(A) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.We know that $|a - b|^2 = |a|^2 + |b|^2 - 2|a||b|\cos\theta$.Substituting the v

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(A) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$.We know that $|a - b|^2 = |a|^2 + |b|^2 - 2|a||b|\cos\theta$.Substituting the values,we get $|a - b|^2 = 1^2 + 1^2 - 2(1)(1)\cos\theta = 2 - 2\cos\theta$.Using the trigonometric identity $1 - \cos\theta = 2\sin^2(\theta/2)$,we have $|a - b|^2 = 2(2\sin^2(\theta/2)) = 4\sin^2(\theta/2)$.Taking the square root on both sides,we get $|a - b| = 2\sin(\theta/2)$.Therefore,$\sin(\theta/2) = \frac{1}{2}|a - b|$.

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

Medium

Let $a$ and $b$ be two unit vectors inclined at an angle $\theta$,then $\sin(\theta/2)$ is equal to

A
$\frac{1}{2}|a-b|$
B
$\frac{1}{2}|a+b|$
C
$|a-b|$
D
$|a+b|$

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Scalar or Dot product of two vectors and its applications

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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

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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

Let $a$ and $b$ be two unit vectors inclined at an angle $\theta$,then $\sin(\theta/2)$ is equal to

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