Let $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0},|\overrightarrow{\mathrm{a}}|=3,|\overrightarrow{\mathrm{b}}|=4,|\overrightarrow{\mathrm{c}}|=5$ Statement-I: Angle between $\vec{\mathrm{a}}$ and $\vec{\mathrm{b}}$ is $\frac{\pi}{2}$ Statement-II: $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^2=|\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{b}}|^2+2|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|$

Statement-I is true and statement-II is false

Statement-I is false and statement-II is true

Statement-I is false and statement-II is false

JEE PHYSICS MCQ: MOTION IN PLANE

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Analyzing +4 | -1

Let

\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0},|\overrightarrow{\mathrm{a}}|=3,|\overrightarrow{\mathrm{b}}|=4,|\overrightarrow{\mathrm{c}}|=5

Statement-I: Angle between

\vec{\mathrm{a}}

and

\vec{\mathrm{b}}

\frac{\pi}{2}

Statement-II:

|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^2=|\overrightarrow{\mathrm{a}}|^2+|\overrightarrow{\mathrm{b}}|^2+2|\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}|

View answer and explanation

Correct option: A

Statement-I is true and statement-II is true

Statement-I is true because the vectors form a right-angled triangle, and Statement-II is true as it represents the law of cosines.