# If → a , → B , → C Are Three Non-coplanar Vectors, Then ( → a + → B + → C ) . [ ( → a + → B ) × ( → a + → C ) ] Equals - Mathematics

MCQ
Sum

If $\vec{a,} \vec{b,} \vec{c}$ are three non-coplanar vectors, then $\left( \vec{a} + \vec{b} + \vec{c} \right) . \left[ \left( \vec{a} + \vec{b} \right) \times \left( \vec{a} + \vec{c} \right) \right]$ equals

#### Options

• 0

• $\left[ \vec{a} \vec{b} \vec{c} \right]$

• $2\left[ \vec{a} \vec{b} \vec{c} \right]$

• $- \left[ \vec{a} \vec{b} \vec{c} \right]$

#### Solution

$- \left[ \vec{a} \vec{b} \vec{c} \right]$

We have

$\left( \vec{a} + \vec{b} + \vec{c} \right) . \left[ \left( \vec{a} + \vec{b} \right) \times \left( \vec{a} + \vec{c} \right) \right]$

$= \left( \vec{a} + \vec{b} + \vec{c} \right) . \left[ \left( \vec{a} + \vec{b} \right) \times \vec{a} + \left( \vec{a} + \vec{b} \right) \times \vec{c} \right] \left(\text { By definition of cross poduct }\right)$

$= \left( \vec{a} + \vec{b} + \vec{c} \right) . \left[ \vec{a} \times \vec{a} + \vec{b} \times \vec{a} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} \right]$

$= \left( \vec{a} + \vec{b} + \vec{c} \right) . \left[ 0 + \vec{b} \times \vec{a} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} \right]$

$= \vec{a} . \left( \vec{b} \times \vec{a} \right) + \vec{a} . \left( \vec{a} \times \vec{c} \right) + \vec{a} . \left( \vec{b} \times \vec{c} \right) + \vec{b} . \left( \vec{b} \times \vec{a} \right) + \vec{b} . \left( \vec{a} \times \vec{c} \right) + \vec{b} . \left( \vec{b} \times \vec{c} \right) + \vec{c} . \left( \vec{b} \times \vec{a} \right) + \vec{c} . \left( \vec{a} \times \vec{c} \right) + \vec{c} . \left( \vec{b} \times \vec{c} \right)$

$= \left[ \vec{a} \vec{b} \vec{a} \right] + \left[ \vec{a} \vec{a} \vec{c} \right] + \left[ \vec{a} \vec{b} \vec{c} \right] + \left[ \vec{b} \vec{b} \vec{a} \right] + \left[ \vec{b} \vec{a} \vec{c} \right] + \left[ \vec{b} \vec{b} \vec{c} \right] + \left[ \vec{c} \vec{b} \vec{a} \right] + \left[ \vec{c} \vec{a} \vec{c} \right] + \left[ \vec{c} \vec{b} \vec{c} \right]$

$= 0 + 0 + \left[ \vec{a} \vec{b} \vec{c} \right] + 0 + \left[ \vec{b} \vec{a} \vec{c} \right] + 0 + \left[ \vec{c} \vec{b} \vec{a} \right] + 0 + 0$

$= \left[ \vec{a} \vec{b} \vec{c} \right] - \left[ \vec{a} \vec{b} \vec{c} \right] - \left[ \vec{a} \vec{b} \vec{c} \right] \left( \because \left[ \vec{b} \vec{a} \vec{c} \right] = - \left[ \vec{c} \vec{a} \vec{b} \right], \left[ \vec{b} \vec{a} \vec{c} \right] = - \left[ \vec{a} \vec{b} \vec{c} \right] \right)$

$= - \left[ \vec{a} \vec{b} \vec{c} \right]$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 26 Scalar Triple Product
MCQ | Q 14 | Page 19