Question

Two vectors A and B are such that |A + B = |A - Bl. The angle between the vectors A and B is
(A) O
(B) π/3 (C) π/2 (D) π
​

Answer 1

Solution :

⏭ Given:

✏ Two vectors A and B are such that |A+B| = |A-B|

⏭ To Find:

✏ Angle between the vectors A and B

⏭ Formula:

✏ As per parallelogram vector addition,

$\bigstar \: \boxed{ \bf{ \pink{ | \vec{A} + \vec{B}| = \sqrt{ { | \vec{A}| }^{2} + { | \vec{B}| }^{2} + 2 | \vec{A}| | \vec{B}| \cos \theta} }}}$

✏ As per parallelogram vector

subtraction,

$\bigstar \: \boxed{ \bf{ \purple{ | \vec{A}- \vec{B}| = \sqrt{ { | \vec{A}| }^{2} + { | \vec{B}| }^{2} - 2 | \vec{A}| | \vec{B}| \cos \theta} }}}$

⏭ Calculation:

$\circ \bf \: \red{ | \vec{A} + \vec{B}| = | \vec{A} - \vec{B}| } \\ \\ \circ \bf \: \sqrt{ { | \vec{A}| }^{2} + { | \vec{B}| }^{2} + 2 | \vec{A}| | \vec{B}| \cos \theta} = \sqrt{ { | \vec{A}| }^{2} + { | \vec{B}| }^{2} - 2 | \vec{A}| | \vec{B}| \cos \theta } \\ \\ \circ \bf \: 2 | \vec{A}| | \vec{B}| \cos \theta + 2 | \vec{A}| | \vec{B}| \cos \theta = 0 \\ \\ \circ \bf \: 4 | \vec{A}| | \vec{B}| \cos \theta = 0 \\ \\ \circ \bf \: \cos \theta = 0 \\ \\ \circ \bf \: \theta = { \cos}^{ - 1} (0) \\ \\ \circ \: \boxed{ \tt{ \green{ \theta = 90 \degree = \dfrac{\pi}{2} \: rad }}} \: \orange{ \bigstar}$

Answer 2

$\tt{ \purple{ \leadsto{ \: \phi \: = 90°}}}$