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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Show that the points A, B, C with position vectors $2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}, \hat{\text{i}} - 3\hat{\text{j}} - 5\hat{\text{k}} \text{ and } 3\hat{\text{i}} - 4\hat{\text{j}} - 4\hat{\text{k}}$ respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle.
✓

Answer

$\vec{\text{AB}} = - \hat{\text{i}} - 2\hat{\text{j}} - 6\hat{\text{k}}, \vec{\text{BC}} = 2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}, \vec{\text{CA}} = -\hat{\text{i}} + 3\hat{\text{j}} + 5\hat{\text{k}}$
Since $\vec{\text{AB}}, \vec{\text{BC}}, \vec{\text{CA}},$ are not parallel vectors, and $\vec{\text{AB}} + \vec{\text{BC}} + \vec{\text{CA}} = \vec{0} \therefore \text{A, B, C}$ form a triangle
$\text{Also} \vec{\text{ BC}}. \vec{\text{CA}} = 0 \text{ }\text{ }\text{ }\text{ }\text{ } \therefore\text{A, B, C}$ form a right triangle
$\text{Area of} \Delta = \frac{1}{2} | \vec{\text{AB}} \times \vec{\text{BC}}| = \frac{1}{2} \sqrt{210}$

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