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Chapter 2: Problem 279

If $\mathrm{A}^{\rightarrow}=2 \mathrm{i} \wedge+5 \mathrm{j} \wedge-\mathrm{k} \wedge\( and \)\mathrm{B}^{\rightarrow}=3 \mathrm{i} \wedge-2 \mathrm{j} \wedge-4 \mathrm{k} \wedge$ the angle between \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) is (A) 0 (B) \((\pi / 2)\) (C) \((\pi / 4)\) (D) \((\pi / 6)\)

Short Answer

Expert verified
(B) \(\frac{\pi}{2}\)

Step by step solution

01

Find the dot product of A and B

To find the angle between A and B, we first need to find the dot product of the two vectors. The dot product of A and B can be found by multiplying corresponding components of the two vectors and adding the results together: \(A \cdot B = (2i + 5j - k) \cdot (3i - 2j - 4k) \)
02

Calculate the dot product

Now we can multiply the components and sum them up: \(A \cdot B = (2 \times 3) + (5 \times -2) + (-1 \times -4) = 6 - 10 + 4 = 0\)
03

Calculate the magnitude of A and B

Next, we need to find the magnitudes of the vectors A and B. The magnitude of a vector is given by the square root of the sum of the squares of its components: \(|A| = \sqrt{(2)^{2} +(5)^{2} +(-1)^{2} } = \sqrt{30} \) \(|B| = \sqrt{(3)^{2} +(-2)^{2} +(-4)^{2} } = \sqrt{29} \)
04

Use the dot product formula to find the angle

Now we can use the dot product formula to find the angle θ between the two vectors. The dot product formula is: \(A \cdot B = |A| |B| \cos(\theta) \) Since we already found that \(A \cdot B = 0\), the formula becomes: \(0 = \sqrt{30} \sqrt{29} \cos(\theta)\)
05

Solve for θ

To find θ, we can divide both sides of the formula by \(\sqrt{30} \sqrt{29}\): \(\cos(\theta) = \frac{0}{\sqrt{30} \sqrt{29}} = 0 \) Angle θ is the angle at which cosine is 0. We know that \(\cos(\frac{\pi}{2}) = 0\), so the angle between A and B is: \(\theta = \frac{\pi}{2}\) Answer: (B) \(\frac{\pi}{2}\)

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Most popular questions from this chapter

$\mathrm{A}^{\boldsymbol{\longrightarrow}}=2 \mathrm{i} \wedge+2 \mathrm{j} \wedge-\mathrm{k} \wedge\( and \)\mathrm{B}^{\rightarrow}=2 \mathrm{i} \wedge-\mathrm{j} \wedge-2 \mathrm{k} \wedge\( Find \)3 \mathrm{~A}^{\rightarrow}-2 \mathrm{~B}^{\rightarrow}$ (A) \(2 \mathrm{i} \wedge+7 \mathrm{j} \wedge+\mathrm{k} \wedge\) (B) \(2 \mathrm{i} \wedge+8 \mathrm{j} \wedge-\mathrm{k} \wedge\) (C) \(2 \mathrm{i} \wedge+8 \mathrm{j} \wedge+\mathrm{k} \wedge\) (D) \(\mathrm{i} \wedge+7 \mathrm{j} \wedge+\mathrm{k} \wedge\)

Two bodies of masses \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) are dropped from heights \(\mathrm{H}\) and \(2 \mathrm{H}\) respectively. The ratio of time taken by the bodies to touch the ground is ... (A) \((1 / 2)\) (B) 2 (C) \((1 / \sqrt{2})\) (D) \((\sqrt{2} / 1)\)

Particle A is projected vertically upward from a top of a tower. At the same time particle \(B\) is dropped from the same point. The graph of distance (s) between the two particle varies with time is.

A body is moving in \(\mathrm{x}\) direction with constant acceleration \(\alpha\). Find the difference of the displacement covered by it in nth second and \((\mathrm{n}-1)\) th second. (A) \(\alpha\) (B) \((\alpha / 2)\) (C) \(3 \alpha\) (D) \((3 / 2) \alpha\)

A balloon is going vertically up with velocity \(12 \mathrm{~m} / \mathrm{s}\). When it is at height of \(65 \mathrm{~m}\) above the ground, it releases a stone. In how much time the stone will reach the ground. (A) \(\sqrt{13} \mathrm{~s}\) (B) \(10 \mathrm{~s}\) (C) \(5 \mathrm{~s}\) (D) \(6 \mathrm{~s}\)
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