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Question: The angle between force \(\vec { F } = ( 3 \hat { i } + 4 \hat { j } - 5 \hat { k } )\) unit and di...

The angle between force F=(3i^+4j^5k^)\vec { F } = ( 3 \hat { i } + 4 \hat { j } - 5 \hat { k } ) unit and displacement d=(5i^+4j^+3k^)\overrightarrow { \mathrm { d } } = ( 5 \hat { \mathrm { i } } + 4 \hat { \mathrm { j } } + 3 \hat { \mathrm { k } } ) unit is :

A

cos1(0.16)\cos ^ { - 1 } ( 0.16 )

B

cos1(0.32)\cos ^ { - 1 } ( 0.32 )

C

cos1(0.24)\cos ^ { - 1 } ( 0.24 )

D

cos1(0.64)\cos ^ { - 1 } ( 0.64 )

Answer

cos1(0.32)\cos ^ { - 1 } ( 0.32 )

Explanation

Solution

Here,

unit

F=(3)2+(4)2+(5)2=50\therefore | \overrightarrow { \mathrm { F } } | = \sqrt { ( 3 ) ^ { 2 } + ( 4 ) ^ { 2 } + ( - 5 ) ^ { 2 } } = \sqrt { 50 } unit

d=(5)2+(4)2+(3)2=50| \overrightarrow { \mathrm { d } } | = \sqrt { ( 5 ) ^ { 2 } + ( 4 ) ^ { 2 } + ( 3 ) ^ { 2 } } = \sqrt { 50 } unit

Let θ\thetabe angle between .

or θ=cos1(0.32)\theta = \cos ^ { - 1 } ( 0.32 )