Question

for the displacement vectors a=(3.0m)i+(4.0m)j and b=(5.0m)i+(-2.0m)j, gave a+b in(a) unit vector notation and as a magnitude and an angle (related to i) now give b-a in unit-vector notation and as magnitude and an angle

Answer 1

Part (a) a + b = (8.0i - 2.0j) and
`\theta = 14.03^o` from the x-axis

Part (b) b - a = (2.0i - 6.0j) and
`\theta = -71.06^o` from the x- axis

Step-by-step explanation:

Given,

• 
  `\vec{a}\ =\ (3.0m)i\ +\ (4.0m)j`
• 
  `\vec{b}\ =\ (5.0m)i\ +\ (-2.0m)j`

From the addition of the two vectors,

`\vec{a}\ +\ \vec{b}\ =\ (3.0i\ +\ 4.0j)\ +\ (5.0i\ -\ 2.0j)\\\Rightarrow \vec{a}\ +\ \vec{b}\ =\ (3.0\ +\ 5.0)i\ +\(4.0\ -\ 2.0)j\\\Rightarrow \vec{a}\ +\ \vec{b}\ =\ 8.0i\ +\ 2.0j`

Let
`\theta` be the angle of the resultant vector of the addition of the vectors with the x-axis (i).

`\therefore Tan\theta\ =\ (2.0)/(8.0)\\\Rightarrow \theta\ =\ Tan^(-1)\left ((2.0)/(8.0)\ \right )\\\Rightarrow \theta\ =\ 14.03^o`

Part (b)

From the subtraction of the two vectors,

`\vec{b}\ -\ \vec{a}\ =\ (5.0i\ -\ 2.0j)\ -\ (3.0i\ +\ 4.0j)\\\Rightarrow \vec{b}\ -\ \vec{a}\ =\ (5.0\ +\ 3.0)i\ +\(-2.0\ -\ 4.0)j\\\Rightarrow \vec{b}\ -\ \vec{a}\ =\ 2.0i\ -\ 6.0j`

Let
`\theta` be the angle of the resultant vector of the addition of the vectors with the x-axis (i).

`\therefore Tan\theta\ =\ (-6.0)/(2.0)\\\Rightarrow \theta\ =\ Tan^(-1)\left ((-6.0)/(2.0)\ \right )\\\Rightarrow \theta\ =\ -71.06^o`