Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 1493

If any vector forms angles \((\pi / 4),(\pi / 3)\) and \((\pi / 6)\) with axis, then such vector with measure 4 unit is (A) \((2,2 \sqrt{3}, 2 \sqrt{2})\) (B) \(\overline{(-2},-2 \sqrt{3}, 2 \sqrt{2})\) (C) \((2,2 \sqrt{3},-2 \sqrt{2})\) (D) \((-2,-2 \sqrt{3},-2 \sqrt{2})\)

Short Answer

Expert verified
The short answer is: There is no exact match in the given options, but option (A) is the closest to the computed vector \((2\sqrt{2}, 2, 2\sqrt{3})\), with the x and z components interchanged.

Step by step solution

01

Find the directional cosines

To determine the components of the vector, we first need to find the directional cosines. These are the cosines of the angles formed by the vector with the x, y, and z axes. They are given by: \[ \alpha = \cos(\pi/4) \\ \beta = \cos(\pi/3) \\ \gamma = \cos(\pi/6) \] Calculating these values, we get: \[ \alpha = \cos(\pi/4) = \frac{\sqrt{2}}{2} \\ \beta = \cos(\pi/3) = \frac{1}{2} \\ \gamma = \cos(\pi/6) = \frac{\sqrt{3}}{2} \]
02

Find the components of the vector

Now that we have the directional cosines, we can use them to find the components of the vector. Since the magnitude of the vector is 4 units, we can simply multiply the directional cosines by the magnitude to obtain the vector components: \[ x = 4\alpha = 4(\frac{\sqrt{2}}{2}) \\ y = 4\beta = 4(\frac{1}{2}) \\ z = 4\gamma = 4(\frac{\sqrt{3}}{2}) \] Calculating these values, we get: \[ x = 2\sqrt{2} \\ y = 2 \\ z = 2\sqrt{3} \] So, the vector components are \((2\sqrt{2}, 2, 2\sqrt{3})\).
03

Compare the vector with the given options

Now, we need to compare our vector \((2\sqrt{2}, 2, 2\sqrt{3})\) with the given options: (A) \((2,2 \sqrt{3}, 2 \sqrt{2})\) (B) \((-2,-2 \sqrt{3}, 2 \sqrt{2})\) (C) \((2,2 \sqrt{3},-2 \sqrt{2})\) (D) \((-2,-2 \sqrt{3},-2 \sqrt{2})\) Our vector doesn't match with any of the given options. This may be due to an error in the exercise. However, the option (A) is the closest to our computed vector, with the x and z components being interchanged.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equation of plane passing through the lines \((\mathrm{x} / 2)=[(\mathrm{y}-1) / 1]=[(\mathrm{z}+2) / 2]\) and \([(2 \mathrm{x}+3) / 4]=[(3-\mathrm{y}) /(-1)]=(\mathrm{z} / 2)\) is (A) \(4 \mathrm{x}+11 \mathrm{y}+14 z=36\) (B) \(\overline{4 x+1} 4 y-11 z=36\) (C) \(4 \mathrm{x}-14 \mathrm{y}-11 \mathrm{z}=36\) (D) \(4 \mathrm{x}-14 \mathrm{y}+11 \mathrm{z}=36\)

If vector \(\underline{x}\) forms an equal angle \(\alpha\) with three axis and \(|\underline{x}|=9\) then \(\alpha=\quad\) where \(0<\alpha<(\pi / 2)\) (A) \(\cos ^{-1}(1 / \sqrt{2})\) (B) \(\cos ^{-1}(1 / 9)\) (C) \(\cos ^{-1}(1 / \sqrt{3})\) (D) \(\cos ^{-1}(1 / 3)\)

The distance of a variable plane from origin to plane is \(\mathrm{p}\) and the Variable plane intersects the axis in \(\mathrm{A}, \mathrm{B}, \mathrm{C}\), then the point of intersection of given plane and the plane parallel to the co-ordinate plane is on \(\left(1 / \mathrm{x}^{2}\right)+\left(1 / \mathrm{y}^{2}\right)+\left(1 / \mathrm{z}^{2}\right)=\) (A) \(\mathrm{p}^{2}\) (B) \(\left(1 / \mathrm{p}^{2}\right)\) (C) \(p\) (D) \((1 / \mathrm{p})\)

If \(\mathrm{m} \angle \mathrm{B}=(\pi / 2)\) in \(\Delta \mathrm{ABC}\) and \(\mathrm{P}, \mathrm{Q}\) are points of trisection of hypotenuse \(\underline{\mathrm{A}} \mathrm{C}\), then \(\mathrm{BP}^{2}+\mathrm{BQ}^{2}=\) (A) \((5 / 9) \mathrm{AC}^{2}\) (B) \((5 / 9) \mathrm{AC}\) (C) \((25 / 81) \mathrm{AC}^{2}\) (D) \((25 / 81) \mathrm{AC}\)

The equation of line passes through \((1,2,3)\) and \((x / 1)=(y / 2)=[z /(-1)]\) and \([(x-1) / 3]=(y / 2)=(z / 6)\) and perpendicular to given two line is (A) \([(\mathrm{x}-1) / 14]=[(\mathrm{y}-2) /(-9)]=[(\mathrm{z}-3) /(-4)]\) (B) \([(\mathrm{x}+1) / 14]=[(\mathrm{y}+2) /(-9)]=[(\mathrm{z}+3) / 4]\) (C) \([(\mathrm{x}-1) / 14]=[(\mathrm{y}-2) / 9]=[(\mathrm{z}-3) /(-4)]\) (D) \([(\mathrm{x}+1) /(-14)]=[(\mathrm{y}+2) / 9]=[(\mathrm{z}+3) / 4]\)
See all solutions

Recommended explanations on English Textbooks

Blog

Read Explanation

Research and Composition

Read Explanation

History of English Language

Read Explanation

5 Paragraph Essay

Read Explanation

Essay Prompts

Read Explanation

Free Response Essay

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free