(C) Let $\overrightarrow{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$.Given that $\overrightarrow{a} \cdot \hat{i} = 1$,we have $a_{1} = 1$.Giv

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(C) Let $\overrightarrow{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$.Given that $\overrightarrow{a} \cdot \hat{i} = 1$,we have $a_{1} = 1$.Given that $\overrightarrow{a} \cdot (\hat{i} + \hat{j}) = 1$,we have $a_{1} + a_{2} = 1$. Substituting $a_{1} = 1$,we get $1 + a_{2} = 1$,which implies $a_{2} = 0$.Given that $\overrightarrow{a} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$,we have $a_{1} + a_{2} + a_{3} = 1$. Substituting $a_{1} = 1$ and $a_{2} = 0$,we get $1 + 0 + a_{3} = 1$,which implies $a_{3} = 0$.Therefore,$\overrightarrow{a} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k} = \hat{i}$.

Class 12 Mathematics Vector Algebra Scalar or Dot product of two vectors and its applications

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

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A
$\hat{i} + \hat{j}$
B
$\hat{i} - \hat{k}$
C
$\hat{i}$
D
$\hat{i} + \hat{j} - \hat{k}$

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Scalar or Dot product of two vectors and its applications

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$OA$ and $OB$ are two vectors of magnitudes $5$ and $6$ respectively. If $\angle BOA = 60^{\circ}$,then $OA \cdot OB$ is equal to

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Classify the following as scalar or vector quantities:
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If $a, b, c$ are the position vectors of the points $A, B, C$ respectively, then match the items of List-$I$ with those of List-$II$.
List-$I$ List-$II$
$A$. $a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = 3\hat{i} + 4\hat{j} + 2\hat{k}, c = 4\hat{i} + 2\hat{j} + 3\hat{k}$ $I$. $\triangle ABC$ is an equilateral triangle
$B$. $a = \hat{i} + 2\hat{j} + 3\hat{k}, b = 3\hat{i} + 4\hat{j} + 7\hat{k}, c = -3\hat{i} - 2\hat{j} - 5\hat{k}$ $II$. $\triangle ABC$ is an isosceles triangle
$C$. $a = 2\hat{i} - \hat{j} + \hat{k}, b = \hat{i} - 3\hat{j} - 5\hat{k}, c = -3\hat{i} - 4\hat{j} - 4\hat{k}$ $III$. $\triangle ABC$ is a right-angled triangle
$D$. $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2\hat{j} + 3\hat{k}, c = 2\hat{i} - \hat{j} + \hat{k}$ $IV$. $A, B, C$ are collinear

The correct match is:

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If $\bar{a}=(2 \hat{i}+2 \hat{j}+3 \hat{k})$,$\bar{b}=(-\hat{i}+2 \hat{j}+\hat{k})$ and $\bar{c}=(3 \hat{i}+\hat{j})$ such that $(\bar{a}+\lambda \bar{b})$ is perpendicular to $\bar{c}$,then the value of $\lambda$ is

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

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List-$I$	List-$II$
$A$. $a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = 3\hat{i} + 4\hat{j} + 2\hat{k}, c = 4\hat{i} + 2\hat{j} + 3\hat{k}$	$I$. $\triangle ABC$ is an equilateral triangle
$B$. $a = \hat{i} + 2\hat{j} + 3\hat{k}, b = 3\hat{i} + 4\hat{j} + 7\hat{k}, c = -3\hat{i} - 2\hat{j} - 5\hat{k}$	$II$. $\triangle ABC$ is an isosceles triangle
$C$. $a = 2\hat{i} - \hat{j} + \hat{k}, b = \hat{i} - 3\hat{j} - 5\hat{k}, c = -3\hat{i} - 4\hat{j} - 4\hat{k}$	$III$. $\triangle ABC$ is a right-angled triangle
$D$. $a = \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2\hat{j} + 3\hat{k}, c = 2\hat{i} - \hat{j} + \hat{k}$	$IV$. $A, B, C$ are collinear

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

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$OA$ and $OB$ are two vectors of magnitudes $5$ and $6$ respectively. If $\angle BOA = 60^{\circ}$,then $OA \cdot OB$ is equal to

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