If overrightarrow{a} cdot hat{i} = overrighta | vedclass

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(D) Let $\overrightarrow{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.Given $\overrightarrow{a} \cdot \hat{i} = 1$, we have $(a_1 \hat{i} + a_2 \hat{

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$\frac{1}{3}(3 \hat{i} - 3 \hat{j} + \hat{k})$

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(D) Let $\overrightarrow{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.Given $\overrightarrow{a} \cdot \hat{i} = 1$, we have $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot \hat{i} = 1$, which implies $a_1 = 1$.Next, $\overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = 1$, so $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (2 \hat{i} + \hat{j}) = 1$.This gives $2a_1 + a_2 = 1$. Substituting $a_1 = 1$, we get $2(1) + a_2 = 1$, so $a_2 = -1$.Finally, $\overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$, so $(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$.This gives $a_1 + a_2 + 3a_3 = 1$. Substituting $a_1 = 1$ and $a_2 = -1$, we get $1 - 1 + 3a_3 = 1$, which implies $3a_3 = 1$, so $a_3 = \frac{1}{3}$.Thus, $\overrightarrow{a} = \hat{i} - \hat{j} + \frac{1}{3} \hat{k} = \frac{1}{3}(3 \hat{i} - 3 \hat{j} + \hat{k})$.

Column-$I$	Column-$II$
$1. m = -2$	$a$. Convex mirror
$2. m = -\frac{1}{2}$	$b$. Concave mirror
$3. m = +2$	$c$. Real Image
$4. m = +\frac{1}{2}$	$d$. Virtual image

If $\overrightarrow{a} \cdot \hat{i} = \overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = \overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$, then $\overrightarrow{a}$ is equal to :

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Match the corresponding entries of Column-$I$ with Column-$II$. [Where $m$ is the magnification produced by the mirror]

Column-$I$ Column-$II$

$1. m = -2$ $a$. Convex mirror

$2. m = -\frac{1}{2}$ $b$. Concave mirror

$3. m = +2$ $c$. Real Image

$4. m = +\frac{1}{2}$ $d$. Virtual image

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