Let vec a =2 hat i +hat j -hat k and vec b =hat i +2 hat j +hat k be two vectors．Consider a vector vec c =α vec a +β vec b , α, β in mathbb R ．If the projection of vec c on the vector (vec a +vec b ) is 3 √(2) ，then the minimum value of (vec c -(vec a × vec b )) × vec c equals \ \ \ \

Mathematics · Vector Algebra

JEE Advanced 2019 — Paper 2 — Question 29

Let $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ be two vectors．Consider a vector $\vec{c}=\alpha \vec{a}+\beta \vec{b}, \alpha, \beta \in \mathbb{R}$ ．If the projection of $\vec{c}$ on the vector $(\vec{a}+\vec{b})$ is $3 \sqrt{2}$ ，then the minimum value of $(\vec{c}-(\vec{a} \times \vec{b})) \cdot \vec{c}$ equals $\_\_\_\_$

Answer: 18

Numerical answer — enter this value.

Step-by-step solution

\frac{\vec{c} \cdot (\vec{a} + \vec{b})}{|\vec{a} + \vec{b}|} = 3\sqrt{2}

\Rightarrow \frac{\alpha \vec{a} \cdot \vec{a} + \beta \vec{b} \cdot \vec{b} + (\alpha + \beta) \vec{a} \cdot \vec{b}}{3\sqrt{2}} = 3\sqrt{2}

\Rightarrow 6\alpha + 6\beta + (\alpha + \beta) 3 = 18

\Rightarrow \alpha + \beta = 2

(\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c} = \alpha^2 |\vec{a}|^2 + \beta^2 |\vec{b}|^2 + 2\alpha \beta \vec{a} \cdot \vec{b} - (\vec{a} \times \vec{b}) \cdot \vec{c}

$= 6\alpha^2 + 6\beta^2 + 6\alpha \beta \quad (\text{as } \vec{c}$ is linearly dependent on $\vec{a} \& \vec{b})$

= 6[(\alpha + \beta)^2 - \alpha \beta], \max \alpha \beta = 1

\Rightarrow \min. (\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c} = 18

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Mathematics
Chapter: Vector Algebra
Topic: Projection & component of a vector along another vector.