Question

The angle between and $\vec { B } = \hat { i } - \hat { j }$ is

• A. $45 ^ { \circ }$
• B. $90 ^ { \circ }$
• C. $- 45 ^ { \circ }$
• D. $180 ^ { \circ }$

Answer 1

Answer: (B)

$90 ^ { \circ }$

Answer 2

Given

$\therefore | \overrightarrow { \mathrm { A } } | = \sqrt { ( 1 ) ^ { 2 } + ( 1 ) ^ { 2 } } = \sqrt { 2 }$and

$| \overrightarrow { \mathrm { B } } | - \sqrt { ( 1 ) ^ { 2 } + ( - 1 ) ^ { 2 } } = \sqrt { 2 }$

Let$\theta$be angle between the vector $\overrightarrow { \mathrm { A } }$ and . Then according to definition of scalar product (or dot product) $\overrightarrow { \mathrm { A } } \cdot \overrightarrow { \mathrm { B } } = | \overrightarrow { \mathrm { A } } | | \overrightarrow { \mathrm { B } } | \cos \theta$

Or $\cos \theta \frac { \vec { A } \cdot \vec { B } } { | \vec { A } | | \vec { B } | } = \frac { ( \hat { i } + \hat { j } ) \cdot ( \hat { i } - \hat { j } ) } { ( \sqrt { 2 } ) ( \sqrt { 2 } ) } = 0$

$\therefore \theta = \cos ^ { - 1 } ( 0 ) = 90 ^ { \circ }$