The resultant of two vectors of magnitudes 2A and ${\sqrt 2 A\;}$ acting at an angle $\theta $ is $\sqrt {10} A$. The correct value of $\theta $ is

NEET UG PHYSICS MCQ: MOTION IN PLANE

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Application +4 | -1

The resultant of two vectors of magnitudes 2A and

{\sqrt 2 A\;}

acting at an angle

\theta

\sqrt {10} A

. The correct value of

\theta

View answer and explanation

Correct option: B

{45^0}

Option B (

{45^0}

) is correct because using the law of cosines,

({\sqrt {10} A})^2 = (2A)^2 + ({\sqrt 2 A})^2 + 2(2A)({\sqrt 2 A})cos\theta

, which simplifies to

10A^2 = 4A^2 + 2A^2 + 4{\sqrt 2 A^2}cos\theta

, leading to

cos\theta = \frac{1}{{\sqrt 2 }}

and

\theta = 45^0

Practice All Chapter MCQs →

\mathrm{\vec a}\,and\,\mathrm{\vec b}

\mathrm{\vec a}\,and\,\mathrm{\vec b}

\overrightarrow P \,\,.\,\,\overrightarrow Q = 0\,and\,\overrightarrow P \,\, \times \,\,\overrightarrow Q = 0

\overrightarrow Q \,and\,\overrightarrow R

\vec{A}=3 \hat{i}+4 \hat{j}

\vec{B}=6 \hat{i}+8 \hat{j}

A

B

\overrightarrow{\mathrm{A}}

\overrightarrow{\mathrm{B}}

{\cot ^{ - 1}}\left( 2 \right)