Let PR→=3i^+j^−2k^ and SQ→=i^−3j^−4k^ determine diagonals of a parallelogram PQRS and PT→=i^+2j^+3k^ be another vector. Then the volume of the parallelopiped determined by the vectors PT→,PQ→ and PS→ is

The side of PS→ and PQ→ are given by PS→=PR→−SQ→2 and PQ→=PR→+SQ→2The volume V is given by V=|[PQ →PS→ PT→]|=14|[PR→+SQ→, PR→−SQ→, PT→]| =12|[PR→,SQ→,PT→]| [Recall that (a→a→b→)=0 and [a→b→c→]=[b→c→a→], etc. ] =1231−21−3−4123=12∣3(−1)−1(7)−2(5)=12(−20)∣=10Remark : One could have also calculated the sides PS→ and PQ→ and then proceeded to evaluate the box product using determinants.

Let PR→=3i^+j^−2k^ and SQ→=i^−3j^−4k^ determine diagonals of a parallelogram PQRS and PT→=i^+2j^+3k^ be another vector. Then the volume of the parallelopiped determined by the vectors PT→,PQ→ and PS→ is

The side of PS→ and PQ→ are given by PS→=PR→−SQ→2 and PQ→=PR→+SQ→2The volume V is given by V=|[PQ →PS→ PT→]|=14|[PR→+SQ→, PR→−SQ→, PT→]| =12|[PR→,SQ→,PT→]| [Recall that (a→a→b→)=0 and [a→b→c→]=[b→c→a→], etc. ] =1231−21−3−4123=12∣3(−1)−1(7)−2(5)=12(−20)∣=10Remark : One could have also calculated the sides PS→ and PQ→ and then proceeded to evaluate the box product using determinants.

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Let $\vec{P R} = 3 \hat{i} + \hat{j} - 2 \hat{k} and \vec{S Q} = \hat{i} - 3 \hat{j} - 4 \hat{k}$ determine diagonals of a parallelogram $P Q R S$ and $\vec{P T} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ be another vector. Then the volume of the parallelopiped determined by the vectors $\vec{P T}, \vec{P Q} and \vec{P S}$ is

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Detailed Solution

The side of $\vec{P S} and \vec{P Q}$ are given by

$\vec{P S} = \frac{\vec{P R} - \vec{S Q}}{2} and \vec{P Q} = \frac{\vec{P R} + \vec{S Q}}{2}$

The volume $V$ is given by

$\begin{array}{l} V = | [\vec{P Q} \vec{P S} \vec{P T}] | = \frac{1}{4} | [\vec{P R} + \vec{S Q}, \vec{P R} - \vec{S Q}, \vec{P T}] | \\ = \frac{1}{2} | [\vec{P R}, \vec{S Q}, \vec{P T}] | \end{array}$

$[Recall that (\vec{a} \vec{a} \vec{b}) = 0 and [\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}], etc. ]$

$= \frac{1}{2} |\begin{array}{ccc} 3 & 1 & - 2 \\ 1 & - 3 & - 4 \\ 1 & 2 & 3 \end{array}| = \frac{1}{2} ∣ [3 (- 1) - 1 (7) - 2 (5) |= \frac{1}{2}| (- 20) ∣= 10$