Question

If \( \vec{PQ} \times \vec{PR} = 4 \hat{i} + 8 \hat{j} - 8 \hat{k} \), then the area \( (\triangle PQR) \) is

• A. \( 2 \) sq units
• B. \( 4 \) sq units
• C. \( 6 \) sq units
• D. \( 12 \) sq units

Hint:

The magnitude of the cross product of two vectors \( \vec{a} \) and \( \vec{b} \), \( |\vec{a} \times \vec{b}| \), represents the area of the parallelogram formed by these two vectors as adjacent sides. The area of the triangle formed by these two vectors as two of its sides is half the area of the parallelogram, i.e., \( \frac{1}{2} |\vec{a} \times \vec{b}| \).

Answer 1

The Correct Option is C

The area of a triangle formed by three points P, Q, and R can be found using the magnitude of the cross product of the vectors representing two of its sides originating from a common vertex. For example, the area of \( \triangle PQR \) is given by half the magnitude of the cross product of \( \vec{PQ} \) and \( \vec{PR} \): $$ \text{Area}(\triangle PQR) = \frac{1}{2} |\vec{PQ} \times \vec{PR}| $$ We are given that \( \vec{PQ} \times \vec{PR} = 4 \hat{i} + 8 \hat{j} - 8 \hat{k} \). We need to find the magnitude of this vector: $$ |\vec{PQ} \times \vec{PR}| = \sqrt{(4)^2 + (8)^2 + (-8)^2} $$ $$ = \sqrt{16 + 64 + 64} $$ $$ = \sqrt{144} $$ $$ = 12 $$ Now, we can find the area of \( \triangle PQR \): $$ \text{Area}(\triangle PQR) = \frac{1}{2} (12) = 6 $$ The area of \( \triangle PQR \) is 6 square units.