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Chapter 2: Problem 290

Find a unit vector in direction of \(\hat{1}+2 \hat{\jmath}-3 \mathrm{k}\) (A) \((1 / \sqrt{7})(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\) (B) \(-(1 / 2)(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\) (C) \((1 / \sqrt{14})(\hat{1}+2 \hat{j}-3 \mathrm{k})\) (D) \((1 / \sqrt{5})(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\)

Short Answer

Expert verified
(C) \((1 / \sqrt{14})(\hat{1}+2 \hat{j}-3 \mathrm{k})\)

Step by step solution

01

Find the magnitude of the given vector

To find the magnitude of the given vector, we can use the formula: \[|\vec{v}| = \sqrt{\vec{v} \cdot \vec{v}}\] where \(\cdot\) represents the dot product. In this case, we have: \[\vec{v} = \hat{1}+2 \hat{\jmath}-3 \mathrm{k}\] Calculate the dot product of \(\vec{v}\) with itself: \[\vec{v} \cdot \vec{v} = (1)^{2} +(2)^{2} +(-3)^{2} = 1 + 4 + 9 = 14\] \squareCalculate the magnitude of \(\vec{v}\) by taking the square root of the dot product: \[|\vec{v}| = \sqrt{14}\]
02

Divide the given vector by its magnitude

Now we will divide the given vector by its magnitude to find the unit vector: \[\frac{\vec{v}}{|\vec{v}|} = \frac{\hat{1}+2 \hat{\jmath}-3 \mathrm{k}}{\sqrt{14}}\] The unit vector in the direction of \(\vec{v}\) is: \[\frac{1}{\sqrt{14}}(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\] Now compare this computed unit vector with the available options: (A) \((1 / \sqrt{7})(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\) (B) \(-(1 / 2)(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\) (C) \((1 / \sqrt{14})(\hat{1}+2 \hat{j}-3 \mathrm{k})\) (D) \((1 / \sqrt{5})(\hat{1}+2 \hat{\jmath}-3 \mathrm{k})\) Option (C) matches our computed unit vector, so the correct answer is: (C) \((1 / \sqrt{14})(\hat{1}+2 \hat{j}-3 \mathrm{k})\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Magnitude
The magnitude of a vector gives us an idea of its size or length. It's like finding the length of a straight line from one point to another in space. For a vector such as \( \vec{v} = \hat{1} + 2\hat{\jmath} - 3\mathrm{k} \), the magnitude is calculated using the formula:\[|\vec{v}| = \sqrt{\vec{v} \cdot \vec{v}}\]This involves the dot product of the vector with itself. Let's break it down:
  • The components of \( \vec{v} \) are 1, 2, and -3.
  • Square each component: \(1^2, 2^2, (-3)^2\) which gives 1, 4, and 9.
  • Add them up: \( 1 + 4 + 9 = 14 \).
  • Take the square root: \( |\vec{v}| = \sqrt{14} \).
This number, \( \sqrt{14} \), represents the vector's magnitude or length.
Dot Product
The dot product is an essential operation to understand in vector mathematics. It provides a way to multiply two vectors to get a scalar (a single number). For vectors \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), their dot product is:\[ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \]This is simply the sum of the products of their corresponding components. In our exercise:
  • \( \vec{v} \cdot \vec{v} = 1 \times 1 + 2 \times 2 + (-3) \times (-3) \).
  • The result is \( 1 + 4 + 9 = 14 \).
This operation helps us find things like the angle between vectors and in our case, the vector's magnitude.
Vector Direction
Finding the direction of a vector involves creating a unit vector. A unit vector has a length of 1 and points in the same direction as the original vector. To find a unit vector for \( \vec{v} = \hat{1} + 2\hat{\jmath} - 3\mathrm{k} \), follow these steps:* Divide each component of \( \vec{v} \) by its magnitude \( |\vec{v}| = \sqrt{14} \).* This gives: \( \frac{1}{\sqrt{14}}(\hat{1} + 2\hat{\jmath} - 3\mathrm{k}) \).This unit vector maintains the direction of \( \vec{v} \) but scales it to a length of 1. It's useful for many applications like determining directions in physics or graphics.

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Most popular questions from this chapter

A particle has initial velocity \((2 \hat{1}+3 \hat{j}) \mathrm{ms}^{-1}\) and has acceleration \((\hat{1}+\hat{j}) \mathrm{ms}^{-2}\). Find the velocity of the particle after 2 second. (A) \((3 \hat{1}+5 \hat{j}) \mathrm{ms}^{-1}\) (B) \((4 \hat{i}+5 \hat{\jmath}) \mathrm{ms}^{-1}\) (C) \((3 \hat{1}+2 \hat{j}) \mathrm{ms}^{-1}\) (D) \((5 \hat{1}+4 \hat{j}) \mathrm{ms}^{-1}\)

\(\mathrm{A}^{-}+\mathrm{B}^{-}\) is perpendicular to \(\mathrm{A}^{-}\) and \(\left|\mathrm{B}^{-}\right|=2\left|\mathrm{~A}^{-}+\mathrm{B}^{-}\right|\) What is the angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) \((\mathrm{A})(\pi / 6)\) (B) \((5 \pi / 6)\) (C) \((2 \pi / 3)\) (D) \((\pi / 3)\)

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) \(6 \mathrm{~s}\)

A particle in \(\mathrm{xy}\) plane is governed by \(\mathrm{x}=\mathrm{A} \cos \omega \mathrm{t}, \mathrm{y}=\mathrm{A}\) \((1-\sin \omega \mathrm{t}) . \mathrm{A}\) and \(\omega\) are constants. What is the speed of the particle. (A) A\omegat (B) \(\mathrm{A} \omega^{2} \mathrm{t}\) (C) \(\mathrm{A} \omega\) (D) \(\mathrm{A}^{2} \omega \sin (\omega \mathrm{t} / 2)\)

Find a unit vector perpendicular to both \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) (A) \(\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) / \mathrm{AB}\right]\) (B) \(\left[\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{-}\right) /(\mathrm{AB} \sin \theta)\right]\) (C) \(\left[\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right) /(\mathrm{AB} \cos \theta)\right]\) (D) \(\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) /(\mathrm{AB} \sin \theta)\right]\)
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