If \[\overset{\to }{\mathop{A}}\,=3i+j+2k\] and \[\overset{\to }{\mathop{B}}\,=2i-2j+4k\], \[\theta \] is the angle between the two vectors then \[\sin \theta \] is equal to
A.\[\dfrac{2}{3}\]
B.\[\dfrac{2}{\sqrt{3}}\]
C.\[\dfrac{2}{\sqrt{7}}\]
D.\[\dfrac{2}{\sqrt{13}}\]

Hint:In this we have to find the angle between the given two vectors and we have to find the value of \[\sin \theta \] from the angle between them. We know that the formula for the angle between two vectors is \[\cos \theta =\dfrac{\overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|}\]. We can find the value of the cosine from the formula, we can then substitute the cosine value in the formula \[\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }\], to get the value of \[\sin \theta \].

Complete step by step answer:
Here we have to find the value of \[\sin \theta \] from the given two vectors,
We know that the given two vectors are,
\[\overset{\to }{\mathop{A}}\,=3i+j+2k\] and \[\overset{\to }{\mathop{B}}\,=2i-2j+4k\]
We can now write the vectors in the form,
\[\overset{\to }{\mathop{a}}\,=3\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\,\]
\[\overset{\to }{\mathop{b}}\,=2\overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+4\overset{\wedge }{\mathop{k}}\,\]
We also know that the formula for the angle between two vectors is
\[\cos \theta =\dfrac{\overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|}\]
We can now substitute the vector values in the above formula and we can write the determinant in denominator as,
\[\Rightarrow \cos \theta =\dfrac{\left( 3\overset{\wedge }{\mathop{i}}\,+\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right)\cdot \left( 2\overset{\wedge }{\mathop{i}}\,-2\overset{\wedge }{\mathop{j}}\,+4\overset{\wedge }{\mathop{k}}\, \right)}{\sqrt{{{3}^{2}}+{{1}^{2}}+{{2}^{2}}}\times \sqrt{{{2}^{2}}+{{\left( -2 \right)}^{2}}+{{4}^{2}}}}\]
\[\Rightarrow \cos \theta =\dfrac{\left( 3\overset{\wedge }{\mathop{i}}\, \right)\left( 2\overset{\wedge }{\mathop{i}}\, \right)\times \left( \overset{\wedge }{\mathop{j}}\, \right)\left( -2\overset{\wedge }{\mathop{j}}\, \right)\times \left( 2\overset{\wedge }{\mathop{k}}\, \right)\left( 4\overset{\wedge }{\mathop{k}}\, \right)}{\sqrt{{{3}^{2}}+{{1}^{2}}+{{2}^{2}}}\times \sqrt{{{2}^{2}}+{{\left( -2 \right)}^{2}}+{{4}^{2}}}}\]
 \[\because \overset{\wedge }{\mathop{i}}\,. \overset{\wedge }{\mathop{i}}\,=\overset{\wedge }{\mathop{j}}\,. \overset{\wedge }{\mathop{j}}\,=\overset{\wedge }{\mathop{k}}\,. \overset{\wedge }{\mathop{k}}\,=1,\overset{\wedge }{\mathop{i}}\,. \overset{\wedge }{\mathop{j}}\,=\overset{\wedge }{\mathop{j}}\,. \overset{\wedge }{\mathop{k}}\,=\overset{\wedge }{\mathop{k}}\,. \overset{\wedge }{\mathop{i}}\,=0\]
We can now simplify the above step, we get
\[\Rightarrow \cos \theta =\dfrac{6-2+8}{\sqrt{9+1+4}\sqrt{4+4+16}}\]
We can now simplify the above step further, we get
\[\Rightarrow \cos \theta =\dfrac{12}{\sqrt{14}\sqrt{24}}=\dfrac{3}{\sqrt{21}}\]
We know that \[\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }\], we can now substitute the above cosine value in this formula, we get
\[\Rightarrow \sin \theta =\sqrt{1-{{\left( \dfrac{3}{\sqrt{21}} \right)}^{2}}}\]
We can now simplify the above step, we get

 \[\Rightarrow \sin \theta =\sqrt{1-\dfrac{9}{21}}=\sqrt{\dfrac{12}{21}}=\sqrt{\dfrac{4}{7}}=\dfrac{2}{\sqrt{7}}\]

So, the correct answer is “Option C”.

Note: We should always remember that the formula to find the angle between two vectors is \[\cos \theta =\dfrac{\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,}{\left| \overset{\to }{\mathop{a}}\, \right|\left| \overset{\to }{\mathop{b}}\, \right|}\], we should also remember that general vector formula such as \[ \overset{\wedge }{\mathop{i}}\,. \overset{\wedge }{\mathop{i}}\,=\overset{\wedge }{\mathop{j}}\,. \overset{\wedge }{\mathop{j}}\,=\overset{\wedge }{\mathop{k}}\,. \overset{\wedge }{\mathop{k}}\,=1,\overset{\wedge }{\mathop{i}}\,. \overset{\wedge }{\mathop{j}}\,=\overset{\wedge }{\mathop{j}}\,. \overset{\wedge }{\mathop{k}}\,=\overset{\wedge }{\mathop{k}}\,. \overset{\wedge }{\mathop{i}}\,=0\] We should also remember some trigonometric conversions such as \[\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }\] to find the required answer.