If vec{a}, vec{b}, vec{c} are non-coplanar ve | vedclass

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Question

(A) Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$,$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$,and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3

Vedclass · Accepted Answer

$[\vec{a} \; \vec{b} \; \vec{c}]^2$

Vedclass · Answer

(A) Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$,$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$,and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$.The scalar triple product squared is given by the Gram determinant:$[\vec{a} \; \vec{b} \; \vec{c}]^2 = \left| {\begin{array}{*{20}{c}} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{array}} ight| 	imes \left| {\begin{array}{*{20}{c}} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{array}} ight|$By the property of matrix multiplication of determinants:$[\vec{a} \; \vec{b} \; \vec{c}]^2 = \left| {\begin{array}{*{20}{c}} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{array}} ight|$Thus,the given determinant is equal to $[\vec{a} \; \vec{b} \; \vec{c}]^2$.

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