Ncert Exercises & Solutions

4.6 Establish the following vector inequalities geometrically or otherwise :

(a) |a+b| ≤ | a| + |b|

(b) |a+b| ≥ ||a| −|b||

(d) |a−b| ≥||a| − |b||

When does the equality sign above apply?

(a) Let two vectors and be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

6.1

Here,

$|\vec{Q R}| = |\vec{a}| . . . . . (i)$
$|\vec{R S}| = |\vec{Q P}| = |\vec{b}| . . . . . (i i)$
$|\vec{Q S}| = |\vec{a} + \vec{b}| . . . . . . (i i i)$

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in $<math xmlns="http://www.w3.org/1998/Math/MathML"></math> Δ Q R S$ ,

QS < (QR + RS)

$|\vec{a} + \vec{b}| < |\vec{a}| + |\vec{b}| . . . . . . (i v)$

If the two vectors $<math xmlns="http://www.w3.org/1998/Math/MathML"></math> a$ and $<math xmlns="http://www.w3.org/1998/Math/MathML"></math> b$ act along a straight line in the same direction, then:

$|\vec{a} + \vec{b}| = |\vec{a}| + |\vec{b}| . . . . . (v)$

Combine equation (iv) and (v),

$|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$

(b) Let two vectors $<math xmlns="http://www.w3.org/1998/Math/MathML"></math> a$ and $<math xmlns="http://www.w3.org/1998/Math/MathML"></math> b$ be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

6.2

Here,

$|\vec{Q R}| = |\vec{a}| . . . . (i)$
$|\vec{R S}| = |\vec{Q P}| = |\vec{b}| . . . . . (i i)$
$|\vec{Q S}| = |\vec{a} + \vec{b}| . . . . . (i i i)$

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in $<math xmlns="http://www.w3.org/1998/Math/MathML"></math> Δ Q R S$ ,

QS + RS > QR

QS + QR > RS

$|\vec{Q S}| > |\vec{Q R} - \vec{Q P}| (a s, Q P = R S)$
$|\vec{a} + \vec{b}| > ||\vec{a}| - |\vec{b}|| . . . . . . (i v)$

If the two vectors $<math xmlns="http://www.w3.org/1998/Math/MathML"></math> a$ and $<math xmlns="http://www.w3.org/1998/Math/MathML"></math> b$ act along a straight line in the same direction, then:

$|\vec{a} + \vec{b}| = ||\vec{a}| - |\vec{b}|| . . . . . . . (v)$

Combine equation (iv) and (v):

$|\vec{a} + \vec{b}| \geq ||\vec{a}| - |\vec{b}||$

(c) Let two vectors $<math xmlns="http://www.w3.org/1998/Math/MathML"></math> a$ and $<math xmlns="http://www.w3.org/1998/Math/MathML"></math> b$ be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

6.3

$\begin{array}{l} or | \vec{a} - \vec{b} | < | \vec{a} | + | - \vec{b} | \\ or | \vec{a} - \vec{b} | < | \vec{a} | + | \vec{b} | \end{array}$ In case, the vectors $\vec{a} a n d \vec{b}$ are along the same straight line but the point in the opposite direction, then

$| \vec{a} - \vec{b} | = | \vec{a} | + | \vec{b} |$ Combining the conditions stated in the equations (v) and (vi), we have

$\begin{matrix} | \vec{a} - \vec{b} | \leq | \vec{a} | + | \vec{b} | \\ (d) To prove | \vec{a} - \vec{b} | \geq | | \vec{a} | - | \vec{b} | \end{matrix}$

In figure (ii) again consider the AOMN. It follows that

$\begin{array}{l} O N + O M > M N \\ or , O N > | M N - O M | \end{array}$

The modulus of MN - OM has been taken for the reason that whereas L.H.S. is positive, R.H.S. may be negative, in case MN is smaller than OM. Since MN= OL, we have

$\begin{array}{l} O N > | O L - O M | \\ | \vec{a} - \vec{b} | > | | \vec{a} | - | - \vec{b} | \\ or , | \vec{a} - \vec{b} | > | | \vec{a} | - | \vec{b} | \end{array}$

In case, the vectors ā and b are along the same straight line and point in the same direction, then

$| \vec{a} - \vec{b} | = | | \vec{a} | - | \vec{b} |$ ..(iv)

Combining the conditions stated in equations (vii) and (vii), we have

$| \vec{a} - \vec{b} | \geq | | \vec{a} | - | \vec{b} |$

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