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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

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Question : 24 of 160

Marks: +1, -0

If

\vec{a}=x\hat{i}+y\hat{j}+z\hat{k},

(\vec{a}\times\hat{i})\cdot(\hat{i}+\hat{j})+(\vec{a}\times\hat{j})\cdot(\hat{j}+\hat{k})+(\vec{a}\times\hat{k})\cdot(\hat{k}+\hat{i})=

$x-y+z$
$x+y+z$
$x+y-z$
$-x+y+z$

Solution:

Simplify the given expression

(a \times \hat{i}) (\hat{i}+ \hat{j}) + (a \times \hat{j}) (\hat{j}+ \hat{k}) + (a \times \hat{k}) (\hat{k}+ \hat{i})

= [a \hat{i} \hat{i}] + [a \hat{i} \hat{j}] + [a \hat{j} \hat{j}]+[a \hat{j} \hat{k}] + [a \hat{k} \hat{k}] + [a \hat{k} \hat{i}]

= a \cdot (\hat{i} \times \hat{j}) + a \cdot (\hat{j} \times \hat{k}) + a \cdot (\hat{k} \times \hat{i})

= a(\hat{i} + \hat{j} + \hat{k})

But according to the given condition,

a = x\hat{i} + y\hat{j} + z\hat{k}

Therefore,

= (x\hat{i} + y\hat{j} + z\hat{k} ) ( \hat{i} + \hat{j} + \hat{k})

= x + y + z

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