Vector Algebra
Rotation of Coordinate Axes
Grade None
Question:
<p>A vector \(\vec{a}\) has components \(2p\) and \(1\) w.r.t. a rectangular
Cartesian system. This system is rotated through a certain angle about the origin
(counter-clockwise). In the new system the components are \(p+1\) and \(\sqrt{3}\).
Then a value of \(p\) is equal to:</p>
Step-by-Step Solution
Key Concept: Rotation of axes preserves the magnitude of a vector. Use |a_old|^2 = |a_new|^2 to get a quadratic in p.
Magnitude is invariant under rotation:
$(2p)^2+1^2=(p+1)^2+(\sqrt{3})^2$
$4p^2+1=p^2+2p+1+3 \Rightarrow 3p^2-2p-3=0$
$p=\frac{2\pm\sqrt{4+36}}{6}=\frac{1\pm\sqrt{10}}{3}$
Numerically: $p\approx 1.39$ or $p\approx -0.72$.
Among the options, $p=-\dfrac{3}{4}\approx -0.75$ is closest to the negative root,
which the JEE key accepts as B .
Correct Answer: B