Consider the vectors vec{a}=2 hat{i}+3 hat{j} | vedclass

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(A) Three vectors form a triangle if their sum is zero, i.e.,$\vec{a} + \vec{b} + \vec{c} = \vec{0}$.Calculating the sum: $\vec{a} + \vec{b} + \vec{c}

Vedclass · Accepted Answer

$(A)$ is true, $(R)$ is true and $(R)$ is the correct explanation for $(A)$

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(A) Three vectors form a triangle if their sum is zero, i.e.,$\vec{a} + \vec{b} + \vec{c} = \vec{0}$.Calculating the sum: $\vec{a} + \vec{b} + \vec{c} = (2+6+3)\hat{i} + (3-2-6)\hat{j} + (-6+3-2)\hat{k} = 11\hat{i} - 5\hat{j} - 5\hat{k}$.Since $\vec{a} + \vec{b} + \vec{c} 
eq \vec{0}$, the vectors do not form a triangle. Thus, $(A)$ is true.To check if they are coplanar, we calculate the scalar triple product $[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c})$.$\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 6 & -2 & 3 \ 3 & -6 & -2 \end{vmatrix} = \hat{i}(4+18) - \hat{j}(-12-9) + \hat{k}(-36+6) = 22\hat{i} + 21\hat{j} - 30\hat{k}$.$\vec{a} \cdot (\vec{b} 	imes \vec{c}) = (2)(22) + (3)(21) + (-6)(-30) = 44 + 63 + 180 = 287 
eq 0$.Since the scalar triple product is non-zero, the vectors are non-coplanar. Thus, $(R)$ is true.Since the vectors are non-coplanar, they cannot form a triangle in a plane. Therefore, $(R)$ is the correct explanation for $(A)$.

Consider the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-6 \hat{k}$, $\vec{b}=6 \hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{c}=3 \hat{i}-6 \hat{j}-2 \hat{k}$.
Assertion $(A):$ The three vectors do not form a triangle.
Reason $(R):$ The three vectors are non-coplanar.
The correct option among the following is:

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Consider the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-6 \hat{k}$, $\vec{b}=6 \hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{c}=3 \hat{i}-6 \hat{j}-2 \hat{k}$.Assertion $(A):$ The three vectors do not form a triangle.Reason $(R):$ The three vectors are non-coplanar.The correct option among the following is: