The position vectors of the points A, B, C ar | vedclass

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(A) Let the position vectors of points $A, B, C$ be $\vec{a} = \hat{i}+2\hat{j}-\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = 2\hat{i}+3

Vedclass · Accepted Answer

$-5\hat{i}+2\hat{j}+\hat{k}$

Vedclass · Answer

(A) Let the position vectors of points $A, B, C$ be $\vec{a} = \hat{i}+2\hat{j}-\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = 2\hat{i}+3\hat{j}+2\hat{k}$.If $A$ is chosen as the origin,the new position vectors of $B$ and $C$ are $\vec{B'} = \vec{b} - \vec{a}$ and $\vec{C'} = \vec{c} - \vec{a}$.$\vec{B'} = (\hat{i}+\hat{j}+\hat{k}) - (\hat{i}+2\hat{j}-\hat{k}) = 0\hat{i} - \hat{j} + 2\hat{k}$.$\vec{C'} = (2\hat{i}+3\hat{j}+2\hat{k}) - (\hat{i}+2\hat{j}-\hat{k}) = \hat{i} + \hat{j} + 3\hat{k}$.The cross product $\vec{B'} 	imes \vec{C'}$ is given by the determinant:$\vec{B'} 	imes \vec{C'} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -1 & 2 \ 1 & 1 & 3 \end{vmatrix}$.$= \hat{i}((-1)(3) - (2)(1)) - \hat{j}((0)(3) - (2)(1)) + \hat{k}((0)(1) - (-1)(1))$.$= \hat{i}(-3 - 2) - \hat{j}(0 - 2) + \hat{k}(0 + 1)$.$= -5\hat{i} + 2\hat{j} + \hat{k}$.

The position vectors of the points $A, B, C$ are $\hat{i}+2\hat{j}-\hat{k}, \hat{i}+\hat{j}+\hat{k}$,and $2\hat{i}+3\hat{j}+2\hat{k}$ respectively. If $A$ is chosen as the origin,then the cross product of the position vectors of $B$ and $C$ is:

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