TS EAMCET 10 May 2024 Shift 2 Solved Paper

Given that the vectors $a, b$, and $c$ each have a magnitude of $\sqrt{2}$ and the angle between any two vectors is $\;\frac{\pi}{3}$, we can analyze the vectors $x$ and $y$ defined as follows:$|a|=|b|=|c|=\sqrt{2}$First, let's express $x$ :$x = a \times (b \times c)$Using the vector triple product identity:$a \times (b \times c) = (a \cdot c) b - (a \cdot b) c$Given that the angle between each pair of vectors is $\;\frac{\pi}{3}$, the dot products are:$a \cdot b = |a||b| \cos \left(\frac{\pi}{3}\right) = \sqrt{2} \times \sqrt{2} \times \frac{1}{2} = 1$Therefore, the expression for $x$ becomes:$x = 1 \cdot b - 1 \cdot c = b - c$Now, consider $y$ :$y = b \times (c \times a)$Using the vector triple product identity again:$b \times (c \times a) = (b \cdot a) c - (b \cdot c) a$This simplifies similarly to: $y = 1 \cdot c - 1 \cdot a = c - a$Hence, the magnitudes of $x$ and $y$ are:$|x|\;=|b-c|$$|y|\;=|c-a|$Since both expressions have the same form, it follows that:$|x|=|y|$