The value of

\(\rm [\vec{a} + 2\vec{b} - \vec{c}, \vec{a} - \vec{b}, \vec{a}-\vec{b}- \vec{c}]=\)   is ?

Concept:

Scalar Triple Product:

A scalar triple product is also called a box product.

It is evident that scalar triple product of vectors means the product of three vectors. It means taking the dot product of one of the vectors with the cross product of the remaining two. It is denoted as [a b c ] = a · [b × c]

The product is cyclic in nature ⇔ [ a b c ] = [ b c a ] = [ c a b ]

Properties of the scalar triple product:

In a scalar triple product, dot and cross can be interchanged without altering the order of occurrences of the vectors ⇔ a · [b × c] = [a × b] ∙ c

Three vectors are coplanar if and only if their Scalar Triple Product is zero.

The Scalar Triple Product of three vectors is zero if any two of them are parallel.

Calculation:

\(\rm [\vec{a} + 2\vec{b} - \vec{c}, \vec{a} - \vec{b}, \vec{a}-\vec{b}- \vec{c}]\\=\rm[\vec {a}\;\vec {b}\;\vec {c}]\begin{vmatrix} 1 & 2 & -1\\ 1 & -1 & 0\\ 1 & -1 & -1 \end{vmatrix}\\=[\vec {a}\;\vec{b}\;\vec {c}][1(1-0)-2(-1-0)-1(-1+1)]\\=3[\vec {a}\;\vec{b}\;\vec {c}]\)

Alternate solution:

\(\rm [\vec{a} + 2\vec{b} - \vec{c}, \vec{a} - \vec{b}, \vec{a}-\vec{b}- \vec{c}]\\=(\vec{a} +2\vec{b} - \vec{c}) \cdot [(\vec{a} - \vec{b}) \times (\vec{a}-\vec{b}- \vec{c})]\)

\(\rm =(\vec{a} - 2\vec{b} - \vec{c}) \cdot [\vec{a} \times \vec{a}-\vec{a} \times \vec{b}-\vec{a} \times \vec{c}-\vec{b} \times \vec{a}+\vec{b} \times \vec{b}+\vec{b} \times \vec{c}]\)

As we know, cross product of parallel vectors is zero.

\(\rm =(\vec{a} + 2\vec{b} - \vec{c}) \cdot [0-\vec{a} \times \vec{b}-\vec{a} \times \vec{c}+\vec{a} \times \vec{b}+0+\vec{b} \times \vec{c}]\)          \((\because \rm \vec{a} \times \vec{b}= -\vec{b} \times \vec{a})\)

\(\rm =(\vec{a} + 2\vec{b} - \vec{c}) \cdot [-\vec{a} \times \vec{c}+\vec{b} \times \vec{c}]\)

\(\rm =-\vec{a}\cdot[\vec{a} \times \vec{c}] - 2\vec{b}\cdot[\vec{a} \times \vec{c}] + \vec{c} \cdot[\vec{a} \times \vec{c}]+\vec{a}\cdot[\vec{b} \times \vec{c}] + 2\vec{b}\cdot[\vec{b} \times \vec{c}] - \vec{c} \cdot[\vec{b} \times \vec{c}]\)

The Scalar Triple Product of three vectors is zero if any two of them are parallel.

\(\rm =0 - 2\vec{b}\cdot[\vec{a} \times \vec{c}] + 0+\vec{a}\cdot[\vec{b} \times \vec{c}] + 0 - 0\)

\(\rm = 2[\vec {a}\;\vec{b}\;\vec {c}] + [\vec {a}\;\vec{b}\;\vec {c}]\\=3[\vec {a}\;\vec{b}\;\vec {c}]\)