Question

The area of the parallelogram whose diagonals are represented by the vectors $\vec{a} = 3\hat{i} - \hat{j} - 2\hat{k}$ and $\vec{b} = -\hat{i} + 3\hat{j} - 3\hat{k}$ is

• A. $\sqrt{266}$ sq. units
• B. $\frac{1}{2}\sqrt{266}$ sq. units
• C. $2\sqrt{266}$ sq. units
• D. $\sqrt{122}$ sq. units

Hint:

Be careful not to mix up the formulas! If the vectors represent adjacent sides, the area is simply $|\vec{a} \times \vec{b}|$. If the vectors represent the diagonals, the area must include a factor of half: $\frac{1}{2}|\vec{a} \times \vec{b}|$. Remembering this subtle definition prevents easy point deductions.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given two spatial vectors, $\vec{a}$ and $\vec{b}$, representing the diagonals of a parallelogram. We need to calculate the area of this parallelogram.

Step 2: Key Formula or Approach:
If the diagonals of a parallelogram are given by vectors $\vec{a}$ and $\vec{b}$, then the area is given by $$\text{Area}=\frac{1}{2}\left|\vec{a}\times\vec{b}\right|.$$ The cross product is computed using $$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_x&a_y&a_z\\b_x&b_y&b_z\end{vmatrix}.$$

Step 3: Detailed Explanation:
For $\vec{a}=3\hat{i}-\hat{j}-2\hat{k}$ and $\vec{b}=-\hat{i}+3\hat{j}-3\hat{k}$, $$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-1&-2\\-1&3&-3\end{vmatrix}.$$ Expanding, $$\vec{a}\times\vec{b}=\hat{i}[(-1)(-3)-(-2)(3)]-\hat{j}[(3)(-3)-(-2)(-1)]+\hat{k}[(3)(3)-(-1)(-1)]$$ $$=9\hat{i}+11\hat{j}+8\hat{k}.$$ Therefore, $$|\vec{a}\times\vec{b}|=\sqrt{9^2+11^2+8^2}=\sqrt{81+121+64}=\sqrt{266}.$$ Hence, $$\text{Area}=\frac{1}{2}\sqrt{266}\ \text{sq. units}.$$

Step 4: Final Answer:
The area of the parallelogram is $\frac{1}{2}\sqrt{266}$ sq. units, matching option (B).