Question

what is the answer ​

Answer 1

Given:

$\longrightarrow\rm{\vec{a}+\vec{b}=\vec{c}}$

$\longrightarrow\rm{\vec{a}^{2}+\vec{b}^{2}=\vec{c}^{2}}---(1)$

To Find:

Angle between the vectors a and b

Solution:

We know that,

• $\rm\pink{(\vec{p}+\vec{q})^{2}=\vec{p}^{2}+\vec{q}^{2}+2\vec{p}.\vec{q}}$

• $\rm\purple{\vec{p}.\vec{q}=\mid\vec{p}\mid\:\mid\vec{q}\mid cos\:\theta}$

where

|a| is magnitude of vector a

|b| is magnitude of vector b

θ is the angle between both vectors

$\rule{190}{1}$

It is given that,

$\longrightarrow\rm{\vec{a}+\vec{b}=\vec{c}}$

On squaring both the sides, we get

$\longrightarrow\rm{(\vec{a}+\vec{b})^{2}=\vec{c}^{2}}$

$\longrightarrow\rm{\red{\vec{a}^{2}+\vec{b}^{2}}+2\vec{a}.\vec{b}=\vec{c}^{2}}$

From (1), we get

$\longrightarrow\rm{\red{\vec{c}^{2}}+2\vec{a}.\vec{b}=\vec{c}^{2}}$

$\longrightarrow\rm{2\vec{a}.\vec{b}=\vec{c}^{2}-\vec{c}^{2}}$

$\longrightarrow\rm{2\vec{a}.\vec{b}=0}$

$\longrightarrow\rm{\vec{a}.\vec{b}=0}$

Let the angle between vectors a and b be θ

So,

$\longrightarrow\rm{\vec{a}.\vec{b}=0}$

$\longrightarrow\rm{\mid\vec{a}\mid\:\mid\vec{b}\mid cos\:\theta=0}$

$\longrightarrow\rm{cos\:\theta=0}$

$\longrightarrow\rm{\theta=cos^{-1}(0)}$

$\longrightarrow\rm{\theta=90^{\circ}}$

Hence, the angle between vectors a and b is 90°.

Answer 2

Given :- a+b=c,,a^2+b^2=c^2

Solution:-Simply square both side

a^2+b^2+2ab=c^2

Putting a^2+b^2=c^2 we get

2ab=c^2-c^2

2ab=0

a.b=0

this can only be possible if angle between these two Vector=90

as

a.b=abcos∅

cos∅=0/ab=0

∅=90

So angle between them=90