If \[\overrightarrow a ,\overrightarrow b ,\overrightarrow c \] form a left-handed orthogonal system and\[\overrightarrow a .\overrightarrow a = 4,\overrightarrow b .\overrightarrow b = 9,\overrightarrow {c.} \overrightarrow c = 16\], then find the value of\[\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right]\]
A. \[24\]
B. \[ - 24\]
C. \[12\]
D. \[ - 12\]

Hint: Initially, we will find the magnitude of each vector. Then using some formulas which are mentioned below, we will find our required answer.

Used formula: \[\overrightarrow a .\overrightarrow a = {\left| {\overrightarrow a } \right|^2}\]
\[\overrightarrow b \times \overrightarrow c = \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin \theta \]
\[\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right] = \overrightarrow a .(\overrightarrow b \times \overrightarrow c )\]

Complete answer:.It is given that, \[\overrightarrow a ,\overrightarrow b ,\overrightarrow c \] form a left-handed orthogonal system.
Also provided that, \[\overrightarrow a .\overrightarrow a = 4,\overrightarrow b .\overrightarrow b = 9,\overrightarrow {c.} \overrightarrow c = 16\]
We know that,
\[\overrightarrow a .\overrightarrow a = {\left| {\overrightarrow a } \right|^2}\],
So, according to the problem using the values given we get, \[\overrightarrow a .\overrightarrow a = {\left| {\overrightarrow a } \right|^2} = 4\]
So, we have, ${\left| {\overrightarrow a } \right|} = 2$
Similarly, we will find ${\left| {\overrightarrow b } \right|} = 3$, ${\left| {\overrightarrow c } \right|} = 4$
Now we know that $\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right] $ is given by the formula,
\[\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right] = \overrightarrow a .(\overrightarrow b \times \overrightarrow c )\]
Here, the angle between the vectors \[\overrightarrow a ,\overrightarrow b \times \overrightarrow c \] is \[{180^ \circ }\]. Since, \[\overrightarrow b \times \overrightarrow c \] is exactly opposite to the vector \[\overrightarrow a ,\]
On simplifying using the angle mentioned above we get,
\[\overrightarrow a .(\overrightarrow b \times \overrightarrow c ) = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b \times \overrightarrow c } \right|\cos {180^ \circ }\]
We know the trigonometric values of \[\cos \theta \]then we get, \[\cos {180^ \circ } = - 1\]
Substitute the value into the above expression we get,
\[\overrightarrow a .(\overrightarrow b \times \overrightarrow c ) = - \left| {\overrightarrow a } \right|.\left| {\overrightarrow b \times \overrightarrow c } \right|\] …. (1)
Now we will consider the value of\[\left| {\overrightarrow b \times \overrightarrow c } \right|\].
\[\left| {\overrightarrow b \times \overrightarrow c } \right| = \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin {90^ \circ }\]
Since, the angle between the vectors \[\overrightarrow b \] and \[\overrightarrow c \] is \[{90^ \circ }\]the value of \[\theta \] is replaced by\[{90^ \circ }\] .
Substitute this value at the expression (1) we get,
\[\overrightarrow a .(\overrightarrow b \times \overrightarrow c ) = - \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin {90^ \circ }\]
Applying the trigonometric value of sine function we get,
\[\overrightarrow a .(\overrightarrow b \times \overrightarrow c ) = - \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\]
Now let us put the values of \[\left| {\overrightarrow a } \right| = 2\],\[\left| {\overrightarrow b } \right| = 3\], \[\left| {\overrightarrow c } \right| = 4\] in the above equation, we get,
\[\overrightarrow a .(\overrightarrow b \times \overrightarrow c ) = - 2 \times 3 \times 4 = - 24\]
Hence,
\[\overrightarrow a .(\overrightarrow b \times \overrightarrow c ) = - 24\]
That is the value of \[\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right]\]is\[ - 24\]

Therefore, the correct option is (B)\[ - 24\].

Note: Let us consider the two vectors \[\overrightarrow b \] and \[\overrightarrow c \].
Then, \[\overrightarrow b \times \overrightarrow c = \left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\sin \theta \]
Since, the given system is left-handed, orthogonal the angle between vectors \[\overrightarrow b \] and \[\overrightarrow c \]\[{90^ \circ }\].
Again,
\[\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right] = \overrightarrow a .(\overrightarrow b \times \overrightarrow c )\]
\[\left[ {\overrightarrow a ,\overrightarrow b ,\overrightarrow c } \right]\]is defined as the box product of the given vectors, the box product is nothing but the combination of dot product with cross product.