JEE Main 2020 — Mathematics Vectors & 3D Geometry

2020numerical valuemedium

Let the position vectors of points ' $A$ ' and ' $B$ ' be $\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}+3\hat{k},$ respectively. A point $'P'$ divides the line segment $AB$ internally in the ratio $\lambda :1(\lambda >0)$ . If $O$ is the origin and $\vec{\mathrm{OB}}\cdot \vec{\mathrm{OP}}-3{|\vec{\mathrm{OA}}\times \vec{\mathrm{OP}}|}^{2}=6$ then $\lambda$ is equal to

Official previous-year question

Held on 2 Sept 2020 · Verified 6 Jul 2026.

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Solution

Position vector of $P$ is $\vec{OP}=\frac{\vec{a}+\lambda \vec{b}}{\lambda +1}$ $\because \vec{\mathrm{OB}}\cdot \vec{\mathrm{OP}}-3|\vec{\mathrm{OA}}\times \vec{\mathrm{OP}}{|}^{2}=6$

$\Rightarrow \vec{b}\cdot (\frac{\vec{a}+\lambda \vec{b}}{\lambda +1})-3{|\vec{a}\times (\frac{\vec{a}+\lambda \vec{b}}{\lambda +1})|}^{2}=6$

$\Rightarrow \frac{\vec{a}\cdot \vec{b}+\lambda |\vec{b}{|}^{2}}{\lambda +1}-\frac{3{\lambda }^{2}}{(\lambda +1{)}^{2}}|\vec{a}\times \vec{b}{|}^{2}=6$

$\Rightarrow \frac{6+\lambda .14}{\lambda +1}-\frac{3{\lambda }^{2}}{(\lambda +1{)}^{2}}\cdot 6=6$

$\Rightarrow \frac{18{\lambda }^{2}}{(\lambda +1{)}^{2}}+6=6+\frac{8\lambda }{\lambda +1}$

$\Rightarrow 18{(\frac{\lambda }{\lambda +1})}^{2}-\frac{8\lambda }{\lambda +1}=0$ $(\frac{\lambda }{\lambda +1}\neq 0)$

$\Rightarrow 10\lambda =8\Rightarrow \lambda =0.8$

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Solution

Position vector of $P$ is $\vec{OP}=\frac{\vec{a}+\lambda \vec{b}}{\lambda +1}$ $\because \vec{\mathrm{OB}}\cdot \vec{\mathrm{OP}}-3|\vec{\mathrm{OA}}\times \vec{\mathrm{OP}}{|}^{2}=6$

$\Rightarrow \vec{b}\cdot (\frac{\vec{a}+\lambda \vec{b}}{\lambda +1})-3{|\vec{a}\times (\frac{\vec{a}+\lambda \vec{b}}{\lambda +1})|}^{2}=6$

$\Rightarrow \frac{\vec{a}\cdot \vec{b}+\lambda |\vec{b}{|}^{2}}{\lambda +1}-\frac{3{\lambda }^{2}}{(\lambda +1{)}^{2}}|\vec{a}\times \vec{b}{|}^{2}=6$

$\Rightarrow \frac{6+\lambda .14}{\lambda +1}-\frac{3{\lambda }^{2}}{(\lambda +1{)}^{2}}\cdot 6=6$

$\Rightarrow \frac{18{\lambda }^{2}}{(\lambda +1{)}^{2}}+6=6+\frac{8\lambda }{\lambda +1}$

$\Rightarrow 18{(\frac{\lambda }{\lambda +1})}^{2}-\frac{8\lambda }{\lambda +1}=0$ $(\frac{\lambda }{\lambda +1}\neq 0)$

$\Rightarrow 10\lambda =8\Rightarrow \lambda =0.8$

JEE Main 2020 — Mathematics Vectors & 3D Geometry

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JEE Main 2020 — Mathematics Vectors & 3D Geometry

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Solution

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JEE Main Vectors & 3D Geometry in other years

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Other Mathematics topics for JEE Main

Algebra

Calculus

Coordinate Geometry

Probability & Statistics

Trigonometry

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