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UPSC NDA I 2018 Mathematics Solved Paper

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Question : 20 of 120

Marks: +1, -0

If

\vec{a}+2\vec{b}+3\vec{c}=\vec{0}

\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}=λ(\vec{b}\times \vec{c }),

then what is the value of λ?

2
3
4
6

Solution:

Concept:

Cross product is anti-commutative:

\vec{u} \times \vec{v} = - \vec{v} \times \vec{u}

.

Cross product of a vector with itself is

\vec{0}

.

Explanation:

Given

\vec{a} + 2\vec{b} + 3\vec{c} = \vec{0}

.

So

\vec{a} = -(2\vec{b} + 3\vec{c})

.

Substitute into

\vec{a}\times \vec{b} + \vec{b}\times \vec{c} + \vec{c}\times \vec{a}

:

= (-(2\vec{b}+3\vec{c})) \times \vec{b} + \vec{b}\times \vec{c} + \vec{c} \times (-(2\vec{b}+3\vec{c}))

= -2\vec{b}\times \vec{b} - 3\vec{c}\times \vec{b} + \vec{b}\times \vec{c} - 2\vec{c}\times \vec{b} - 3\vec{c}\times \vec{c}

.

Simplify:

\vec{b}\times \vec{b}=\vec{0}

,

\vec{c}\times \vec{c}=\vec{0}

.

Also,

\vec{c}\times \vec{b} = - \vec{b}\times \vec{c}

so

-3\vec{c}\times \vec{b}=3\vec{b}\times \vec{c}

and

-2\vec{c}\times \vec{b}=2\vec{b}\times \vec{c}

.

Thus the expression becomes:

0 + 3(\vec{b}\times \vec{c}) + (\vec{b}\times \vec{c}) + 2(\vec{b}\times \vec{c}) + 0 = 6(\vec{b}\times \vec{c})

.

Comparing with

\lambda (\vec{b}\times \vec{c})

, we get

\lambda = 6

.

Answer:

D. 6

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