For a triangle ABC, let p = B C , q = C A a n d r = B A . If | p | = 2 3 , | q | = 2 a n d cos = 1 3 , where is the angle between p a n d q , t h e n | p × ( q - 3 r ) | 2 + 3 | r | 2 is equal to: [2026]

Question

For a triangle ABC, let $\vec{p} = \vec{B C}, \vec{q} = \vec{C A} and \vec{r} = \vec{B A}$ . If $| \vec{p} | = 2 \sqrt{3}, | \vec{q} | = 2 and \cos θ = \frac{1}{\sqrt{3}},$

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: [2026]

Smart Achievers · Accepted Answer

(3)

$\vec{p} + \vec{q} = \vec{r}$

$\cos (π - θ) = \frac{| \vec{p} |^{2} + | \vec{q} |^{2} - | \vec{r} |^{2}}{2 | \vec{p} | | \vec{q} |}$

$\frac{- 1}{\sqrt{3}} = \frac{12 + 4 - | \vec{r} |^{2}}{2 \cdot 2 \sqrt{3} \cdot 2}$

$| \vec{r} |^{2} = 24$

$∴$ $| \vec{p} \times (\vec{q} - 3 \vec{r}) |^{2} + 3 | \vec{r} |^{2}$

$= | \vec{p} \times (\vec{q} - 3 \vec{p} - 3 \vec{q}) |^{2} + 72$

$= | \vec{p} \times (- 3 \vec{p} - 2 \vec{q}) |^{2} + 72$

$= | - 2 \vec{p} \times \vec{q} |^{2} + 72$

$= 4 | \vec{p} |^{2} | \vec{q} |^{2} \sin^{2} θ + 72$

$= 4 \cdot 12 \cdot 4 \cdot \frac{2}{3} + 72$

$= 200$

Solution

1 340

2 220

3 200

4 410

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