For a triangle ABC, let vecp=overrightarrowBC,vecq=overrightarrowCA$\vec{p}=\overrightarrow{BC},\vec{q}=\overrightarrow{CA}$ and vecr=overrightarrowBA$\vec{r}=\overrightarrow{BA}$. If |vecp|=2sqrt3,|vecq|=2$|\vec{p}|=2\sqrt{3},|\vec{q}|=2$ and costheta=frac1sqrt3$\cos\theta=\frac{1}{\sqrt{3}}$, where theta$\theta$ is the angle between vecp$\vec{p}$ and vecq$\vec{q}$, then left|vecptimes(vecq-3vecr)right|^2+3left|vecrright|^2$\left|\vec{p}\times(\vec{q}-3\vec{r})\right|^{2}+3\left|\vec{r}\right|^{2}$ is equal to:

### Related Formula textCosine Rule in triangle: cos(pi-theta) = frac|vecp|^2+|vecq|^2-|vecr|^22|vecp||vecq|$$\text{Cosine Rule in triangle: } \cos(\pi-\theta) = \frac{|\vec{p}|^2+|\vec{q}|^2-|\vec{r}|^2}{2|\vec{p}||\vec{q}|}$$ textCross product identity: vecA times vecA = 0 quad ; quad |vecA times vecB| = |A||B|sintheta$$\text{Cross product identity: } \vec{A} \times \vec{A} = 0 \quad ; \quad |\vec{A} \times \vec{B}| = |A||B|\sin\theta$$ ### Core Logic

From triangle law of addition, overrightarrowBC + overrightarrowCA = overrightarrowBA implies vecp + vecq = vecr$\overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{BA} \implies \vec{p} + \vec{q} = \vec{r}$. The angle between vectors vecp$\vec{p}$ and vecq$\vec{q}$ extended head-to-tail is theta$\theta$. The internal angle of the triangle is pi - theta$\pi - \theta$. ### Step 1: Calculate magnitude of r Using cosine rule for the side |vecr|$|\vec{r}|$: cos(pi - theta) = frac|vecp|^2 + |vecq|^2 - |vecr|^22|vecp||vecq|$$\cos(\pi - \theta) = \frac{|\vec{p}|^2 + |\vec{q}|^2 - |\vec{r}|^2}{2|\vec{p}||\vec{q}|}$$ Since costheta = 1/sqrt3$\cos\theta = 1/\sqrt{3}$, cos(pi - theta) = -1/sqrt3$\cos(\pi - \theta) = -1/\sqrt{3}$. -frac1sqrt3 = frac(2sqrt3)^2 + (2)^2 - |vecr|^22(2sqrt3)(2)$$-\frac{1}{\sqrt{3}} = \frac{(2\sqrt{3})^2 + (2)^2 - |\vec{r}|^2}{2(2\sqrt{3})(2)}$$ -frac1sqrt3 = frac12 + 4 - |vecr|^28sqrt3$$-\frac{1}{\sqrt{3}} = \frac{12 + 4 - |\vec{r}|^2}{8\sqrt{3}}$$ -8 = 16 - |vecr|^2 implies |vecr|^2 = 24$$-8 = 16 - |\vec{r}|^2 \implies |\vec{r}|^2 = 24$$ ### Step 2: Simplify the Cross Product term We need to evaluate |vecp times (vecq - 3vecr)|^2$|\vec{p} \times (\vec{q} - 3\vec{r})|^2$. Substitute vecr = vecp + vecq$\vec{r} = \vec{p} + \vec{q}$: = |vecp times (vecq - 3(vecp + vecq))|^2$$= |\vec{p} \times (\vec{q} - 3(\vec{p} + \vec{q}))|^2$$ = |vecp times (-3vecp - 2vecq)|^2$$= |\vec{p} \times (-3\vec{p} - 2\vec{q})|^2$$ Since vecp times vecp = 0$\vec{p} \times \vec{p} = 0$: = |-2 (vecp times vecq)|^2 = 4 |vecp times vecq|^2$$= |-2 (\vec{p} \times \vec{q})|^2 = 4 |\vec{p} \times \vec{q}|^2$$ = 4 ( |vecp|^2 |vecq|^2 sin^2theta )$$= 4 ( |\vec{p}|^2 |\vec{q}|^2 \sin^2\theta )$$ ### Step 3: Final Calculation Given costheta = 1/sqrt3 implies cos^2theta = 1/3 implies sin^2theta = 2/3$\cos\theta = 1/\sqrt{3} \implies \cos^2\theta = 1/3 \implies \sin^2\theta = 2/3$. 4 |vecp times vecq|^2 = 4 (12)(4)left(frac23right) = 4(4)(4)(2) = 128$$4 |\vec{p} \times \vec{q}|^2 = 4 (12)(4)\left(\frac{2}{3}\right) = 4(4)(4)(2) = 128$$ The required expression is: |vecp times (vecq - 3vecr)|^2 + 3|vecr|^2 = 128 + 3(24) = 128 + 72 = 200$$|\vec{p} \times (\vec{q} - 3\vec{r})|^2 + 3|\vec{r}|^2 = 128 + 3(24) = 128 + 72 = 200$$ ### Pattern Recognition Always convert secondary vectors back to base components using the triangle condition vecp + vecq = vecr$\vec{p} + \vec{q} = \vec{r}$. It zeroes out self-cross products instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra