The perimeter of a triangle with sides repres | vedclass

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Question

(A) The perimeter of a triangle is the sum of the magnitudes of its sides.Let the sides be $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$,$\vec{b} = 4\hat

Vedclass · Accepted Answer

$\sqrt{450}$

Vedclass · Answer

(A) The perimeter of a triangle is the sum of the magnitudes of its sides.Let the sides be $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$,$\vec{b} = 4\hat{i} - 3\hat{j} - 5\hat{k}$,and $\vec{c} = 7\hat{i} + \hat{j}$.Magnitude of side $\vec{a}$ is $|\vec{a}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$.Magnitude of side $\vec{b}$ is $|\vec{b}| = \sqrt{4^2 + (-3)^2 + (-5)^2} = \sqrt{16 + 9 + 25} = \sqrt{50} = 5\sqrt{2}$.Magnitude of side $\vec{c}$ is $|\vec{c}| = \sqrt{7^2 + 1^2 + 0^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}$.Perimeter = $|\vec{a}| + |\vec{b}| + |\vec{c}| = 5\sqrt{2} + 5\sqrt{2} + 5\sqrt{2} = 15\sqrt{2}$.Since $15\sqrt{2} = \sqrt{225 	imes 2} = \sqrt{450}$,the correct option is $A$.

The perimeter of a triangle with sides represented by vectors $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$,$\vec{b} = 4\hat{i} - 3\hat{j} - 5\hat{k}$,and $\vec{c} = 7\hat{i} + \hat{j}$ is:

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