Vector `\vec {F}` having magnitude 5.5 N makes 10° with x-xis and vector `\vec {r}` with magnitude 4.3 m makes 80°with x-xis. What is the magnitude of the dot and cross products?

(Ans. 8.1 Nm and 22.2 Nm)


Given:

Vector `\vec {F}` having magnitude 5.5 N makes 10° with x-xis
Vector `\vec {r}` with magnitude 4.3 m makes 80°with x-xis
Hence the angle between two vector = θ =  80° - 10° = 70°
 

To Find:

|`\vec {F}` . `\vec {r}`|= ?
|`\vec {F}` ｘ `\vec {r}`|= ?

Solution:

We know that the dot product of two vectors is

|`\vec {F}` . `\vec {r}`|= F r cos θ 

by putting values

|`\vec {F}` . `\vec {r}`|= 5.5 N ｘ 4.3 m ｘ cos 70° 

|`\vec {F}` . `\vec {r}`|= 5.5 N ｘ 4.3 m ｘ 0.342

|`\vec {F}` . `\vec {r}`|= 8.08 N m

|`\vec {F}` . `\vec {r}`|= 8.1 N m -------------Ans. 1


And the cross-product of two vectors is

|`\vec {F}` ｘ `\vec {r}`|= F r sin θ 

by putting values

|`\vec {F}` ｘ `\vec {r}`|= 5.5 N ｘ 4.3 m ｘ sin 70° 

|`\vec {F}` ｘ `\vec {r}`|= 5.5 N ｘ 4.3 m ｘ 0.940

|`\vec {F}` ｘ `\vec {r}`|= 22.2 N m -------------Ans. 2


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