Question

If $\vec{a}⋅\vec{b}=1,\vec{b}⋅\vec{c}=2 and$ $\vec{c}⋅\vec{a}=3,$
then the value of
[$[\vec{a}×(\vec{b}×\vec{c}),\vec{b}×(\vec{c}×\vec{a}),\vec{c}×(\vec{b}×\vec{a})]$
is

• A. 0
• B. $-6\vec{a}⋅×(\vec{b}×\vec{c})$
• C. $12\vec{c}⋅×(\vec{a}×\vec{b})$
• D. $-12\vec{b}⋅×(\vec{c}×\vec{a})$

Answer 1

Answer: (A)

0

Answer 2

The correct answer is (A) : 0
$∵\vec{a}×(\vec{b}×\vec{c})=3\vec{b}−\vec{c}=\vec{u}$
$\vec{b}×(\vec{c}×\vec{a})=\vec{c}−2\vec{a}=\vec{v}$
$\vec{c}×(\vec{b}×\vec{a})=3\vec{b}−2\vec{a}=\vec{w}$
∴ $\vec{u}+\vec{v}=\vec{w}$
So vectors $\vec{u},\vec{v} and$ $\vec{w}$ are coplanar, hence their Scalar triple product will be zero.