Question

Consider the three displacement vectors = ( 4î − 3ĵ) m, = (3î − 6ĵ) m, and = (−6î + 5ĵ) m. Use the component method to determine the following. (Take the +x direction to be to the right.)

(a) the magnitude and direction of the vector = Darrowbold = A with arrow + B with arrow + C with arrow

magnitude=____ m

direction=____ ° counterclockwise from the +x axis(b) the magnitude and direction of E with arrow = −A with arrow − B with arrow + C with arrow magnitude

Answer 1

(a).The magnitude and direction of the vector is 4.12 m and 284°

(b). The magnitude and direction of the vector is 19.10 m and 313°

Step-by-step explanation:

Given that,

The three displacement are

`A=(4i-3j)\ m`

`B=(3i-6j)\ m`

`C=(-6i+5j)\ m`

We need to calculate the magnitude of the vector

`\vec{D}=\vec{A}+\vec{B}+\vec{C}`

Put the value into the formula

`\vec{D}=(4i-3j)+(3i-6j)+(-6i+5j)`

`\vec{D}=(i-4j)`

`|\vec{D}|=√((1)^2+(4)^2)`

`|\vec{D}|=√(17)`

`|\vec{D}|=4.12\ m`

We need to calculate the direction of the vector

Using formula of direction

`\tan\theta=(j)/(i)`

`\theta=\tan^(-1)((j)/(i))`

Put the value into the formula

`\theta=\tan^(-1)((-4)/(1))`

`\theta=360^(\circ)-76^(\circ)`

`\theta=284^(\circ)`

(b). We need to calculate the magnitude of the vector

`\vec{D}=-\vec{A}-\vec{B}+\vec{C}`

Put the value into the formula

`\vec{D}=-(4i-3j)-(3i-6j)+(-6i+5j)`

`\vec{D}=(-13i+14j)`

`|\vec{D}|=√((13)^2+(14)^2)`

`|\vec{D}|=19.10\ m`

We need to calculate the direction of the vector

Using formula of direction

`\tan\theta=(j)/(i)`

`\theta=\tan^(-1)((j)/(i))`

Put the value into the formula

`\theta=\tan^(-1)((14)/(-13))`

`\theta=360^(\circ)-47^(\circ)`

`\theta=313^(\circ)`

Hence, (a).The magnitude and direction of the vector is 4.12 m and 284°

(b). The magnitude and direction of the vector is 19.10 m and 313°