Let a=2 i+j-k and b=i+2 j+k be two vectors. Consider a vector c= … - Vidyadip

MCQ

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 $\vec{c}=\alpha \vec{a}+\beta \vec{b}, \alpha, \beta \in R$. If the projection of $\vec{c}$ on the vector $(\vec{a}+\vec{b})$ is $3 \sqrt{2}$, then the minimum value of $(\vec{c}-(\vec{a} \times \vec{b}))$. $\vec{c}$ equals

✓
$18$
B
$20$
C
$25$
D
$30$

✓

Answer

Correct option: A.

$18$

a
$\overrightarrow{ c }=(2 \alpha+\beta) \hat{ i }+\hat{ j }(\alpha+2 \beta)+\hat{ k }(\beta-\alpha)$

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

$\Rightarrow \quad \alpha+\beta=2$

$(\overrightarrow{ c }-(\overrightarrow{ a } \times \overrightarrow{ b })) \cdot(\alpha \overrightarrow{ a }+\beta \overrightarrow{ b })$

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

$=6\left(\alpha^2+\beta^2+\alpha \beta\right)$

$=6\left(\alpha^2+(2-\alpha)^2+\alpha(2-\alpha)\right)$

$=6\left((\alpha-1)^2+3\right)$

$\Rightarrow \quad \text { Min. value }=18$

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