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PART - 1 CH - 3 Motion in a Plane — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsPART - 1 CH - 3 Motion in a Plane5 Marks
Question
Establish the following vector in equalities geometrically or otherwise :
(a) $|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$
(b) $|\vec{a}+\vec{b}| \geq|\vec{a}|-|\vec{b}| \mid$
(c) $|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|$
(d) $|\vec{a}-\vec{b}| \geq|\vec{a}|-|\vec{b}| \mid$
When does the equality sign above apply?
✓

Answer

Consider two vectors $\vec{a}$ and $\vec{b}$ which are represented by $\overrightarrow{ OP }$ and $\overrightarrow{ OQ }$ two sides of parallelogram OPSQ. From the vector addition rule of parallelogram $(\vec{a}+\vec{b})$ is represented by the $\overrightarrow{ OS }$ vector.
Image
In the given figure, $OP =|\vec{a}|, OQ = PS =|\vec{b}|$
and $\quad OS =|\vec{a}+\vec{b}|$
(a) To prove : $|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$
Proof : We know that in a triangle the length of one side is always less than the sum of lengths of the remaining two sides. Therefore in $\triangle OPS$ :
$OS < OP+PS$
$OS < OP+OQ$
$|\vec{a}+\vec{b}| <|\vec{a}|+|\vec{b}|\quad\quad...(1)$
If both the vectors are acting in the same direction along the same straight line, that is, they are collinear then,
$|\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}|\quad\quad...(2)$
From equations (1) and (2)
$|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$
The equal sign applies only when the vectors are collinear and acting in the same direction.
Hence Proved.
(b) To prove : $|\vec{a}+\vec{b}| \geq \mid |\vec{a}|-|\vec{b}| \mid$
Proof : In a triangle, one side is greater than the difference between the other two sides. In $\triangle OPS$,
$OS+PS >OP $
$OS > |OP-PS| $
$OS >|OP-OQ| \quad\quad...(3)$
$\quad\quad\quad\therefore PS=OQ$
The modulus of ($OP - OQ$) has been taken, hence LHS will always be positive. But the value of RHS can also be engative if $OP < PS$.
Thus from equation (3),
$|\vec{a}+\vec{b}|> \mid |\vec{a}|-|\vec{b}| \mid\quad\quad...(4)$
If two vectors $\vec{a}$ and $\vec{b}$ are working in opposite directions along the same straight line, then
$|\vec{a}+\vec{b}|= \mid |\vec{a}|-|\vec{b}| \mid\quad\quad...(5)$
From equations (4) and (5)
$|\vec{a}+\vec{b}| \geq \mid |\vec{a}|-|\vec{b}| \mid$
Hence Proved.
(c) To prove : $|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|$
Proof : In $\triangle OPR$,
$OP=|\vec{a}|, PR=|-\vec{b}|=\vec{b}, OR=|\vec{a}-\vec{b}|$
From $\triangle OPR$,
$OR < OP+PR $
$|\vec{a}-\vec{b}| < |\vec{a}|+|\vec{b}|\quad\quad...(6)$
Now, vector $\vec{a}$ and $\vec{b}$ are in opposite directions along the same line then,
$|\vec{a}-\vec{b}|=|\vec{a}|+|\vec{b}|\quad\quad...(7)$
From equations (6) and (7)
$|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|$
Hence Proved.
(d) To prove : $|\vec{a}-\vec{b}| \geq \mid|\vec{a}|-|\vec{b}| \mid$
Proof : Since in any triangle one side is greater than the difference between the two sides then, in $\triangle OPR$,
$OP+PR > OP $
$OR > |OP-PR| $
$OR > |OP-OT|\quad\quad...(8)$
$\quad\quad\quad\because OT=PR$
Hence, modulus of (OP - OT) will always be positive. This means that the value of LHS will always be positive, but the value of RHS can also be negative if
$OP < OT$
Then from equation (8),
$|\vec{a}-\vec{b}|> \mid|\vec{a}|-|\vec{b}|\mid$ $\quad\quad...(9)$
If both vectors are in straight line and in same direction then,
$|\vec{a}-\vec{b}| = \mid|\vec{a}|-|\vec{b}|\mid$ $\quad\quad...(10)$
From equations (9) and (10)
$|\vec{a}-\vec{b}| \geq \mid|\vec{a}|-|\vec{b}|\mid$
Hence Proved.

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