If -a-b- - -a-b- then -a-b- -6-4-3-2

The correct option is D π/2|a+b|=|a b|⇒ |a+b|^2=|a b|^2 ⇒ (a+b)^2 =(a b)^2 ⇒ a^2+b^2+2a.b =a^2+b^2 2a.b ⇒ 4a.b=0 ⇒ a.b=0 ⇒ (a,b)=90^∘ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Equal Vectors

If |a+b| = |a...

Question

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

\frac{π}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

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

Suggest Corrections

Similar questions

If $s i n A = \frac{1}{2},$ $c o s B = \frac{12}{13},$ where $\frac{π}{2} < A < π$ and $\frac{3 π}{2} < B < 2 π,$ find tan(A-B)

If $t a n A = \frac{3}{4}, c o s B = \frac{9}{41},$ where $π < A < \frac{3 π}{2}$ and $0 < B < \frac{π}{2},$ find $t a n (A + B)$ .

If $c o s A = - \frac{24}{25}$ and $c o s B = \frac{3}{5}$ , where $π < A < \frac{3 π}{2}$ and $\frac{3 π}{2} < B < 2 π$ , find the following:
(i)sin(A+B) (ii)cos(A+B)

Q. If f (x) = e^x sin x in [0, π], then c in Rolle's theorem is

(a) π/6

(b) π/4

(c) π/2

(d) 3π/4

I

: If

| ¯ ¯ ¯ a . ¯ ¯ b | = | ¯ ¯ ¯ a \times ¯ ¯ b |

, then

(¯ ¯ ¯ a . ¯ ¯ b) = \frac{π}{4}

II

| ¯ ¯ ¯ a . ¯ ¯ b |^{2} = | ¯ ¯ ¯ a \times ¯ ¯ b |^{2}

, then

(¯ ¯ ¯ a, ¯ ¯ b) = \frac{3 π}{4}

Join BYJU'S Learning Program

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

The correct option is D π2|a+b|=|a−b|⇒|a+b|2=|a−b|2⇒(a+b)2=(a−b)2⇒a2+b2+2a.b=a2+b2−2a.b⇒4a.b=0⇒a.b=0⇒(a,b)=90∘

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