If A = a_1 hat{i} + b_1 hat{j} and B = a_2 ha | vedclass

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(A) Two vectors are perpendicular if their dot product is zero.Given $A = a_1 \hat{i} + b_1 \hat{j}$ and $B = a_2 \hat{i} + b_2 \hat{j}$.The dot produ

Vedclass · Accepted Answer

$\frac{a_1}{b_1} = -\frac{b_2}{a_2}$

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(A) Two vectors are perpendicular if their dot product is zero.Given $A = a_1 \hat{i} + b_1 \hat{j}$ and $B = a_2 \hat{i} + b_2 \hat{j}$.The dot product $A \cdot B = (a_1 \hat{i} + b_1 \hat{j}) \cdot (a_2 \hat{i} + b_2 \hat{j}) = a_1 a_2 + b_1 b_2$.For perpendicular vectors,$A \cdot B = 0$,so $a_1 a_2 + b_1 b_2 = 0$.This implies $a_1 a_2 = -b_1 b_2$,or $\frac{a_1}{a_2} = -\frac{b_2}{b_1}$ (which is not directly listed) or $a_1 b_2 + a_2 b_1 = 0$.Looking at option $C$,$\frac{a_1}{a_2} = -\frac{b_1}{b_2}$ is equivalent to $a_1 b_2 = -a_2 b_1$,which is incorrect.Wait,let's re-evaluate: $a_1 a_2 + b_1 b_2 = 0 \implies a_1 a_2 = -b_1 b_2 \implies \frac{a_1}{b_2} = -\frac{b_1}{a_2}$.Actually,option $A$ is $\frac{a_1}{b_1} = -\frac{b_2}{a_2} \implies a_1 a_2 = -b_1 b_2$,which is $a_1 a_2 + b_1 b_2 = 0$. Thus,option $A$ is correct.

If $A = a_1 \hat{i} + b_1 \hat{j}$ and $B = a_2 \hat{i} + b_2 \hat{j}$,the condition that they are perpendicular to each other is

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