Question

The route followed by a hiker consists of three displacement vectors, X, Y and Z. Vector X is along a measured trail and is 1430m in a direction 42 degrees north of east. Vector Y is along a measured trail that goes 2200m due south. The hiker then follows vector Z and ends up back where they started, so that X + Y + Z = 0. Find the magnitude of Z and an angle that specifies its direction.

Answer 1

• magnitude : 1635.43 m
• Angle: 130°28'20'' north of east

Step-by-step explanation:

First, we will find the Cartesian Representation of the
`\vec{X}` and
`\vec{Y}` vectors. We can do this, using the formula

`\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )`

where
`| \vec{A} |` its the magnitude of the vector and θ the angle. For
`\vec{X}` we have:

`\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )`

`\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )`

where the unit vector
`\hat{i}` points east, and
`\hat{j}` points north. Now, the
`\vec{Y}` will be:

`\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )`

Now, taking the sum:

`\vec{X} + \vec{Y} + \vec{Z} = 0`

This is

`\vec{Z} = - \vec{X} - \vec{Y}`

`(Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )`

`(Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 2200 m \ - \ 956.86 m \ )`

`(Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 1243.14 m\ )`

Now, for the magnitude, we just have to take its length:

`|\vec{Z}| = √(Z_x^2 + Z_y^2)`

`|\vec{Z}| = √((- 1062.70 m)^2 + (1243.14 m)^2)`

`|\vec{Z}| = 1635.43 m`

For its angle, as the vector lays in the second quadrant, we can use:

`\theta = 180\° - arctan((1243.14 m)/( - 1062.70 m))`

`\theta = 180\° - arctan( -1.1720)`

`\theta = 180\° - 45\°31'40''`

`\theta = 130\°28'20''`