QJEEM21Mab0c9: JEE Main 2021 Physics - Class 11 JEEM21Mab0c9 Solution: Let and be two vectors such that and the angle between and is…

Question

Verified Solution • 2025–26 Session moderate

Let

\overrightarrow a

and

\overrightarrow b

be two vectors
such that

\left| {2\overrightarrow a + 3\overrightarrow b } \right| = \left| {3\overrightarrow a + \overrightarrow b } \right|

and the angle between

\overrightarrow a

and

\overrightarrow b

is 60

^\circ

. If

{1 \over 8}\overrightarrow a

is a unit vector, then

\left| {\overrightarrow b } \right|

is equal to :

Accepted Answer

{\left| {3\overrightarrow a + \overrightarrow b } \right|^2} = {\left| {2\overrightarrow a + 3\overrightarrow b } \right|^2}

\left( {3\overrightarrow a + \overrightarrow b } \right).\left( {3\overrightarrow a + \overrightarrow b } \right) = \left( {2\overrightarrow a + 3\overrightarrow b } \right).\left( {2\overrightarrow a + 3\overrightarrow b } \right)

9\overrightarrow a .\,\overrightarrow a + 6\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow b = 4\overrightarrow a \,.\,\overrightarrow a + 12\overrightarrow a \,.\,\overrightarrow b + 9\overrightarrow b \,.\,\overrightarrow b

5{\left| {\overrightarrow a } \right|^2} - 6\overrightarrow a \,.\,\overrightarrow b = 8{\left| {\overrightarrow b } \right|^2}

5{(8)^2} - 6.8\,.\,\left| {\overrightarrow b } \right|\cos 60^\circ = 8{\left| {\overrightarrow b } \right|^2}

\because

(\begin{matrix} \frac{1}{8} | \vec{a} | = 1 \\ \Rightarrow | \vec{a} | = 8 \end{matrix})

40 - 3\left| {\overrightarrow b } \right| = {\left| {\overrightarrow b } \right|^2}

\Rightarrow {\left| {\overrightarrow b } \right|^2} + 3\left| {\overrightarrow b } \right| - 40 = 0

\left| {\overrightarrow b } \right| = - 8

,

\left| {\overrightarrow b } \right| = 5

(rejected)

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