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Unit Vectors

Let a=2î+ĵ-k̂...

Question

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 $\vec{c} = α \vec{a} + β \vec{b} α, β \in R$ . If the projection of vector $c$ 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

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Solution

Explanation for the correct answer:
Given:
The two vectors are $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} + \hat{k}$ .
Step 1: Projection of vector $\vec{c} o n (\vec{a} + \vec{b})$ :
$\frac{\vec{c} . (\vec{a} + \vec{b})}{| \vec{a} + \vec{b} |} = 3 \sqrt{2} \dots (i)$
Now let us consider, $|\vec{a} + \vec{b}|$ and substitute the values of $\vec{a}$ and $\vec{b}$ .
$\begin{array}{rcl} |\vec{a} + \vec{b}| & = & \sqrt{9 + 9} \\ = & \sqrt{18} \\ = & 3 \sqrt{2} \end{array}$
Also we get
$|\vec{a}| = \sqrt{6}, |\vec{b}| = \sqrt{6} \vec{a} . \vec{b} = 3$
Now let us consider the given condition and substitute the values of $\vec{a}$ and $\vec{b}$ .
$\begin{array}{rcl} \frac{\vec{c} . (\vec{a} + \vec{b})}{3 \sqrt{2}} & = & 3 \sqrt{2} \\ (α \vec{a} + β \vec{b}) \cdot (\vec{a} + \vec{b}) & = & 3 \sqrt{2} \times 3 \sqrt{2} \\ α (\vec{a} \cdot \vec{a}) + β (\vec{a} \cdot \vec{b}) + α (\vec{b} \cdot \vec{a}) + β (\vec{b} \cdot \vec{b}) & = & 9 \times 2 \\ α {|\vec{a}|}^{2} + β (\vec{a} \cdot \vec{b}) + α (\vec{b} \cdot \vec{a}) + β {|\vec{b}|}^{2} & = & 18 \end{array}$
Substitute the value of $|\vec{a}| = \sqrt{6}, |\vec{b}| = \sqrt{6} a n d \vec{a} . \vec{b} = 3$ in the above equation.
$\begin{array}{rcl} α {(\sqrt{6})}^{2} + β (3) + α (3) + β {(\sqrt{6})}^{2} & = & 18 \\ 6 α + 3 β + 3 α + 6 β & = & 18 \\ 9 α + 9 β & = & 18 \\ 9 (α + β) & = & 18 \\ α + β & = & \frac{18}{9} \\ α + β & = & 2 \end{array}$
Step 2: Finding minimum value of $(\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c}$
Now consider
$\begin{array}{rcl} (\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c} & = & (α \vec{a} + β \vec{b} - \vec{a} \times \vec{b}) \cdot (α \vec{a} + β \vec{b}) \\ = & α^{2} (\vec{a} \cdot \vec{a}) + α β (\vec{a} \cdot \vec{b}) + α β (\vec{a} \cdot \vec{b}) + β^{2} (\vec{b} \cdot \vec{b}) \\ = & α^{2} {(\vec{a})}^{2} + α β (\vec{a} \cdot \vec{b}) + α β (\vec{a} \cdot \vec{b}) + β^{2} {(\vec{b})}^{2} \end{array}$
Substitute the value of $|\vec{a}| = \sqrt{6}, |\vec{b}| = \sqrt{6} a n d \vec{a} . \vec{b} = 3$ in the above equation.
$\begin{array}{rcl} α^{2} (6) + α β (3) + α β (3) + β^{2} (6) & = & 6 α^{2} + 6 α β + 6 β^{2} \\ = & 6 (α^{2} + α β + β^{2}) \\ = & 6 [{(α + β)}^{2} - α β] [∵ {(a + b)}^{2} = a^{2} + b^{2} + 2 a b] \\ = & 6 [4 - α β] \\ = & 6 [4 - α (2 - α)] \\ = & 6 [4 - 2 α + α^{2}] \\ = & 6 [3 + 1 - 2 α + α^{2}] \\ = & 6 [3 + {(α - 1)}^{2}] \end{array}$
To find minimum value, take $α = 1$ Now the minimum value of
$\begin{array}{rcl} 6 (4 - α + α^{2}) & = & 6 (3) \\ = & 18 \end{array}$
Therefore, the minimum value is $18$ .
Hence, the minimum value of $(\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c}$ is equal to $18$ .

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