Question

A person walks from her home at position O to position F by taking a path that is comprised of three displacement vectors traveled in succession. Vector A with rightwards harpoon with barb upwards on top has a magnitude of 155 m and points 25 degree North of West (above the -x-direction), vector B with rightwards harpoon with barb upwards on top has a magnitude of 92.0 m and points due North, and vector C with rightwards harpoon with barb upwards on top points 50 degree East of North (to the right of the +y-direction) and has a magnitude of 64.7 m.

Calculate A with rightwards harpoon with barb upwards on top times C with rightwards harpoon with barb upwards on top using the components definition.

and C with rightwards harpoon with barb upwards on top cross times A with rightwards harpoon with barb upwards on top

Answer 1

Answer:

Explanation:

A = 155 m at 25° North of west

B = 92 m due north

C = 64.7 m at an angle 50° East of north

Write the displacements in the vector form

$\overrightarrow{A} = 155\left ( Cos25\widehat{i}-Sin25\widehat{j} \right )$

$\overrightarrow{A}=140.5\widehat{i}-65.5\widehat{j}$

$\overrightarrow{B}=92\widehat{j}$

$\overrightarrow{C} = 64.7\left ( Cos50\widehat{i}+Sin50\widehat{j} \right )$

$\overrightarrow{C} = 41.6\widehat{i}+49.6\widehat{j}$

(a)

$\overrightarrow{A}.\overrightarrow{C}=\left ( 140.5\widehat{i}-65.5\widehat{j} \right ).\left ( 41.6\widehat{i}+49.5\widehat{j} \right )$

$\overrightarrow{A}.\overrightarrow{C} = 2602.55$

(b)

$\overrightarrow{A}\times \overrightarrow{C}=\left ( 140.5\widehat{i}-65.5\widehat{j} \right )\times \left ( 41.6\widehat{i}+49.5\widehat{j} \right )$

$\overrightarrow{A}\times \overrightarrow{C}=9679.55 \widehat{k}$