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Motion in a Plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane3 Marks
Question
Find the angle between force $\vec{\text{F}}=(3\vec{\text{i}}+4\vec{\text{j}}-5\vec{\text{k}})$ unit and displacement $\vec{\text{d}}=(5\vec{\text{i}}+4\vec{\text{j}}+3\vec{\text{k}})$ unit. Also find the projection of $\vec{\text{F}}$ on $\vec{\text{d}}.$
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Answer

$\vec{\text{F}}=(3\vec{\text{i}}+4\vec{\text{j}}-5\vec{\text{k}})$$\vec{\text{d}}=(5\vec{\text{i}}+4\vec{\text{j}}+3\vec{\text{k}})$
$\text{F}=\sqrt{9+16+25}=\sqrt{50}$
$\text{d}=\sqrt{25+16+9}=\sqrt{50}$
$\vec{\text{F}}.\vec{\text{d}}=(3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}).(5\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}})$
$=15+16-15=16$
Let Q be the angle between $\vec{\text{F}}$ and $\vec{\text{d}}$
$\text{W}=\vec{F}.\vec{\text{d}}=\text{Fd}\cos\theta$
$16=\sqrt{50}\sqrt{50}\cos\theta$
$\frac{16}{50}=\cos\theta$
$\Rightarrow\ \theta=\cos^{-1}(0.32)=71.3^\circ$
Unit vector along $\vec{\text{d}}$ is
$\hat{\text{d}}=\frac{5\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}}{\sqrt{5^2+4^2+3^2}}=\frac{5\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}}{\sqrt{50}}$ $\vec{\text{F}}.\vec{\text{d}}=\frac{16}{\sqrt{50}}$ Component vector of $\vec{\text{F}}$ along $\vec{\text{d}}$ is $(\vec{\text{F}}.\vec{\text{d}})\hat{\text{d}}=\frac{16}{\sqrt{50}}\Big(\frac{5\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}}{\sqrt{50}}\Big)$ $=0.32(5\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}})$

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