Question

two vectors of magnitude 4 u and 3 u are such that their scalar product is zero. Find their resultant.​

Answer 1

SOLUTION

GIVEN

• Two vectors of magnitude 4 u and 3 u

• Their scalar product is zero

TO DETERMINE

The resultant

FORMULA TO BE IMPLEMENTED

In vector scalar product

1. Vector scalar product is commutative

$\vec{a}. \vec{b} = \vec{b}. \vec{a}$

$2. \: \: { | \vec{a}| }^{2} = \vec{a}. \vec{a}$

EVALUATION

Let the given two vectors are

$\vec{a} \: \: and \: \: \vec{b}$

Such that

$| \vec{a}| = 4 \: \: and \: \: | \vec{b}| = 3$

Also the scalar product is zero

$\implies \vec{a}. \vec{b} = 0$

We have to determine thier resultant i.e

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

Now

${ | \: \vec{a} + \vec{b} \: | }^{2}$

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

$= \vec{a}. \vec{a} + \vec{a}. \vec{b} + \vec{b}. \vec{a} + \vec{b}. \vec{b}$

$= { | \: \vec{a} \: | }^{2} + \vec{a}. \vec{b} + \vec{a}. \vec{b} + { | \: \vec{b} \: | }^{2}$

$= { | \: \vec{a} \: | }^{2} +2 \vec{a}. \vec{b} + { | \: \vec{b} \: | }^{2}$

$= {(4)}^{2} + (2 \times 0) + {(3)}^{2}$

$= 16 + 9$

$= 25$

$\implies \: { | \: \vec{a} + \vec{b} \: | }^{2} = 25$

$\therefore \: \: \: { | \: \vec{a} + \vec{b} \: | } = 5$

FINAL ANSWER

The resultant is 5 unit

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