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in Vector algebra by (15 points)
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Let \( \overrightarrow{ a }=2 \hat{ i }+\hat{ j }-\hat{ k } \) and \( \overrightarrow{ b }=\hat{ i }+2 \hat{ j }+\hat{ k } \) be two vectors. Consider a vector \( \overrightarrow{ c }=\alpha \overrightarrow{ a }+\beta \overrightarrow{ b }, \alpha, \beta \in R \). If the projection of \( \overrightarrow{ c } \) on the vector \( (\overrightarrow{ a }+\overrightarrow{ b }) \) is \( 3 \sqrt{2} \), then the minimum value of \( (\overrightarrow{ c }-(\overrightarrow{ a } \times \overrightarrow{ b })) \cdot \overrightarrow{ c } \) equals

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\(\vec a = 2\hat i + \hat j - \hat k\)

\(\vec b = \hat i + 2\hat j - \hat k\)

\(\vec c = \alpha \,\vec a + \beta\,\vec b, \alpha, \beta \in\,R\) 

Projection of \(\vec c\)on the vector \((\vec a + \vec b)\)

\(= \frac{\vec c . (\vec a + \vec b)}{|\vec a + \vec b|}\)

\(= \frac{(\alpha \,\vec a+ \beta\, \vec b).(\vec a + \vec b)}{|\vec a + \vec b|}\) 

\(= \frac{\alpha (\vec a)^2 + (\alpha + \beta ) \,\vec a . \vec b + \beta |\vec b|^2}{|\vec a + \vec b |}\)

\(= \frac{6 \alpha + 3(\alpha + \beta ) + 6\beta}{|3\hat i + 3\hat j|}\)          \(\begin{pmatrix}\therefore |\vec a|^2 = 2^2 + 1^2 + (-1)^2 = 6\\ |\vec b|^2 = 1^2 + 2^2 + 1^2 = 6\\\vec a .\vec b = (2\hat i+ \hat j - \hat k).(\hat i + 2\hat j+ \hat k)\\= 2+2-1 =3\end {pmatrix}\) 

\(= \frac{9(\alpha + \beta)}{\sqrt{9+9}}\)

\(=\frac{3}{\sqrt{2}}(\alpha + \beta)\) 

Given that projection of \(\vec c\)on \((\vec a + \vec b)\) is 3√2.

\(\therefore \frac{3}{\sqrt2}(\alpha + \beta) = 3\sqrt2\)

⇒ \(\alpha + \beta = 2 \)     .......(i)

Now,

\([\vec c - (\vec a \times \vec b )]. \vec c = \vec c . \vec c - (\vec a \times \vec b). \vec c \)

\( = |\vec c|^2 - [\vec c \,\,\vec a\,\,\vec b] \)         \((\because \vec a.(\vec b \times \vec c) = [\vec a \,\, \vec b\,\, \vec c])\) 

\(= |\alpha \vec a + \beta \vec b|^2 - [\alpha \vec a \,+\beta \vec b \,\,\,\,\vec a \,\,\,\, \vec b]\)

\( = \alpha^2 |\vec a|^2 + \beta^2 |\vec b|^2 + 2 \alpha \beta \, \vec a. \vec b - \alpha[\vec a \,\,\,\, \vec a \,\,\,\,\vec b] - \beta[\vec b \,\,\,\,\vec a\,\,\,\,\vec b]\)

\(= 6\alpha^2+ 6\beta ^2 + 6\alpha \beta - 0\)      \((\because [\vec a \,\,\,\,\vec a\,\,\,\,\vec b]= 0) \) 

\( = 6 (\alpha ^2 + \beta ^2 + \alpha\beta)\)

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

\( = 6 (\alpha + \beta )^2 - 6 \alpha\beta\)

\(= 6 (2)^2 - 6\alpha \beta\)           \((\because \alpha + \beta = 2 [\text{from (i)}])\) 

\( = 24 - 6\alpha \beta\) 

\(\because \alpha + \beta = 2\)

⇒ \((\alpha + \beta)^2 = 4\) 

⇒ \(\alpha ^2 + \beta^2 + 2\alpha\beta = 4\) 

⇒ \(2 \alpha \beta = 4 -(\alpha^2 + \beta^2)\)

⇒ \(2\alpha \beta \le 4 \)         \((\because \alpha^2 + \beta^2\ge0)\)

⇒ \(\alpha \beta \le 2 \)

⇒ \(-\alpha \beta \ge -2 \)

\(\therefore [\vec c - (\vec a \times \vec b)].\vec c = 24 - 6 \alpha \beta\) 

\(\ge 24 - 12\)

⇒ \([\vec c - (\vec a \times \vec b)].\vec c \ge 12\) 

\(\therefore \) Minimum value is 12.

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