Angle between force F =(3 hat{ i }+4 hat{ j }-5 hat{ k }) unit and displacement d =(5 hat{ i }+4 hat

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3-1.Vectors

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Find the angle between force $F =(3 \hat{ i }+4 \hat{ j }-5 \hat{ k })$ unit and displacement $d =(5 \hat{ i }+4 \hat{ j }+3 \hat{ k })$ unit. Also find the projection of $F$ on $d.$

Option A

Option B

Option C

Option D

Solution

$F \cdot d =F_{x} d_{x}+F_{y} d_{y}+F_{z} d_{z}$
$=3(5)+4(4)+(-5)(3)$
$=16$ unit
Hence $F \cdot d =F d \cos \theta=16$ unt
Now $F \cdot F \quad=F^{2}=F_{x}^{2}+F_{y}^{2}+F_{z}^{2}$
$=9+16+25$
$=50$ unit
and $d \cdot d \quad=d^{2}=d_{x}^{2}+d_{y}^{2}+d_{z}^{2}$
$=25+16+9$
$=50$ unit
$\therefore \cos \theta=\frac{16}{\sqrt{50} \sqrt{50}}=\frac{16}{50}=0.32$
$\theta=\cos ^{-1} 0.32$

Standard 11

Physics

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