Skip to main contentSkip to solution

The area (in sq.units) of a triangle formed by vertices O, A and B where OA⃗=i^+2j^+3k^\vec{OA} = \hat{i} + 2\hat{j} + 3\hat{k} and OB⃗=−3i^−2j^+k^\vec{OB} = -3\hat{i} - 2\hat{j} + \hat{k} is

Solution

✅ Correct Option: 1

The area of a triangle with adjacent sides represented by vectors OA⃗\vec{OA} and OB⃗\vec{OB} is given by:

Area=12∣OA⃗×OB⃗∣\text{Area} = \frac{1}{2} |\vec{OA} \times \vec{OB}|

  1. Find the cross product OA⃗×OB⃗\vec{OA} \times \vec{OB}:

OA⃗×OB⃗=∣i^j^k^123−3−21∣\vec{OA} \times \vec{OB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1 \end{vmatrix}

=i^(2−(−6))−j^(1−(−9))+k^(−2−(−6))= \hat{i}(2 - (-6)) - \hat{j}(1 - (-9)) + \hat{k}(-2 - (-6))

=8i^−10j^+4k^= 8\hat{i} - 10\hat{j} + 4\hat{k}

  1. Calculate the magnitude:

∣OA⃗×OB⃗∣=82+(−10)2+42=64+100+16=180|\vec{OA} \times \vec{OB}| = \sqrt{8^2 + (-10)^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180}

  1. Calculate the Area:

Area=12180=1804=45 sq. units\text{Area} = \frac{1}{2} \sqrt{180} = \sqrt{\frac{180}{4}} = \sqrt{45}\text{ sq. units}

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question