Three vectors \(\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}}\) and \(\overrightarrow{\mathrm{r}}\) are given as

\(\overrightarrow{\mathrm{p}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \)

\(\overrightarrow{\mathrm{q}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \)

\(\overrightarrow{\mathrm{r}}=2 \hat{i}+3 \hat{j}+4 \hat{k}\)

Which of the following is/are CORRECT?

Explanation:

(a) \(\vec{p} \times(\vec{q} \times \vec{r})=(\vec{p} \cdot \vec{r}) \vec{q}-(\vec{p} \cdot \vec{q}) \vec{r}\)

(This is always true for any three given vectors)

(b) We know that \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\) is always true but \(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\) because \(\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}\)

This can be true only when \(\vec{a} \times \vec{b}=0\)

So, \(\quad \vec{r} \cdot(\vec{p} \times \vec{q})=\vec{r} .(\vec{q} \times \vec{p})\)

\(\vec{r} \cdot(\vec{p} \times \vec{q})=-\vec{r} \cdot(\vec{p} \times \vec{q})\)

This can be true if

\(\vec{r} \cdot(\vec{p} \times q) =0\)

\(\vec{p} \times \vec{q} =\hat{i}-2 \hat{j}+j\)

\((\vec{r} \cdot \vec{p} \times \vec{q})=0\) this is true in this case

(c) \(\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{r}})\) can't be equal to \((\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}) \times \overrightarrow{\mathrm{r}}\) because \(\vec{p} \times(\vec{q} \times \vec{r}) \perp \vec{p}\) and \((\vec{p} \times \vec{q}) \times \vec{r}) \perp \vec{r}\)

So, \((\vec{p} \times(\vec{q} \times \vec{r}) \neq(\vec{p} \times \vec{q})) \times \vec{r}\)

(d) \(\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{r}})+\overrightarrow{\mathrm{q}} \times(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}})+\overrightarrow{\mathrm{r}} \times(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})=\overrightarrow{0}\)

\((\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}) \overrightarrow{\mathrm{q}}-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{r}}+(\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{p}}) \overrightarrow{\mathrm{p}}-(\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{r}}) \overrightarrow{\mathrm{p}}+(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{p}}-(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{q}} \)

0 = 0

\(\because \quad \vec{p} \cdot \vec{r} =\vec{r} \cdot \vec{p} \)

\(\vec{p} \cdot \vec{q} =\vec{q} \cdot \vec{p} \)

\(\vec{q} \cdot \vec{r} =\vec{r} \cdot \vec{q}\)

(Hence this is proved)