Vector Cross Product of \(4\hat i – 2\hat j + 6\hat k\) and \(7\hat i + \hat j – 12\hat k\): Find \(\alpha + \beta + \gamma\)

Question Statement

Let \(\vec a = 4\hat i – 2\hat j + 6\hat k\) and \(\vec b = 7\hat i + \hat j – 12\hat k\).
If \(\vec a \times \vec b = \alpha \hat i + \beta \hat j + \gamma \hat k\), find \(\alpha + \beta + \gamma\).

Answer: The value of \(\alpha + \beta + \gamma\) is \(126\).

Concept of Cross Product

The vector cross product of two vectors \(\vec a\) and \(\vec b\) in three dimensions is a vector
\(\vec a \times \vec b\) perpendicular to both, with magnitude \(\|\vec a\|\,\|\vec b\|\sin\theta\),
where \(\theta\) is the angle between them.

Algebraically, for \(\vec a = (a_1,a_2,a_3)\) and \(\vec b = (b_1,b_2,b_3)\),
the cross product is computed using the determinant

\[
\vec a \times \vec b =
\begin{vmatrix}
\hat i & \hat j & \hat k \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{vmatrix}.
\]

Step 1: Components of Vectors

Write the vectors in component form:

• \(\vec a = (4,\,-2,\,6)\)
• \(\vec b = (7,\,1,\,-12)\)

The cross product will again be a three‑dimensional vector with components along
\(\hat i\), \(\hat j\) and \(\hat k\).

Step 2: Determinant Setup

Set up the determinant for \(\vec a \times \vec b\):

\[
\vec a \times \vec b =
\begin{vmatrix}
\hat i & \hat j & \hat k \\
4 & -2 & 6 \\
7 & 1 & -12
\end{vmatrix}.
\]

Expand this determinant along the first row:

\[
\vec a \times \vec b =
\hat i
\begin{vmatrix}
-2 & 6 \\
1 & -12
\end{vmatrix}
–
\hat j
\begin{vmatrix}
4 & 6 \\
7 & -12
\end{vmatrix}
+
\hat k
\begin{vmatrix}
4 & -2 \\
7 & 1
\end{vmatrix}.
\]

Step 3: Evaluate 2×2 Determinants

Compute each minor determinant to obtain the components of the cross product:

• For the \(\hat i\) component:
  \[
  \begin{vmatrix}
  -2 & 6 \\
  1 & -12
  \end{vmatrix}
  = (-2)(-12) – (6)(1) = 24 – 6 = 18.
  \]
• For the \(\hat j\) component:
  \[
  \begin{vmatrix}
  4 & 6 \\
  7 & -12
  \end{vmatrix}
  = (4)(-12) – (6)(7) = -48 – 42 = -90.
  \]
  Because of the minus sign in the expansion, the \(\hat j\) component is \(-(-90) = 90\).
• For the \(\hat k\) component:
  \[
  \begin{vmatrix}
  4 & -2 \\
  7 & 1
  \end{vmatrix}
  = (4)(1) – (-2)(7) = 4 + 14 = 18.
  \]

Therefore,
\[
\vec a \times \vec b = 18\hat i + 90\hat j + 18\hat k.
\]

Identifying coefficients, \(\alpha = 18\), \(\beta = 90\), and \(\gamma = 18\).

Step 4: Find \(\alpha + \beta + \gamma\)

Finally, add the three components:

\[
\alpha + \beta + \gamma = 18 + 90 + 18 = 126.
\]

So the required value of \(\alpha + \beta + \gamma\) for the cross product of
\(4\hat i – 2\hat j + 6\hat k\) and \(7\hat i + \hat j – 12\hat k\) is \(126\).