JEE Main 2026 21 January Evening Shift
Exam JEE Main
Year 2026
Date 21 January
Shift Evening Shift
Subject Mathematics
Chapter Vector Algebra JEE Main 2026 · 21 January · Evening Shift · Mathematics · Vector Algebra
For a triangle ABC, let p = BC , q = CA and r = BA . If | p |=2 3...
For a triangle ABC, let
p → = B C → , q → = C A → \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{CA}} p = BC , q = CA and
r → = B A → \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{BA}} r = BA . If
∣ p → ∣ = 2 3 , ∣ q → ∣ = 2 |\overrightarrow{\mathrm{p}}|=2 \sqrt{3},|\overrightarrow{\mathrm{q}}|=2 ∣ p ∣ = 2 3 , ∣ q ∣ = 2 and
cos θ = 1 3 \cos \theta=\frac{1}{\sqrt{3}} cos θ = 3 1 , where
θ \theta θ is the angle between
p ⃗ \vec{p} p and
q ⃗ \vec{q} q , then
∣ p ⃗ × ( q ⃗ − 3 r ⃗ ) ∣ 2 + 3 ∣ r ⃗ ∣ 2 |\vec{p} \times(\vec{q}-3 \vec{r})|^{2}+3|\vec{r}|^{2} ∣ p × ( q − 3 r ) ∣ 2 + 3∣ r ∣ 2 is equal to :
Step-by-step solution
Given
∣ p ⃗ ∣ = 2 3 |\vec{p}| = 2\sqrt{3} ∣ p ∣ = 2 3 ,
∣ q ⃗ ∣ = 2 |\vec{q}| = 2 ∣ q ∣ = 2 , and
cos θ = 1 3 \cos\theta = \frac{1}{\sqrt{3}} cos θ = 3 1 where
θ \theta θ is the angle between
p ⃗ \vec{p} p and
q ⃗ \vec{q} q .
Since
p ⃗ = B C → \vec{p} = \overrightarrow{BC} p = BC ,
q ⃗ = C A → \vec{q} = \overrightarrow{CA} q = C A , and
r ⃗ = B A → \vec{r} = \overrightarrow{BA} r = B A , from triangle closure
r ⃗ = p ⃗ + q ⃗ \vec{r} = \vec{p} + \vec{q} r = p + q .
p ⃗ ⋅ q ⃗ = ∣ p ⃗ ∣ ∣ q ⃗ ∣ cos θ = 2 3 ⋅ 2 ⋅ 1 3 = 4 \vec{p} \cdot \vec{q} = |\vec{p}||\vec{q}|\cos\theta = 2\sqrt{3} \cdot 2 \cdot \frac{1}{\sqrt{3}} = 4 p ⋅ q = ∣ p ∣∣ q ∣ cos θ = 2 3 ⋅ 2 ⋅ 3 1 = 4 ∣ r ⃗ ∣ 2 = ∣ p ⃗ ∣ 2 + ∣ q ⃗ ∣ 2 + 2 p ⃗ ⋅ q ⃗ = 12 + 4 + 8 = 24 |\vec{r}|^2 = |\vec{p}|^2 + |\vec{q}|^2 + 2\vec{p} \cdot \vec{q} = 12 + 4 + 8 = 24 ∣ r ∣ 2 = ∣ p ∣ 2 + ∣ q ∣ 2 + 2 p ⋅ q = 12 + 4 + 8 = 24 q ⃗ − 3 r ⃗ = q ⃗ − 3 ( p ⃗ + q ⃗ ) = − 3 p ⃗ − 2 q ⃗ \vec{q} - 3\vec{r} = \vec{q} - 3(\vec{p} + \vec{q}) = -3\vec{p} - 2\vec{q} q − 3 r = q − 3 ( p + q ) = − 3 p − 2 q p ⃗ × ( q ⃗ − 3 r ⃗ ) = − 2 ( p ⃗ × q ⃗ ) \vec{p} \times (\vec{q} - 3\vec{r}) = -2(\vec{p} \times \vec{q}) p × ( q − 3 r ) = − 2 ( p × q ) sin 2 θ = 1 − 1 3 = 2 3 \sin^2\theta = 1 - \frac{1}{3} = \frac{2}{3} sin 2 θ = 1 − 3 1 = 3 2 ∣ p ⃗ × q ⃗ ∣ 2 = 12 ⋅ 4 ⋅ 2 3 = 32 |\vec{p} \times \vec{q}|^2 = 12 \cdot 4 \cdot \frac{2}{3} = 32 ∣ p × q ∣ 2 = 12 ⋅ 4 ⋅ 3 2 = 32 ∣ p ⃗ × ( q ⃗ − 3 r ⃗ ) ∣ 2 = 4 ⋅ 32 = 128 |\vec{p} \times (\vec{q} - 3\vec{r})|^2 = 4 \cdot 32 = 128 ∣ p × ( q − 3 r ) ∣ 2 = 4 ⋅ 32 = 128 Result:
128 + 3 ( 24 ) = 128 + 72 = 200 128 + 3(24) = 128 + 72 = 200 128 + 3 ( 24 ) = 128 + 72 = 200
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