnitude of To that of ?
A)1.8
B)2.2
C)3.2
D)1.3
E)1.6

Question

On occasion,the notation  = [A,θ] will be a shorthand notation for  .
-If vector Is added to vector ,the result is a third vector that is perpendicular toAnd has a magnitude equal to 3 What is the ratio of the magnitude of  To that of ?
A)1.8 
B)2.2 
C)3.2 
D)1.3 
E)1.6

Quizplus · Accepted Answer

Let's denote the three vectors as A, B, and C, with magnitudes |A|, |B|, and |C| respectively.

From the problem statement, we know that:

A + B = C   ...(1)

|C| = 3|A|   ...(2)

A and C are perpendicular, which means that their dot product is 0:

A · C = 11ea7ff9_2a41_a261_a768_2706bd9700df_TB2295_11 · 11ea7ff9_2a41_c972_a768_95e1aa404918_TB2295_11 = 0

We can expand the dot product using the scalar product formula:

A · C = |A||C| cos θ

Since A and C are perpendicular, cos θ = 0, which means that:

A · C = 0   ...(3)

Using the dot product formula again, we can express A · C in terms of its components:

A · C = 11ea7ff9_2a41_a261_a768_2706bd9700df_TB2295_11 * 11ea7ff9_2a41_c972_a768_95e1aa404918_TB2295_11 + θ

θ = -11ea7ff9_2a41_a261_a768_2706bd9700df_TB2295_11 * 11ea7ff9_2a41_c972_a768_95e1aa404918_TB2295_11

Substituting A + B for C from equation (1), we get:

A · (A + B) = 0

Expanding the dot product using distributivity:

A · A + A · B = 0

But A · A = |A|^2, so we have:

|A|^2 + A · B = 0   ...(4)

From equation (2), we know that |C| = 3|A|. Substituting C = A + B from equation (1) into this expression, we get:

|A + B| = 3|A|

Expanding the magnitude using the scalar product formula:

√(A·A + 2A·B + B·B) = 3|A|

Squaring both sides:

A·A + 2A·B + B·B = 9|A|^2

Substituting equation (4) into this expression, we get:

|A|^2 - A · B + B·B = 9|A|^2

Simplifying:

8|A|^2 = A · B + B·B   ...(5)

Now we have two equations with two unknowns, A·B and |A|:

A·B + B·B = 8|A|^2   ...(5)

A·C = 0   ...(3)

We can solve for A·B from equation (5) by substituting B = C - A from equation (1):

A·B + (C - A)·(C - A) = 8|A|^2

Expanding the dot product and simplifying:

A·C - |A|^2 + |C|^2 - 2A·C = 8|A|^2

Substituting equation (2) and cancelling the A·C terms:

|C|^2 - 9|A|^2 = |A|^2 - A·B

Substituting equation (3) and simplifying:

3|A|^2 = |A|^2 - A·B

Solving for A·B:

A·B = -2|A|^2

Substituting this expression into equation (5), we get:

B·B - 2|A|^2 = 8|A|^2

Simplifying and solving for |A|:

|A| = sqrt(B·B / 10)

We want the ratio of |C| / |B|:

|C| / |B| = 3|A| / sqrt(B·B)

Substituting our expression for |A|:

|C| / |B| = 3sqrt(B·B / 10) / sqrt(B·B)

Simplifying:

|C| / |B| = 3sqrt(10) / 10

Approximating this value to one decimal place:

|C| / |B| ≈ 1.6

Therefore, the answer is (C) 1.6.

On Occasion,the Notation = [A,θ] Will Be a Shorthand