JEE Mains · Maths · STD 12 - 10. vector algebra
For a triangle ABC, let \( \vec{p}=\vec{BC}, \vec{q}=\vec{CA} \) and \( \vec{r}=\vec{BA} \). If \( |\vec{p}|=2\sqrt{3}, |\vec{q}|=2 \) and \( cos\hat{\theta}=\frac{1}{\sqrt{3}} \) where θ 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 :
- A 340
- B 220
- C 410
- D 200
Answer & Solution
Correct Answer
(D) 200
Step-by-step Solution
Detailed explanation
\( \vec{p}+\vec{q}=\vec{r} \) \( cos(\pi-\theta)=\frac{|\vec{p}|^{2}+|\vec{q}|^{2}-|\vec{r}|^{2}}{2|\vec{p}||\vec{q}|} \) \( \frac{-1}{\sqrt{3}}=\frac{12+4-|\vec{r}|^{2}}{2\cdot2\sqrt{3}\cdot2} \) \(|\vec{r}|^2=24\) \( |\vec{p}\times(\vec{q}-3\vec{r})|^{2}+3|\vec{r}|^{2} \)…
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