Question

If \[\overrightarrow A = 3\widehat i + \widehat j + 2\hat k\] and $\vec B = 2\hat i - 2\hat j + 4\hat k$ then value of $\left| {\vec A \times \vec B} \right|$ will be
(A) $8\sqrt 2 $
(B) $8\sqrt 3 $
(C) $8\sqrt 5 $
(D) $5\sqrt 8 $

Answer 1

Hint: Cross product of two vectors is defined as a binary operation on the two vectors in three-dimensional space. The resultant vector formed by the cross product is perpendicular to both vectors. Cross product is also commonly known as the vector product. The Vector product of two vectors, $\vec a$ and $\vec b$, is denoted by $\vec a \times \vec b$. The resultant vector obtained from $\vec a \times \vec b$ is perpendicular to both $\vec a$ and $\vec b$.

FORMULA USED: We can determine the cross-product of two vectors using the determinant of the matrix represented by the vectors. Let $\vec P$ and $\vec Q$ be two vectors such that $\vec P = a\hat i + b\hat j + c\hat k$ and $\vec Q = x\hat i + y\hat j + z\hat k$ then the cross product $\vec P \times \vec Q$ is given by
$\vec P \times \vec Q = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  a&b&c \\
  x&y&z
\end{array}} \right|$
$ \Rightarrow \vec P \times \vec Q = \left( {bz - cy} \right)\hat i - \left( {az - cx} \right)\hat j + \left( {ay - bx} \right)\hat k$
$ \Rightarrow \vec P \times \vec Q = \left( {bz - cy} \right)\hat i + \left( {cx - az} \right)\hat j + \left( {ay - bx} \right)\hat k$

Complete step-by-step answer:
We are given two vectors and we need to find the cross-product of the two vectors.
We just have to replace the values of vectors $\vec A$ and $\vec B$ in the above formula to find the cross-product of those vectors using the determinant formula.
Replacing the values by vectors \[\overrightarrow A = 3\widehat i + \widehat j + 2\hat k\] and $\vec B = 2\hat i - 2\hat j + 4\hat k$ in the formula we get:
$\vec A \times \vec B = \left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  3&1&2 \\
  2&{ - 2}&4
\end{array}} \right|$
$ \Rightarrow \vec A \times \vec B = \left( {1 \times 4 - 2 \times \left( { - 2} \right)} \right)\hat i + \left( {2 \times 2 - 3 \times 4} \right)\hat j + \left( {3 \times \left( { - 2} \right) - 1 \times 2} \right)\hat k$
$ \Rightarrow \vec A \times \vec B = \left( {4 + 4} \right)\hat i + \left( {4 - 12} \right)\hat j + \left( { - 6 - 2} \right)\hat k$
$ \Rightarrow \vec A \times \vec B = 8\hat i - 8\hat j - 8\hat k$
Therefore, the required cross product of $\vec A$ and $\vec B$ is $8\hat i - 8\hat j - 8\hat k$.
Now, we have to find the magnitude of the cross-product of two vectors.
Therefore, the required magnitude of the two vectors $\vec A$ and $\vec B$ is given by $\left| {\vec A \times \vec B} \right|$.
$\therefore \left| {\vec A \times \vec B} \right| = \sqrt {{{\left( 8 \right)}^2} + {{\left( { - 8} \right)}^2} + {{\left( { - 8} \right)}^2}} $
$ = 8 \times \sqrt {1 + 1 + 1} $
$ = 8\sqrt 3 $
Hence, the required value of $\left| {\vec A \times \vec B} \right|$ is $8\sqrt 3 $ units.

So, the correct answer is “Option B”.

Note: The magnitude of the cross product of two vectors $\vec P = a\hat i + b\hat j + c\hat k$ and $\vec Q = x\hat i + y\hat j + z\hat k$ can also be calculated by $\left| {\vec P \times \vec Q} \right| = \left| {\vec P} \right|\left| {\vec Q} \right|\left| {\sin \theta } \right|$ where $\theta $ is the angle between two vectors $\vec P$ and $\vec Q$ and $\left| {\vec P} \right| = \sqrt {{a^2} + {b^2} + {c^2}} $ and $\left| {\vec Q} \right| = \sqrt {{x^2} + {y^2} + {z^2}} $. But this formula does not apply here because we do not have any information regarding the angle between two vectors. While calculating the determinant of the matrix one must be very careful with the signs of the numbers.