Question

In a triangle ABC, if $\vec{BC}=\hat{i}-2\hat{j}+2\hat{k}$ and $\vec{CA}=6\hat{i}+3\hat{j}-2\hat{k}$, then the perimeter of the triangle is

• A. $5(2+\sqrt{3})$
• B. $5(2+\sqrt{2})$
• C. $\sqrt{10}(3+\sqrt{10})$
• D. $10(2+\sqrt{5})$

Hint:

Remember the triangle law of vector addition: for a triangle with vertices A, B, C, the vectors forming a closed loop sum to zero, i.e., $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$. This allows you to find one side vector if the other two are known.

Answer 1

The Correct Option is B

The perimeter of a triangle is the sum of the lengths of its three sides. The lengths of the sides are the magnitudes of the vectors representing them.
Perimeter = $|\vec{AB}| + |\vec{BC}| + |\vec{CA}|$.
We are given the vectors for two sides:
$\vec{BC}=\hat{i}-2\hat{j}+2\hat{k}$
$\vec{CA}=6\hat{i}+3\hat{j}-2\hat{k}$
For any triangle, the sum of the vectors representing the sides in order is the zero vector: $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$.
From this, we can find the third side vector, $\vec{AB}$:
$\vec{AB} = -(\vec{BC} + \vec{CA})$.
$\vec{BC} + \vec{CA} = (\hat{i}-2\hat{j}+2\hat{k}) + (6\hat{i}+3\hat{j}-2\hat{k}) = (1+6)\hat{i} + (-2+3)\hat{j} + (2-2)\hat{k} = 7\hat{i}+\hat{j}$.
$\vec{AB} = -(7\hat{i}+\hat{j}) = -7\hat{i}-\hat{j}$.
Now, we calculate the magnitude of each side vector:
$|\vec{BC}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.
$|\vec{CA}| = \sqrt{6^2 + 3^2 + (-2)^2} = \sqrt{36+9+4} = \sqrt{49} = 7$.
$|\vec{AB}| = \sqrt{(-7)^2 + (-1)^2 + 0^2} = \sqrt{49+1} = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$.
The perimeter is the sum of these magnitudes:
Perimeter = $3 + 7 + 5\sqrt{2} = 10 + 5\sqrt{2}$.
Factoring out 5, we get: Perimeter = $5(2+\sqrt{2})$.