Question

Two vectors A and B are such that, A + B = C and A^2 + B^2 = C^2. If ‘phi’ is the angle between positive direction A and B then the correct statement is, *​

Answer 1

$\large\rm { \vec {A} + \vec {B} = \vec {C} \ and \ </p><p>A² + B² = C²}$

$\large\rm { ( \vec{A} + \vec{B} ) \cdot ( \vec{A} + \vec{B} ) = | \vec{A} |² + \vec {A} \cdot \vec {B} + \vec {B} \cdot \vec{A} + | \vec{B} |² }$

$\large\rm { = C² + 2 \vec {A} \cdot \vec {B}}$

Where , $\large\rm { \vec {A} \ and \ \vec{B} }$ should be zero.

so, $\large\rm { \vec {A} \propto \vec {B} }$

so, third option is correct. (A is perpendicular to B).

Kindly scroll the answer upto right side.

Answer 2

Answer:

\large\rm { \vec {A} + \vec {B} = \vec {C} \ and \ A² + B² = C²}

A

+

B

=

C

and A²+B²=C²

\large\rm { ( \vec{A} + \vec{B} ) \cdot ( \vec{A} + \vec{B} ) = | \vec{A} |² + \vec {A} \cdot \vec {B} + \vec {B} \cdot \vec{A} + | \vec{B} |² }(

A

+

B

)⋅(

A

+

B

)=∣

A

∣²+

A

⋅

B

+

B

⋅

A

+∣

B

∣²

\large\rm { = C² + 2 \vec {A} \cdot \vec {B}}=C²+2

A

⋅

B

Where , \large\rm { \vec {A} \ and \ \vec{B} }

A

and

B

should be zero.

so, \large\rm { \vec {A} \propto \vec {B} }

A

∝

B

so, third option is correct. (A is perpendicular to B).

Kindly scroll the answer upto right side.