If vec{A}=3 hat{i}-2 hat{j}+hat{k},vec{B}=hat | vedclass

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(B) First,calculate the magnitudes squared of the vectors:$A^2 = |\vec{A}|^2 = 3^2 + (-2)^2 + 1^2 = 9 + 4 + 1 = 14$$B^2 = |\vec{B}|^2 = 1^2 + (-3)^2 +

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$\vec{A}=\vec{B}+\vec{C}$ and $B^2=A^2+C^2$

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(B) First,calculate the magnitudes squared of the vectors:$A^2 = |\vec{A}|^2 = 3^2 + (-2)^2 + 1^2 = 9 + 4 + 1 = 14$$B^2 = |\vec{B}|^2 = 1^2 + (-3)^2 + 5^2 = 1 + 9 + 25 = 35$$C^2 = |\vec{C}|^2 = 2^2 + 1^2 + (-4)^2 = 4 + 1 + 16 = 21$We observe that $B^2 = A^2 + C^2$ $(35 = 14 + 21)$,which satisfies the Pythagorean theorem for a right-angled triangle.Next,check the vector sum relation:$\vec{A} + \vec{C} = (3+2)\hat{i} + (-2+1)\hat{j} + (1-4)\hat{k} = 5\hat{i} - \hat{j} - 3\hat{k} 
eq \vec{B}$$\vec{B} = \vec{A} + \vec{C}$ is not satisfied. Let us check $\vec{A} = \vec{B} + \vec{C}$:$\vec{B} + \vec{C} = (1+2)\hat{i} + (-3+1)\hat{j} + (5-4)\hat{k} = 3\hat{i} - 2\hat{j} + \hat{k} = \vec{A}$Thus,$\vec{A} = \vec{B} + \vec{C}$ and $B^2 = A^2 + C^2$ is the correct condition.

If $\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k}$,$\vec{B}=\hat{i}-3 \hat{j}+5 \hat{k}$ and $\vec{C}=2 \hat{i}+\hat{j}-4 \hat{k}$ form a right-angled triangle,then which of the following is satisfied?

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