If |a+b| = |a-b| then (a,b) =

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is D $\frac{π}{2}$
$| a + b | = | a - b | \Rightarrow | a + b |^{2} = | a - b |^{2} \Rightarrow (a + b)^{2} = (a - b)^{2} \Rightarrow a^{2} + b^{2} + 2 a . b = a^{2} + b^{2} - 2 a . b \Rightarrow 4 a . b = 0 \Rightarrow a . b = 0 \Rightarrow (a, b) = 90^{\circ}$

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State, Whether the following statements are true of false.

(i) If $a < b, t h e n a - c < b - c$

(ii) If $a > b, t h e n a + c > b + c$

(iii) If $a < b, t h e n a c > b c$

(iv) If $a > b, t h e n \frac{a}{c} < \frac{b}{c}$

(v) If $a - c > b - d; t h e n a + d > b + c$

(vi) If $a < b, a n d c > 0, t h e n a - c > b - c where a, b, c and are real numbers and c \neq 0.$

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