The area of the triangle formed by the vertices whose position vectors are \(3\widehat{i}+\widehat{j}\) , \(5\widehat{i}+2\widehat{j}+\widehat{k}\) , \(\widehat{i}-2\widehat{j}+3\widehat{k}\) is
Step-by-step Solution:
Suppose the three vertices are: \[ A = 3\hat{i} + \hat{j}, \quad B = 5\hat{i} + 2\hat{j} + \hat{k}, \quad C = \hat{i} - 2\hat{j} + 3\hat{k} \] The sides are: \[ AB = 2\hat{i} + \hat{j} + \hat{k}, \quad AC = -2\hat{i} - 3\hat{j} + 3\hat{k} \] The area of the triangle is: \[ \frac{1}{2} \left| AB \times AC \right| = \frac{1}{2} \left| 6\hat{i} - 8\hat{j} - 4\hat{k} \right| = \sqrt{29} \]