For a triangle A B C, let vec p=->B C, vec q=->C A and -> r=-> BA. If |-> p|=2 sqrt(3),|-> q|=2 and cos theta=frac1sqrt(3), where theta is the angle between vec P and vec q, then |-> p x (-> q-3 -> r)|^2+3|-> r|^2 is equal to:

-> p+-> q=-> r cos (pi-theta)=frac|-> p|^2+|-> q|^2-|-> r|^22|-> p \| -> q| frac-1sqrt(3)=frac12+4-|-> r|^22 * 2 sqrt(3) * 2 |-> r|^2=24 Therefore |-> p x (-> q-3 -> r)|^2+3|-> r|^2 =|vec p x (vec q-3 vec p-3 vec q)|^2+72 =|vec p x (-3 vec p-2 vec q)|^2+72 =|-2 -> p x -> q|^2+72 =4|-> p|^2|-> q|^2 x sin ^2 theta+72 =4 * 12.4 * (2 / 3)+72 =200

For a triangle A B C let p B C q C A and r BA If p 2 sq... | null (Mathematics)

JEE Mains 2026 21st January Shift 2

Mathematics

For a triangle

A B C

, let

\vec{p}=\overrightarrow{B C}, \vec{q}=\overrightarrow{C A}

and

\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{BA}}

. If

|\overrightarrow{\mathrm{p}}|=2 \sqrt{3},|\overrightarrow{\mathrm{q}}|=2

and

\cos \theta=\frac{1}{\sqrt{3}}

, where

\theta

is the angle between

\vec{P}

and

\vec{q}

, then

|\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}}-3 \overrightarrow{\mathrm{r}})|^{2}+3|\overrightarrow{\mathrm{r}}|^{2}

is equal to:

340

220

410

200

Prev Next

Correct Answer:

Explanation

\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{r}}\\

\cos (\pi-\theta)=\frac{|\overrightarrow{\mathrm{p}}|^{2}+|\overrightarrow{\mathrm{q}}|^{2}-|\overrightarrow{\mathrm{r}}|^{2}}{2|\overrightarrow{\mathrm{p}} \| \overrightarrow{\mathrm{q}}|}\\

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

|\overrightarrow{\mathrm{r}}|^{2}=24\\

\therefore|\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}}-3 \overrightarrow{\mathrm{r}})|^{2}+3|\overrightarrow{\mathrm{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 \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|^{2}+72\\

=4|\overrightarrow{\mathrm{p}}|^{2}|\overrightarrow{\mathrm{q}}|^{2} \times \sin ^{2} \theta+72\\

=4 \cdot 12.4 \cdot \frac{2}{3}+72\\

=200

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