The three vectors vec A = 3hat i - 2hat j + h | vedclass

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(C) First,check if the vectors form a triangle by checking if their sum is zero. $\vec A + \vec B + \vec C = (3+1+2)\hat i + (-2-3+1)\hat j + (1+5-4)\

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$A$ right-angled triangle

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(C) First,check if the vectors form a triangle by checking if their sum is zero. $\vec A + \vec B + \vec C = (3+1+2)\hat i + (-2-3+1)\hat j + (1+5-4)\hat k = 6\hat i - 4\hat j + 2\hat k 
eq 0$. Since the sum is not zero,these vectors do not form a closed triangle in the standard sense of vector addition. However,if we treat the magnitudes of these vectors as the sides of a triangle,we calculate the magnitudes:$|\vec A| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$$|\vec B| = \sqrt{1^2 + (-3)^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$$|\vec C| = \sqrt{2^2 + 1^2 + (-4)^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$Observe that $|\vec A|^2 + |\vec C|^2 = 14 + 21 = 35 = |\vec B|^2$. Since the square of one side equals the sum of the squares of the other two sides,these lengths satisfy the Pythagorean theorem,forming a right-angled triangle.

The three vectors $\vec A = 3\hat i - 2\hat j + \hat k$,$\vec B = \hat i - 3\hat j + 5\hat k$,and $\vec C = 2\hat i + \hat j - 4\hat k$ form:

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