$\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{B} = 5\hat{i} - 4\hat{k}$
The magnitude of area of the parallelogram is ________ (Rounded off to two decimal places)
Parallelogram Area Calculation
The area of a parallelogram formed by two adjacent vectors $\vec{A}$ and $\vec{B}$ is equal to the magnitude of their cross product, denoted as $|\vec{A} \times \vec{B}|$.
Vector Cross Product Calculation
Given vectors are:
- $\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$
- $\vec{B} = 5\hat{i} - 4\hat{k}$ (which can be written as $5\hat{i} + 0\hat{j} - 4\hat{k}$)
The cross product $\vec{A} \times \vec{B}$ is calculated using a determinant:
$ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 5 & 0 & -4 \end{vmatrix} $Expanding the determinant:
$ = \hat{i}((3)(-4) - (-1)(0)) - \hat{j}((2)(-4) - (-1)(5)) + \hat{k}((2)(0) - (3)(5)) $ $ = \hat{i}(-12 - 0) - \hat{j}(-8 - (-5)) + \hat{k}(0 - 15) $ $ = -12\hat{i} - \hat{j}(-3) - 15\hat{k} $ $ = -12\hat{i} + 3\hat{j} - 15\hat{k} $Magnitude of Cross Product
The resulting vector is $-12\hat{i} + 3\hat{j} - 15\hat{k}$. The magnitude of this vector is calculated as:
$ |\vec{A} \times \vec{B}| = \sqrt{(-12)^2 + (3)^2 + (-15)^2} $ $ = \sqrt{144 + 9 + 225} $ $ = \sqrt{378} $Final Area Calculation
Calculating the square root:
$ \sqrt{378} \approx 19.44155 $Rounding off to two decimal places gives $19.44$.
The magnitude of the area of the parallelogram is approximately 19.44.
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 \(\sqrt{(F_1^2 + F_2^2)}\)
