QJEEM21Ma5d33: JEE Main 2021 Physics - Class 11 JEEM21Ma5d33 Solution: If is perpendicular to and is perpendicular to , then the angle …

Question

Verified Solution • 2025–26 Session moderate

If

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

is perpendicular to

\left( {7\overrightarrow a - 5\overrightarrow b } \right)

and

\left( {\overrightarrow a - 4\overrightarrow b } \right)

is perpendicular to

\left( {7\overrightarrow a - 2\overrightarrow b } \right)

, then the angle between

\overrightarrow a

and

\overrightarrow b

(in degrees) is _______________.

Accepted Answer

\left( {\overrightarrow a + 3\overrightarrow b } \right) \bot \left( {7\overrightarrow a - 5\overrightarrow b } \right)

\therefore

\left( {\overrightarrow a + 3\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 5\overrightarrow b } \right) = 0

\Rightarrow

7{\left| {\overrightarrow a } \right|^2} - 15{\left| {\overrightarrow b } \right|^2} + 16\overrightarrow a \,.\,\overrightarrow b = 0

....(1)

Also,

\left( {\overrightarrow a - 4\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 2\overrightarrow b } \right) = 0

\Rightarrow

7{\left| {\overrightarrow a } \right|^2} + 8{\left| {\overrightarrow b } \right|^2} - 30\overrightarrow a \,.\,\overrightarrow b = 0

.....(2)

Equation (1) × 30

210{\left| {\overrightarrow a } \right|^2} - 450{\left| {\overrightarrow b } \right|^2} + 480\overrightarrow a \,.\,\overrightarrow b = 0

....(3)

Equation (2) × 16

112{\left| {\overrightarrow a } \right|^2} + 128{\left| {\overrightarrow b } \right|^2} - 480\overrightarrow a \,.\,\overrightarrow b = 0

.....(4)

from (3) & (4)

322{\left| {\overrightarrow a } \right|^2} = 322{\left| {\overrightarrow b } \right|^2}

\Rightarrow

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

\Rightarrow

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

From equation (2),

15\left| {\overrightarrow a } \right| = 30\overrightarrow a .\overrightarrow b

\Rightarrow

15{\left| {\overrightarrow a } \right|^2} = 30\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \theta

\cos \theta = {{15} \over {30}} = {1 \over 2}

\therefore

\theta = 60^\circ

Step-by-Step Verified Solution