The value of -a-bb-cc-a- is -

Explanation for the correct option: We can write [a bb cc a] as (a b).((b c) ×(c a))Now to solve it we will do the cross product of two vectors b c and c a.=(a ...

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Standard XII

Mathematics

Condition for Two Lines to be on the Same Plane

The value of ...

Question

The value of $[\vec{a} - \vec{b} \vec{b} - \vec{c} \vec{c} - \vec{a}]$ is ?

$0$

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$1$

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$2$

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None of the above.

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Solution

The correct option is A
$0$

Explanation for the correct option:
We can write $[\vec{a} - \vec{b} \vec{b} - \vec{c} \vec{c} - \vec{a}]$ as $(\vec{a} - \vec{b}) . ((\vec{b} - \vec{c}) \times (\vec{c} - \vec{a}))$
Now to solve it we will do the cross product of two vectors $\vec{b} - \vec{c}$ and $\vec{c} - \vec{a} .$
$= (\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a}) = (\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a}) [∵ \vec{c} \times \vec{c} = 0] = (\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a})$
Now we will be finding the dot product of all the terms inside the bracket with $\vec{a} - \vec{b} .$
$= \vec{a} \cdot \vec{b} \times \vec{c} + \vec{a} \cdot \vec{a} \times \vec{c} + \vec{a} \cdot \vec{c} \times \vec{a} - \vec{b} \cdot \vec{b} \times \vec{c} - \vec{b} \cdot \vec{a} \times \vec{b} + \vec{b} \cdot \vec{c} \times \vec{a}$
Now we know that $\vec{a} \times \vec{c}$ is perpendicular to $\vec{a}, \vec{b} \times \vec{c}$ is perpendicular to $\vec{b}$ and $\vec{a} \times \vec{b}$ is perpendicular to $\vec{b}$ and taking dot products of the perpendicular vectors gives $\vec{0} .$
$\Rightarrow \vec{a} \cdot \vec{b} \times \vec{c} - \vec{b} \cdot \vec{c} \times \vec{a}$
We can rewrite it as
$\Rightarrow \vec{a} \cdot \vec{b} \times \vec{c} - \vec{a} \cdot \vec{b} \times \vec{c} = 0$
Therefore, $[\vec{a} - \vec{b} \vec{b} - \vec{c} \vec{c} - \vec{a}]$ is equal to $0 .$
Hence, the correct answer is option (A).

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