Question

Two adjacent sides of a triangle are represented by the vectors $2\vec{i}+\vec{j}-2\vec{k}$ and $2\sqrt{3}\vec{i}-2\sqrt{3}\vec{j}+\sqrt{3}\vec{k}$. Then the least angle of the triangle and perimeter of the triangle are respectively

• A. $\frac{\pi}{3}; 3(3+\sqrt{3})$
• B. $\frac{\pi}{12}; 6+3\sqrt{2}$
• C. $\frac{\pi}{2}; 12$
• D. $\frac{\pi}{6}; 9+3\sqrt{3}$

Hint:

When given two vectors representing adjacent sides of a triangle, always calculate their dot product first. If it is zero, the triangle is right-angled, which simplifies finding the other sides and angles significantly.

Answer 1

The Correct Option is D

Step 1: Compute magnitudes of vectors.
\[ \vec{a} = 2\vec{i}+\vec{j}-2\vec{k}, \vec{b} = 2\sqrt{3}\vec{i}-2\sqrt{3}\vec{j}+\sqrt{3}\vec{k} \] \[ |\vec{a}| = 3, |\vec{b}| = 3\sqrt{3} \]
Step 2: Compute dot product and check angle.
\[ \vec{a} \cdot \vec{b} = 0 \implies \text{vectors perpendicular} \implies \text{angle } = \frac{\pi}{2} \]
Step 3: Compute third side.
\[ |\vec{c}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2} = \sqrt{9 + 27} = 6 \]
Step 4: Determine least angle.
\[ \sin \alpha = \frac{3}{6} = \frac{1}{2} \implies \alpha = \frac{\pi}{6} \]
Step 5: Compute perimeter.
\[ \text{Perimeter} = 3 + 3\sqrt{3} + 6 = 9 + 3\sqrt{3} \]