JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 20)
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 _______________.
Answer
60
Explanation
$$\left( {\overrightarrow a + 3\overrightarrow b } \right) \bot \left( {7\overrightarrow a - 5\overrightarrow b } \right)$$
$$ \therefore $$ $$\left( {\overrightarrow a + 3\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 5\overrightarrow b } \right) = 0$$
$$ \Rightarrow $$ $$7{\left| {\overrightarrow a } \right|^2} - 15{\left| {\overrightarrow b } \right|^2} + 16\overrightarrow a \,.\,\overrightarrow b = 0$$ ....(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),
$$15\left| {\overrightarrow a } \right| = 30\overrightarrow a .\overrightarrow b $$
$$ \Rightarrow $$ $$15{\left| {\overrightarrow a } \right|^2} = 30\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \theta $$
$$\cos \theta = {{15} \over {30}} = {1 \over 2}$$
$$\therefore$$ $$\theta = 60^\circ $$
$$ \therefore $$ $$\left( {\overrightarrow a + 3\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 5\overrightarrow b } \right) = 0$$
$$ \Rightarrow $$ $$7{\left| {\overrightarrow a } \right|^2} - 15{\left| {\overrightarrow b } \right|^2} + 16\overrightarrow a \,.\,\overrightarrow b = 0$$ ....(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),
$$15\left| {\overrightarrow a } \right| = 30\overrightarrow a .\overrightarrow b $$
$$ \Rightarrow $$ $$15{\left| {\overrightarrow a } \right|^2} = 30\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \theta $$
$$\cos \theta = {{15} \over {30}} = {1 \over 2}$$
$$\therefore$$ $$\theta = 60^\circ $$
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