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TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

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Question : 33 of 160

Marks: +1, -0

\overset{-}{a}, \overset{-}{b}, \text{ and } \overset{-}{c}

be three non-coplanar vectors and

\overset{-}{p}, \overset{-}{q} \text{ and } \overset{-}{r}

\overset{-}{p} = \frac{\overset{-}{b} \times \overset{-}{c}}{\overset{-}{a}.\left(\overset{-}{b} \times \overset{-}{c}\right)}

\overset{-}{q} = \frac{\overset{-}{c} \times \overset{-}{a}}{\overset{-}{b}.\left(\overset{-}{c} \times \overset{-}{a}\right)}

\overset{-}{r} = \frac{\overset{-}{a} \times \overset{-}{b}}{\overset{-}{c}.\left(\overset{-}{a} \times \overset{-}{b}\right)}

\alpha = \left(\overset{-}{a} + \overset{-}{b}\right).\overset{-}{p} + \left(\overset{-}{b} + \overset{-}{c}\right).\overset{-}{q} + \left(\overset{-}{c} + \overset{-}{a}\right).\overset{-}{r}

\beta = \frac{\left(\overset{-}{a} + \overset{-}{b}\right).\left(\overset{-}{b} + \overset{-}{c}\right) \times \left(\overset{-}{a} + \overset{-}{b} + \overset{-}{c}\right)}{\overset{-}{b}.\left(\overset{-}{a} \times \overset{-}{c}\right)}

\alpha + \beta =

2
3
4
0

Solution:

We have,

(a+b) \cdot p = (a+b) \cdot \frac{b \times c}{a \cdot (b \times c)}

= \frac{a \cdot (b \times c) + b \cdot (b \times c)}{a \cdot (b \times c)} = \frac{[a b c] + [b b c]}{[a b c]}

= \frac{[a \cdot b c]}{[a b c]} \quad [ \because [b b c] = 0 ]

=1

Similarly,

(b+c) \cdot q = (c+a) \cdot r = 1

\therefore \quad \alpha = 1 + 1 + 1 = 3

Now,

(a+b) \cdot (b+c) \times (a+b+c)

= [a+b \quad b+c \quad a+b+c]

= \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} [a \quad b \quad c].

= [ I(l-1) - 1(0-1) ] [a b c] = [a b c]

and

b \cdot (a \times c) = [b a c] = -[a b c]

\therefore \quad \beta = \frac{[a b c]}{-[a b c]} = -1

\therefore \quad \alpha + \beta = 3 - 1 = 2

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