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 (integer)
60
Solution
$$\left( {\overrightarrow a + 3\overrightarrow b } \right) \bot \left( {7\overrightarrow a - 5\overrightarrow b } \right)$$<br><br>$\therefore$ $$\left( {\overrightarrow a + 3\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 5\overrightarrow b } \right) = 0$$<br><br>$\Rightarrow$ $$7{\left| {\overrightarrow a } \right|^2} - 15{\left| {\overrightarrow b } \right|^2} + 16\overrightarrow a \,.\,\overrightarrow b = 0$$ ....(1)<br><br>Also, $$\left( {\overrightarrow a - 4\overrightarrow b } \right)\,.\,\left( {7\overrightarrow a - 2\overrightarrow b } \right) = 0$$<br><br>$\Rightarrow$ $$7{\left| {\overrightarrow a } \right|^2} + 8{\left| {\overrightarrow b } \right|^2} - 30\overrightarrow a \,.\,\overrightarrow b = 0$$ .....(2)
<br><br>Equation (1) × 30
<br><br>$$210{\left| {\overrightarrow a } \right|^2} - 450{\left| {\overrightarrow b } \right|^2} + 480\overrightarrow a \,.\,\overrightarrow b = 0$$ ....(3)
<br><br>Equation (2) × 16
<br><br>$$112{\left| {\overrightarrow a } \right|^2} + 128{\left| {\overrightarrow b } \right|^2} - 480\overrightarrow a \,.\,\overrightarrow b = 0$$ .....(4)
<br><br>from (3) & (4)<br><br>$$322{\left| {\overrightarrow a } \right|^2} = 322{\left| {\overrightarrow b } \right|^2}$$
<br><br>$\Rightarrow$ $${\left| {\overrightarrow a } \right|^2} = {\left| {\overrightarrow b } \right|^2}$$
<br><br>$\Rightarrow$ $\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$
<br><br>From equation (2),
<br><br>$15\left| {\overrightarrow a } \right| = 30\overrightarrow a .\overrightarrow b$
<br><br>$\Rightarrow$ $$15{\left| {\overrightarrow a } \right|^2} = 30\left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \theta $$
<br><br>$\cos \theta = {{15} \over {30}} = {1 \over 2}$<br><br>$\therefore$ $\theta = 60^\circ$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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