JEE MAIN - Mathematics (2012 - No. 6)
Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two unit vectors. If the vectors $$\,\overrightarrow c = \widehat a + 2\widehat b$$ and $$\overrightarrow d = 5\widehat a - 4\widehat b$$ are perpendicular to each other, then the angle between $$\overrightarrow a $$ and $$\overrightarrow b $$ is :
$${\pi \over 6}$$
$${\pi \over 2}$$
$${\pi \over 3}$$
$${\pi \over 4}$$
Explanation
Let $$\overrightarrow c = \widehat a + 2\widehat b$$ and $$\overrightarrow d = 5\widehat a - 4\widehat b$$
Since $$\overrightarrow c $$ and $$\overrightarrow d $$ are perpendicular to each other
$$\therefore$$ $$\overrightarrow c .\overrightarrow d = 0 \Rightarrow \left( {\widehat a + 2\widehat b} \right).\left( {5\widehat a - 4\widehat b} \right) = 0$$
$$ \Rightarrow 5 + 6\widehat a.\widehat b - 8 = 0$$ $$\,\,\,\,\,\,$$ (as $$\widehat a.\widehat a = 1$$)
$$ \Rightarrow \widehat a.\widehat b = {1 \over 2} \Rightarrow \theta = {\pi \over 3}$$
Since $$\overrightarrow c $$ and $$\overrightarrow d $$ are perpendicular to each other
$$\therefore$$ $$\overrightarrow c .\overrightarrow d = 0 \Rightarrow \left( {\widehat a + 2\widehat b} \right).\left( {5\widehat a - 4\widehat b} \right) = 0$$
$$ \Rightarrow 5 + 6\widehat a.\widehat b - 8 = 0$$ $$\,\,\,\,\,\,$$ (as $$\widehat a.\widehat a = 1$$)
$$ \Rightarrow \widehat a.\widehat b = {1 \over 2} \Rightarrow \theta = {\pi \over 3}$$
Comments (0)
