NIMCET 2015 — Mathematics PYQ

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Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

If $a$ and $b$ are vectors in space, given by $a = \frac{i ^ - 2 j ^}{5}$ and $b = \frac{2 i ^ + j ^ + 3 k ^}{14}$ , then the value of $(2 a + b) \cdot [(a \times b) \times (a - 2 b)]$ is:

Choose the correct answer:

Explanation

Let $E = (2\vec{a} + \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} - 2\vec{b})]$

Using property: $\vec{u} \cdot (\vec{v} \times \vec{w}) = [\vec{u} \,\, \vec{v} \,\, \vec{w}]$

E = [ (2\vec{a} + \vec{b}) \quad (\vec{a} \times \vec{b}) \quad (\vec{a} - 2\vec{b}) ]

Cyclic rearrangement:

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

Solving the cross product:

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

= \vec{a} \times \vec{b} + 4(\vec{a} \times \vec{b}) = 5(\vec{a} \times \vec{b})

Substituting back:

E = (\vec{a} \times \vec{b}) \cdot 5(\vec{a} \times \vec{b})

E = 5 |\vec{a} \times \vec{b}|^2

Magnitude calculations:

$|\vec{a}|^2 = \frac{1+4}{5} = 1$

$|\vec{b}|^2 = \frac{4+1+9}{14} = 1$

$\vec{a} \cdot \vec{b} = \frac{(1)(2) + (-2)(1) + 0}{\sqrt{70}} = 0 \implies \vec{a} \perp \vec{b}$

Since $\theta = 90^\circ$ :

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin 90^\circ = 1$

Final Result:

E = 5(1)^2 = 5

Correct Option: (c)

Choose the correct answer:

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