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Chapter 2: Problem 283

The angle between \(i \wedge+j \wedge\) and \(z\) axis is \(\ldots\) (A) 0 (B) 45 (C) 90 (D) 180

Short Answer

Expert verified
The angle between $i \wedge +j \wedge$ and $z$ axis is \(0\) degrees. (A)

Step by step solution

01

Find the cross product vector

Start by finding the cross product of the two given vectors: \(i \wedge +j \wedge\), which will give a new vector. The cross product is given by: (i, j, k) |(1, 0, 0)| |(0, 1, 0)| The cross product vector is \(k \wedge\).
02

Calculate the dot product

Now, we need to find the dot product between given cross product vector, \(k \wedge\), and the unit vector in z-direction, which is also \(k \wedge\). The dot product is given by: \(k \wedge \cdot k \wedge = (0, 0, 1) \cdot (0, 0, 1) = 0 \cdot 0 + 0 \cdot 0 + 1 \cdot 1 = 1\)
03

Find the magnitudes of the vectors

In order to find the angle between the given cross product vector and the z-axis, we will first need to find the magnitudes of these two vectors. In this case, since the cross product vector is \(k \wedge\) and the unit vector along the z-axis is also \(k \wedge\), the magnitudes of both vectors are 1. Magnitude of \(k \wedge = |\ k \wedge\ | = \sqrt{0^2 + 0^2 + 1^2} = 1\)
04

Apply the formula for the angle between two vectors

Now we can use the dot product, the magnitudes of the vectors, and the formula for the angle between two vectors to find the required angle. The formula is: \(\cos \theta = \dfrac{(k \wedge \cdot k \wedge)}{|\ k \wedge\ ||\ k \wedge\ |}\) Plugging in the values, we get: \(\cos \theta = \dfrac{1}{1 \cdot 1} = 1\) θ = inverse cosine of 1 = \((0)\) radians or \(0\) degrees.
05

Choose the correct answer

From our calculations, we found that the angle between the given cross product vector and the z-axis is \(0\) degrees. So, the correct answer is: (A) 0

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Most popular questions from this chapter

Which from the following is true? (A) $\cos \theta=\left[\left(\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|\right) / \mathrm{AB}\right]$ (B) $\sin \theta=\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) / \mathrm{AB}\right]$ (C) $\tan \theta=\left[\left(\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|\right) /\left(\mathrm{A}^{-}-\mathrm{B}^{-}\right)\right]$ (D) $\cot \theta=\left[\mathrm{AB} /\left(\left|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right|\right)\right]$

The resultant of two vectors \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) (A) can be smaller than \(\mathrm{A}-\mathrm{B}\) in magnitude (B) can be greater than \(\mathrm{A}+\mathrm{B}\) in magnitude (C) can't be greater than \(\mathrm{A}+\mathrm{B}\) or smaller than \(\mathrm{A}-\mathrm{B}\) in magnitude (D) none of above is true

A particle goes from point \(\mathrm{A}\) to \(\mathrm{B}\). Its displacement is \(\mathrm{X}\) and path length is \(\mathrm{y}\). So $\mathrm{x} / \mathrm{y} \ldots$ \((\mathrm{A})>1\) (B) \(<1\) (C) \(\geq 1\) (D) \(\leq 1\)

Linear momentum of a particle is $(3 \mathrm{i} \wedge+2 \mathrm{j} \wedge-\mathrm{k} \wedge) \mathrm{kg} \mathrm{ms}^{-1}$. Find its magnitude. (A) \(\sqrt{14}\) (B) \(\sqrt{12}\) (C) \(\sqrt{15}\) (D) \(\sqrt{11}\)

The area under acceleration versus time graph for any time interval represents... (A) Initial velocity (B) final velocity (C) change in velocity in the time interval (D) Distance covered by the particle
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