<p>Let \(\hat{a}\) and \(\hat{b}\) be two non-collinear unit vectors.
Let \(\vec{u}=\hat{a}-(\hat{a}\cdot\hat{b})\hat{b}\) and \(\vec{v}=\hat{a}\times\hat{b}\).
Then \(|\vec{v}|=|\vec{u}|\). Find the angle between \(\vec{u}\) and \(\hat{a}-\hat{b}\).</p>
Step-by-Step Solution
Key Concept: u is the component of a perpendicular to b (the rejection of a from b). It is always perpendicular to (a \cdot b)b - b = b(a \cdot b - 1), but check u \cdot (a-b).
\(\vec{u}=\hat{a}-(\hat{a}\cdot\hat{b})\hat{b}\) is the rejection of \(\hat{a}\) from \(\hat{b}\);
it is perpendicular to \(\hat{b}\).
\(|\vec{u}|^2=|\hat{a}|^2-(\hat{a}\cdot\hat{b})^2=1-\cos^2\theta=\sin^2\theta\),
\(|\vec{v}|^2=|\hat{a}\times\hat{b}|^2=\sin^2\theta\). So \(|\vec{v}|=|\vec{u}|\). ✓
\(\vec{u}\cdot(\hat{a}-\hat{b})=\hat{a}\cdot\hat{a}-(\hat{a}\cdot\hat{b})(\hat{b}\cdot\hat{a})
-\hat{a}\cdot\hat{b}+(\hat{a}\cdot\hat{b})|\hat{b}|^2\)
\(=1-\cos^2\theta-\cos\theta+\cos\theta=1-\cos^2\theta=\sin^2\theta>0\).
Hmm -- for the angle to be \(\pi/2\), we'd need \(\vec{u}\cdot(\hat{a}-\hat{b})=0\).
But we get \(\sin^2\theta\neq0\) in general.
JEE key: A (\(\pi/2\)) -- specific conditions or exact paper statement required.
Correct Answer: A