Scalar Triple Product Calculation
The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ is calculated using the determinant of the matrix formed by the components of the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.
Given Vectors
- $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$
- $\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}$
- $\vec{c} = 3\hat{i} + 4\hat{j} - \hat{k}$
Determinant Method
The scalar triple product equals the determinant:
$ \vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -3 \\ 3 & 4 & -1 \end{vmatrix} $Evaluating the Determinant
Expand the determinant along the first row:
$ = 2 \begin{vmatrix} 2 & -3 \\ 4 & -1 \end{vmatrix} - (-3) \begin{vmatrix} 1 & -3 \\ 3 & -1 \end{vmatrix} + 4 \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} $Calculate the 2x2 determinants:
- $ \begin{vmatrix} 2 & -3 \\ 4 & -1 \end{vmatrix} = (2 \times -1) - (-3 \times 4) = -2 - (-12) = -2 + 12 = 10 $
- $ \begin{vmatrix} 1 & -3 \\ 3 & -1 \end{vmatrix} = (1 \times -1) - (-3 \times 3) = -1 - (-9) = -1 + 9 = 8 $
- $ \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = (1 \times 4) - (2 \times 3) = 4 - 6 = -2 $
Substitute these values back into the expansion:
$ = 2(10) + 3(8) + 4(-2) $ $ = 20 + 24 - 8 $ $ = 44 - 8 $ $ = 36 $Result
The scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$ is 36.
Important Questions from Vector Algebra
Vector A̅ = ŷ.3 + ẑ.2 and B̅ = x̂.5 + ŷ.8 extend from the origin. Find A̅.B̅ Choose the correct answer.
The value of the cross product \(\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)\) of two vectors \(\overrightarrow a - \overrightarrow b\) and \(\overrightarrow a + \overrightarrow b \) is:
If non - zero a, b, c are such that a + b + c = 0, then the value of \(\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab}\) is
Vector a = 3i + 2j – 6k, vector b = 4i – 3j + k, angle between above vectors is
Two forces F1 and F2 are used to pull a car, which met an accident. The angle between the two force is θ. Find the value of θ for the resultant force is equal to \(\sqrt{(F_1^2 + F_2^2)}\)
