Question:

Let \( \mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \) and \( \mathbf{b} = \mathbf{i} - 2\mathbf{j} - 3\mathbf{k} \) be two vectors. If \( A_1 \) is the area of the quadrilateral having \( \mathbf{a}, \mathbf{b} \) as its diagonals and \( A_2 \) is the area of the parallelogram having \( \mathbf{a}, \mathbf{b} \) as its adjacent sides, then \( A_1 : A_2 = \)

Show Hint

Recall the formulas for the area of a quadrilateral with diagonals \( \mathbf{a} \) and \( \mathbf{b} \) (\( A_1 = \frac{1}{2} |\mathbf{a} \times \mathbf{b}| \)) and the area of a parallelogram with adjacent sides \( \mathbf{a} \) and \( \mathbf{b} \) (\( A_2 = |\mathbf{a} \times \mathbf{b}| \)). Calculate the cross product \( \mathbf{a} \times \mathbf{b} \) and its magnitude to find \( A_1 \) and \( A_2 \), then find their ratio.
Updated On: May 12, 2025
  • \( 1 : 2 \)
  • \( 2 : 7 \)
  • \( 2 : 5 \)
  • \( 2 : 7 \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

The area of a quadrilateral having diagonals \( \mathbf{a} \) and \( \mathbf{b} \) is given by \( A_1 = \frac{1}{2} |\mathbf{a} \times \mathbf{b}| \).
The area of a parallelogram having adjacent sides \( \mathbf{a} \) and \( \mathbf{b} \) is given by \( A_2 = |\mathbf{a} \times \mathbf{b}| \).
First, calculate the cross product \( \mathbf{a} \times \mathbf{b} \): $$ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
1 & 2 & 3
1 & -2 & -3 \end{vmatrix} = \mathbf{i}(-6 - (-6)) - \mathbf{j}(-3 - 3) + \mathbf{k}(-2 - 2) $$ $$ \mathbf{a} \times \mathbf{b} = \mathbf{i}(0) - \mathbf{j}(-6) + \mathbf{k}(-4) = 0\mathbf{i} + 6\mathbf{j} - 4\mathbf{k} = 6\mathbf{j} - 4\mathbf{k} $$ The magnitude of \( \mathbf{a} \times \mathbf{b} \) is: $$ |\mathbf{a} \times \mathbf{b}| = \sqrt{0^2 + 6^2 + (-4)^2} = \sqrt{0 + 36 + 16} = \sqrt{52} $$ Now, calculate the areas \( A_1 \) and \( A_2 \): $$ A_1 = \frac{1}{2} |\mathbf{a} \times \mathbf{b}| = \frac{1}{2} \sqrt{52} $$ $$ A_2 = |\mathbf{a} \times \mathbf{b}| = \sqrt{52} $$ The ratio \( A_1 : A_2 \) is: $$ A_1 : A_2 = \frac{1}{2} \sqrt{52} : \sqrt{52} = \frac{1}{2} : 1 = 1 : 2 $$
Was this answer helpful?
0
0

Top Questions on Geometry and Vectors

View More Questions