Vector Algebra
Parallelogram Diagonal Identity
Grade 12
Question:
<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>
<li>\(\dfrac{\pi}{2}\)</li>
<li>\(\dfrac{\pi}{3}\)</li>
<li>\(\dfrac{\pi}{4}\)</li>
<li>\(\dfrac{\pi}{6}\)</li>
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