Question

Find the angle between force (3i + 4j - 5k) and displacement (5i + 4j + 3k).

Answer 1

Solution:

Given:

$\sf{\implies Force,\;\overrightarrow{f} = 3\widehat{i}+4\widehat{j}-5\widehat{k}}$

$\sf{\implies Displacement,\; \overrightarrow{d} = 5\widehat{i}+4\widehat{j}-3\widehat{k}}$

To find:

=> Angle between force and displacement (Ф)

So,

$\sf{Let\;\overrightarrow{A} = 3\widehat{i}+4\widehat{j}-5\widehat{k}}$

$\sf{\overrightarrow{B} = 5\widehat{i}+4\widehat{j}-3\widehat{k}}$

Then,

$\sf{\overrightarrow{A}.\overrightarrow{B}=(3\times 5) + (4 \times 4) + (-5 \times 3)}$

$\sf{\implies 15+16-15}$

$\sf{\implies 16}$

And,

$\sf{A = \sqrt{3^{2}+4^{2}+5^{2}}}$

$\sf{= \sqrt{9+16+25}}$

$\sf{A = 5\sqrt{2}}$

And,

$\sf{B = \sqrt{5^{2}+4^{2}+3^{2}}}$

$\sf{= \sqrt{25+16+9}}$

$\sf{B = 5\sqrt{2}}$

Then,

$\sf{\overrightarrow{A}.\overrightarrow{B}= AB \cos \theta}$

We know,

A & B = 5√2,

$\sf{\overrightarrow{A}.\overrightarrow{B } = 16}$

So,

$\sf{\overrightarrow{A}.\overrightarrow{B}= AB \cos \theta}$

$\sf{16=5\sqrt{2} \times 5\sqrt{2}\cos \theta}$

$\sf{16 = 50 \cos \theta}$

$\sf{\cos \theta = \dfrac{16}{50}}$

So,

${\boxed{\boxed{\sf{\implies \theta = \cos^{-1}\bigg[\dfrac{8}{25}\bigg]}}}}$

Answer 2

Answer:-

$\red{ \boxed{\boxed{ \green{ \bf{ \theta = {cos}^{ - 1} \bigg( \frac{8}{25} \bigg)}}}}}$

Explanation:-

To find:-

Angle between force and displacement.

Solution:-

Given:-

f =( 3i + 4j - 5k )N

d= ( 5i + 4j + 3k) m

we know that ,

Angle between two vector is →

$\bf{cos \: \theta = \frac{ \vec{f}. \vec{d}}{fd} }$

$\bf{ \vec{f}. \vec{d} = (3i + 4j - 5k).(5i + 4j +3k)} \\ \\ \: \: \: \: \: \: \: \: = 15 + 16 - 15 = 16 \\ \\ \\ \bf{fd = \big( \sqrt{ {3}^{2} + {4}^{2} + {5}^{2} } \big) \big( \sqrt{ {5}^{2} + {4}^{2} + {3}^{2} } \big) }\\ \\ \: \: \: \: \: \: = \sqrt{50}. \sqrt{50} = 50$

Hence,

$\bf{cos \: \theta = \frac{16}{50} } \\ \\ \bf{ cos \: \theta \: = \frac{8}{25} } \\ \\ \boxed{ \red{\theta = {cos}^{ - 1} \bigg( \frac{8}{25} \bigg)}}$