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Chapter 2: Problem 284

$\mathrm{A} \overrightarrow{\mathrm{r}}=2 \mathrm{i} \wedge-\mathrm{j} \wedge+2 \mathrm{k} \wedge\( and \)\mathrm{B}^{\rightarrow}=-\mathrm{i} \wedge-2 \mathrm{j} \wedge+4 \mathrm{k} \wedge$ what is the angle between \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) (A) \(\cos ^{-1} 0.8888\) (B) \(\cos ^{-1} 0.4444\) (C) \(\sin ^{-1} 0.4444\) (D) \(\sin ^{-1} 0.8888\)

Short Answer

Expert verified
The short answer to the given question is: (A) cos^(-1) 0.8888

Step by step solution

01

Find the dot product A⃗ • B⃗

We can find the dot product of A⃗ and B⃗ by multiplying their corresponding components and adding them together. The formula is: A⃗ • B⃗ = AixBi + AjyBj + AkzBk Given: A⃗ = 2i - j + 2k B⃗ = -i - 2j + 4k A⃗ • B⃗ = (2 × -1) + (-1 × -2) + (2 × 4) = -2 + 2 + 8 = 8
02

Find the magnitudes of A⃗ and B⃗

We can find the magnitude of a vector using the formula: |A⃗| = √(Aix^2 + Ajy^2 + Akz^2) So for A⃗ and B⃗, we have: |A⃗| = √(2^2 + (-1)^2 + 2^2) = √(4 + 1 + 4) = √9 = 3 |B⃗| = √((-1)^2 + (-2)^2 + 4^2) = √(1 + 4 + 16) = √21
03

Find the angle between A⃗ and B⃗

Now we will use the dot product formula to find the angle θ between the two vectors: A⃗ • B⃗ = |A⃗| |B⃗| cos(θ) We can rearrange the formula to find cos(θ): cos(θ) = (A⃗ • B⃗) / (|A⃗| |B⃗|) cos(θ) = 8 / (3 * √21) ≈ 0.8888 Finally, we can find the angle θ: θ = cos^(-1)(0.8888) Comparing the result with the options given in the exercise, we find that the correct answer is: (A) cos^(-1) 0.8888

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