If \(\vec a, \vec b\:and \: \vec c\) are coplanar, then what is \((2\vec a\times 3\vec b)\cdot4\vec c+(5\vec b\times 3\vec c)\cdot6\vec a\) equal to?

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NDA (Held On: 18 Apr 2021) Maths Previous Year paper

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Answer 1

CONCEPT:

Properties of Scalar Triple Product

[a b c] = [b c a] = [c a b]
[a b c] = - [b a c] = - [c b a] = - [a c b]
[(a + b) c d] = [a c d] + [b c d]
[λa, b c] = λ [a b c]
Three non-zero vectors \(\vec a,\;\vec b\;and\;\vec c\) are coplanar if and only if [a b c] = 0

CALCULATION:

Given: \(\vec a, \vec b\:and \: \vec c\) are coplanar i.e \([\vec a \ \vec b \ \vec c] = 0\)

⇒ \((2\vec a\times 3\vec b)\cdot4\vec c = [2\vec a \ 3\vec b \ 4\vec c]\) and \((5\vec b\times 3\vec c)\cdot6\vec a = [5\vec b \ 3\vec c \ 6\vec a]\)

⇒ \((2\vec a\times 3\vec b)\cdot4\vec c+(5\vec b\times 3\vec c)\cdot6\vec a = [2\vec a \ 3\vec b \ 4\vec c] + [5\vec b \ 3\vec c \ 6\vec a]\)

As we know that, [a b c] = [b c a] = [c a b]

⇒ \((2\vec a\times 3\vec b)\cdot4\vec c+(5\vec b\times 3\vec c)\cdot6\vec a = [2\vec a \ 3\vec b \ 4\vec c] + [6\vec a \ 5\vec b \ 3\vec c]\)

As we know that, [λa, b c] = λ [a b c]

⇒ \( [2\vec a \ 3\vec b \ 4\vec c] + [6\vec a \ 5\vec b \ 3\vec c] = 24 [\vec a \ \vec b \ \vec c] + 90 [\vec a \ \vec b \ \vec c]\)

As we know that, vectors \(\vec a,\;\vec b\;and\;\vec c\) are coplanar if and only if [a b c] = 0

⇒ \( [2\vec a \ 3\vec b \ 4\vec c] + [6\vec a \ 5\vec b \ 3\vec c] = 0\)

⇒ \((2\vec a\times 3\vec b)\cdot4\vec c+(5\vec b\times 3\vec c)\cdot6\vec a = 0\)

Hence, correct option is 3.