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Model Paper 7 — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsModel Paper 75 Marks
Question
Establish the following vector inequalities:
i. $|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$
ii. $|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|$
When does the equality sign apply?
✓

Answer

i. If $\theta$ be the angle between $\vec{a}$ and $\vec{b}$, then
$
|\vec{a}+\vec{b}|=\sqrt{|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}||\vec{b}| \cos \theta}
$
Now $|\vec{a}+\vec{b}|$ will be maximum when
$
\begin{aligned}
& \cos \theta=1 \text { or } \theta=0 \\
& \therefore|\vec{a}+\vec{b}|_{\max }=\sqrt{|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}||\vec{b}| \cos 0^{\circ}} \\
& =\sqrt{(|\vec{a}|+|\vec{b}|)^2}=|\vec{a}|+|\vec{b}|
\end{aligned}
$
Hence $|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$
The equality sign in applicable when $\theta=0^{\circ}$ i.e., when $\vec{a}$ and $\vec{b}$ are in the same direction.
ii. If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then the angle between $\vec{a}$ and $-\vec{b}$ will be $\left(180^{\circ}-\theta\right)$, as shown in figure.
Image
$
\begin{aligned}
& \therefore|\vec{a}-\vec{b}|=|\vec{a}+(-\vec{b})| \\
& =\sqrt{|\vec{a}|^2+|-\vec{b}|^2+2|\vec{a}||-\vec{b}| \cos \left(180^{\circ}-\theta\right)} \\
& =\sqrt{|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}| \cos \theta} \\
& {\left[\because|-\vec{b}|=|\vec{b}|, \cos \left(180^{\circ}-\theta\right)=-\cos \theta\right]}
\end{aligned}
$
$|\vec{a}-\vec{b}|$ will be maximum when $\cos \theta=-1$ or $\theta=180^{\circ}$
$
\begin{aligned}
& \therefore|\vec{a}-\vec{b}|_{\max }=\sqrt{|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}| \cos 180^{\circ}} \\
& =\sqrt{(|\vec{a}|+|\vec{b}|)^2}=|\vec{a}|+|\vec{b}|
\end{aligned}
$
Hence $|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|$
The equality sign is applicable when $\theta=180^{\circ}$

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