Find the vector, not with determinants, but by using properties of cross products. k \times (i-2j)

Alyce Wilkinson

Alyce Wilkinson

Answered question

2021-06-12

Find the vector, not with determinants, but by using properties of cross products. kΓ—(iβˆ’2j)

Answer & Explanation

odgovoreh

odgovoreh

Skilled2021-06-13Added 107 answers

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madeleinejames20

madeleinejames20

Skilled2023-05-14Added 165 answers

To find the vector resulting from the cross product of 𝐀 with (π’βˆ’2𝐣) without using determinants but by utilizing the properties of cross products, we can employ the following steps:
Let's denote 𝐀 as 𝐚 and (π’βˆ’2𝐣) as 𝐛. Thus, we need to find the vector resulting from πšΓ—π›.
First, we can express 𝐚 and 𝐛 in terms of their components:
𝐚=ax𝐒+ay𝐣+az𝐀
𝐛=bx𝐒+by𝐣+bz𝐀
Substituting the given values, we have:
𝐚=0𝐒+0𝐣+1𝐀
𝐛=1π’βˆ’2𝐣+0𝐀
Next, we can use the properties of cross products to calculate the resulting vector:
πšΓ—π›=(aybzβˆ’azby)𝐒+(azbxβˆ’axbz)𝐣+(axbyβˆ’aybx)𝐀
Plugging in the values, we obtain:
πšΓ—π›=(0Β·0βˆ’1Β·(βˆ’2))𝐒+(1Β·0βˆ’0Β·1)𝐣+(0Β·(βˆ’2)βˆ’0Β·1)𝐀
Simplifying the expression further, we get:
πšΓ—π›=2𝐒+0𝐣+0𝐀
Therefore, the vector resulting from the cross product of 𝐀 with (π’βˆ’2𝐣) is 2i.
nick1337

nick1337

Expert2023-05-14Added 777 answers

Let's first write the given vectors in component form:
𝐀=0𝐒+0𝐣+1𝐀=0k
𝐒=1𝐒+0𝐣+0𝐀=𝐒
βˆ’2𝐣=0𝐒+(βˆ’2)𝐣+0𝐀=βˆ’2𝐣
Now, we can use the properties of cross products, specifically the fact that 𝐒×𝐣=𝐀, to simplify the expression:
𝐀×(π’βˆ’2𝐣)=0kΓ—(π’βˆ’2𝐣)
Expanding the cross product, we get:
0kΓ—(π’βˆ’2𝐣)=0kΓ—π’βˆ’0kΓ—2𝐣
Since the cross product is distributive, we can further simplify:
0kΓ—π’βˆ’0kΓ—2𝐣=0kβˆ’2(0k×𝐣)
Finally, we use the property that 𝐀×𝐣=βˆ’π£Γ—π€ to rewrite the expression:
0kβˆ’2(0k×𝐣)=0kβˆ’2(βˆ’π£Γ—0k)
Since 𝐣×0k is zero, we have:
0kβˆ’2(βˆ’π£Γ—0k)=0kβˆ’20k=0
Therefore, the vector resulting from the cross product 𝐀×(π’βˆ’2𝐣) is 0.
Eliza Beth13

Eliza Beth13

Skilled2023-05-14Added 130 answers

Answer:
2𝐒+𝐣
Explanation:
πšΓ—π›=(a2b3βˆ’a3b2)𝐒+(a3b1βˆ’a1b3)𝐣+(a1b2βˆ’a2b1)𝐀
Given 𝐀×(π’βˆ’2𝐣), we can substitute the values into the formula:
𝐀×(π’βˆ’2𝐣)=(k2(βˆ’2)βˆ’(βˆ’2)(1))𝐒+((1)(1)βˆ’k1(βˆ’2))𝐣+(k1(2)βˆ’(βˆ’2)(βˆ’2))𝐀
Simplifying further:
𝐀×(π’βˆ’2𝐣)=(βˆ’2k2+2)𝐒+(1+2k1)𝐣+(2k1+4)𝐀
Since we want the answer in the form *2\mathbf{i} + \mathbf{j}*, we can rewrite the vector as:
𝐀×(π’βˆ’2𝐣)=2𝐒+(1+2k1)𝐣+(2k1+4)𝐀
Hence, the vector is 2𝐒+𝐣.

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