Vector Algebra
Triangle Law — Cross Product Magnitude
nta_pyq_2026_jan
Grade 12
Question:
For a triangle ABC, let $\vec{p}=\overrightarrow{BC}$, $\vec{q}=\overrightarrow{CA}$ and $\vec{r}=\overrightarrow{BA}$. If $|\vec{p}|=2\sqrt{3}$, $|\vec{q}|=2$ and $\cos\theta=\dfrac{1}{\sqrt{3}}$, where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p}\times(\vec{q}-3\vec{r})|^2+3|\vec{r}|^2$ is equal to:
Step-by-Step Solution
Key Concept: From triangle closure $\vec{r}=\vec{p}+\vec{q}$. Compute $\vec{p}\cdot\vec{q}=2\sqrt{3}\cdot2\cdot\frac{1}{\sqrt{3}}=4$. Then $|\vec{r}|^2=12+4+8=24$. Simplify $\vec{q}-3\vec{r}=-3\vec{p}-2\vec{q}$, so $\vec{p}\times(\vec{q}-3\vec{r})=-2(\vec{p}\times\vec{q})$.
$128+72=200$.
Correct Answer: 3