Question

A vector is given by vector R = 2.10 i hat + 1.80 j + 3.03 k. (a) Find the magnitudes of the x, y, and z components. x = 2.1 Correct: Your answer is correct. y = 1.8 Correct: Your answer is correct. z = 3.03 Correct: Your answer is correct. (b) Find the magnitude of vector R. 4.102 Correct: Your answer is correct. (c) Find the angle between vector R and the x axis. 58.99 Correct: Your answer is correct. ° Find the angle between vector R and the y axis. 49.398 Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations.° Find the angle between vector R and the z axis.

Answer 1

Answer:

part (a)  x-component = 2.1

              y-component = 1.8

              z-component = 3.03

part (b) R = 4.102

part (c) Angle between R and x-axis = $\theta\ =\ 59.2^o$

            Angle between R and y-axis = $\theta\ =\ 63.97^o$

            Angle between R and z-axis = $\theta\ =\ 42.38^o$

Explanation:

Given,

• $\vec{R}\ =\ (2.10i\ +\ 1.80j\ +\ 3.03k)$
• x-axis as a vector representation = $\hat{i}$
• y-axis as a vector representation = $\hat{j}$

• z-axis as a vector representation = $\hat{k}$

part (a)

Let any vector is represented by with (xi\ +\ yj\ +\ zk}

By comparing $\vec{R}$ with the the above represented vector, we get,

x-component of  R = x = 2.10

y-component of R = y = 1.80

z-component of R = z = 3.03

part (b)

The resultant of the $\vec{R}\ =\ |\vec{R}|\ =\ \sqrt{2.10^2\ +\ 1.80^2\ +\ 3.03^2}\ =\ 4.102$

part (c)

From the dot product between the two vectors

$\vec{a}.\vec{b}\ =\ |\vec{a}|.|\vec{b}|cos\theta$

Where \theta be the angel between the two vectors

Case - 1

$\theta$ be the angle between the vector R and x- axis

From the dot product,

$\vec{R}.\hat{i}\ =\ |\vec{R}|.|\hat{i}|cos\theta\\\Rightarrow (2.10i\ +\ 1.80j\ +\ 3.03k).(i)\ =\ 4.102\times 1\times cos\theta\\\Rightarrow \theta\ =\ cos^{-1}\left ( \dfrac{2.10}{4.102}\ \right )\\\Rightarrow \theta\ =\ 59.2^o$

Case - 2

$\theta$ be the angle between the vector R and y- axis

From the dot product,

$\vec{R}.\hat{j}\ =\ |\vec{R}|.|\hat{k}|cos\theta\\\Rightarrow (2.10i\ +\ 1.80j\ +\ 3.03k).(j)\ =\ 4.102\times 1\times cos\theta\\\Rightarrow \theta\ =\ cos^{-1}\left ( \dfrac{1.80}{4.102}\ \right )\\\Rightarrow \theta\ =\ 63.97^o$

Case - 4

$\theta$ be the angle between the vector R and z- axis

From the dot product,

$\vec{R}.\hat{k}\ =\ |\vec{R}|.|\hat{k}|cos\theta\\\Rightarrow (2.10i\ +\ 1.80j\ +\ 3.03k).(k)\ =\ 4.102\times 1\times cos\theta\\\Rightarrow \theta\ =\ cos^{-1}\left ( \dfrac{3.03}{4.102}\ \right )\\\Rightarrow \theta\ =\ 42.38^o$