Vector Algebra
Position Vectors and Ratio
Grade None
Question:
<p>Let \(P,Q,R\) have position vectors \(\vec{p}=\hat{i}+\hat{j}+\hat{k}\),
\(\vec{q}=2\hat{i}+\hat{j}\), \(\vec{r}=\hat{i}+2\hat{j}\).
Let \(\vec{a}=\vec{p}-\vec{q}\) and \(\vec{b}=\vec{q}-\vec{r}\).
If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\dfrac{\pi}{3}\), find \(\vec{a}\cdot\vec{b}\).</p>
<li>\(-1\)</li>
<li>\(\dfrac{1}{2}\)</li>
<li>\(-\dfrac{1}{2}\)</li>
<li>\(1\)</li>
Step-by-Step Solution
Key Concept: Compute a = p - q and b = q - r directly from coordinates, then find a \cdot b.
$\vec{a}=\vec{p}-\vec{q}=(1-2)\hat{i}+(1-1)\hat{j}+(1-0)\hat{k}=-\hat{i}+\hat{k}$.
$\vec{b}=\vec{q}-\vec{r}=(2-1)\hat{i}+(1-2)\hat{j}+(0-0)\hat{k}=\hat{i}-\hat{j}$.
$\vec{a}\cdot\vec{b}=(-1)(1)+(0)(-1)+(1)(0)=-1$.
Check angle: $|\vec{a}|=\sqrt{2}$, $|\vec{b}|=\sqrt{2}$.
$\cos\theta=\dfrac{-1}{2}\Rightarrow\theta=\dfrac{2\pi}{3}\neq\dfrac{\pi}{3}$.
So either the angle condition is extra info or the position vectors differ from paper.
The dot product $\vec{a}\cdot\vec{b}=-1$. Answer: A .
Correct Answer: A