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evaluate the magnitude of (šŸš’– + š’—)

Ɨ(3š’– āˆ’ šŸ“š’—), knowing that A is defined as š“ = |š’– Ɨ š’—|. (Hint: The cross product is distributive over addition and substraction, e.g., š’– Ɨ (š’— + š’˜) = š’– Ɨ š’— + š’– Ɨ š’˜)
A) 10A B) 10āˆš2š“ C) 12A D) 13A E) none

I hope you're not p, for whom I did almost this very problem ...

since the cross-product is distributive and anti-commutative,
(2u+v)Ɨ(3uāˆ’5v)
= 2uƗ3u - 2uƗ5v + vƗ3u - vƗ5v
= -10uƗv + 3vƗu
= 13vƗu
= 13A

To evaluate the magnitude of (2š’– + š’—) Ɨ (3š’– āˆ’ 5š’—) using the given expression š“ = |š’– Ɨ š’—|, we will follow these steps:

Step 1: Expand the given expression using the distributive property of the cross product:

(2š’– + š’—) Ɨ (3š’– āˆ’ 5š’—) = (2š’– Ɨ 3š’–) + (2š’– Ɨ -5š’—) + (š’— Ɨ 3š’–) + (š’— Ɨ -5š’—)

Step 2: Simplify each term of the expanded expression:

(2š’– Ɨ 3š’–) = 6(š’– Ɨ š’–) (Using the property š’‚ Ɨ š’ƒ = -š’ƒ Ɨ š’‚ for š’– Ɨ š’–)
(2š’– Ɨ -5š’—) = -10(š’– Ɨ š’—)
(š’— Ɨ 3š’–) = -3(š’– Ɨ š’—) (Using the property š’‚ Ɨ š’ƒ = -š’ƒ Ɨ š’‚ for š’– Ɨ š’—)
(š’— Ɨ -5š’—) = -5š’– Ɨ š’— (Using the property š’‚ Ɨ š’ƒ = -š’ƒ Ɨ š’‚ for š’– Ɨ š’—)

So, the expanded expression becomes:
6(š’– Ɨ š’–) - 10(š’– Ɨ š’—) - 3(š’– Ɨ š’—) - 5š’– Ɨ š’—

Step 3: Simplify further by combining like terms:
(6š’– Ɨ š’–) = 6|š’– Ɨ š’–| (Magnitude of a cross product is equal to magnitude of |š’– Ɨ š’–|)
(-10š’– Ɨ š’— - 3š’– Ɨ š’—) = -13(š’– Ɨ š’—) (Combine like terms)
(-5š’– Ɨ š’—) = -5(š’– Ɨ š’—) (Combining like terms)

The expression now becomes:
6|š’– Ɨ š’–| - 13(š’– Ɨ š’—) - 5(š’– Ɨ š’—)

Step 4: Substitute š“ = |š’– Ɨ š’—| into the expression:
6š“ - 13(š’– Ɨ š’—) - 5(š’– Ɨ š’—)

Step 5: Since we know that A is defined as š“ = |š’– Ɨ š’—|, we can further simplify the expression as:
6š“ - 13(š’– Ɨ š’—) - 5(š’– Ɨ š’—)
= 6š“ - (13 + 5)(š’– Ɨ š’—)
= 6š“ - 18(š’– Ɨ š’—)

Step 6: We are asked to evaluate the magnitude of the expression, so we need to find |6š“ - 18(š’– Ɨ š’—)|. However, we can further simplify the expression:
|6š“ - 18(š’– Ɨ š’—)|
= |6š“ - 18A|
= |6A(1 - 3)|
= |-12A|
= 12A

Hence, the magnitude of (2š’– + š’—) Ɨ (3š’– āˆ’ 5š’—) is equal to 12A.

Therefore, the correct answer is C) 12A.

To evaluate the magnitude of (2š’– + š’—) Ɨ (3š’– āˆ’ 5š’—), we can use the given hint: "The cross product is distributive over addition and subtraction, e.g., š’– Ɨ (š’— + š’˜) = š’– Ɨ š’— + š’– Ɨ š’˜."

Let's expand the expression:
(2š’– + š’—) Ɨ (3š’– āˆ’ 5š’—)

Using the distributive property:
= (2š’– Ɨ 3š’–) + (2š’– Ɨ -5š’—) + (š’— Ɨ 3š’–) + (š’— Ɨ -5š’—)

Simplifying further:
= 6(š’– Ɨ š’–) - 10(š’– Ɨ š’—) + 3(š’— Ɨ š’–) - 5(š’— Ɨ š’—)

We know that the cross product of two vectors is orthogonal to both vectors. So, š’– Ɨ š’– = š’— Ɨ š’— = 0.

Therefore:
(2š’– + š’—) Ɨ (3š’– āˆ’ 5š’—)
= -10(š’– Ɨ š’—) + 3(š’— Ɨ š’–)
= -10A + 3A

Simplifying further:
= -7A

Now, we need to find the magnitude of -7A. The magnitude of a vector is given by |š‘Ø| = āˆš(š€.š€).

Therefore:
|-7A| = āˆš((-7A).(-7A))
= āˆš(49A^2)
= 7A

So, the magnitude of (2š’– + š’—) Ɨ (3š’– āˆ’ 5š’—) is 7A.

The correct answer is E) none.