If overrightarrow{A} = a_{1} hat{imath} + a_{ | vedclass

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(D) Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are perpendicular if their dot product is zero,i.e.,$\overrightarrow{A} \cdot \overright

Vedclass · Accepted Answer

$\frac{a_{1}}{b_{2}} = -\frac{a_{2}}{b_{1}}$

Vedclass · Answer

(D) Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are perpendicular if their dot product is zero,i.e.,$\overrightarrow{A} \cdot \overrightarrow{B} = 0$.Given $\overrightarrow{A} = a_{1} \hat{\imath} + a_{2} \hat{\jmath}$ and $\overrightarrow{B} = b_{1} \hat{\imath} + b_{2} \hat{\jmath}$.Calculating the dot product: $(a_{1} \hat{\imath} + a_{2} \hat{\jmath}) \cdot (b_{1} \hat{\imath} + b_{2} \hat{\jmath}) = a_{1}b_{1} + a_{2}b_{2} = 0$.This implies $a_{1}b_{1} = -a_{2}b_{2}$.Rearranging the terms to match the options: $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$ is not correct,but looking at $a_{1}b_{1} = -a_{2}b_{2}$,we can write $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$ or $\frac{b_{2}}{a_{1}} = -\frac{a_{2}}{b_{1}}$ is incorrect. Let's re-examine: $a_{1}b_{1} + a_{2}b_{2} = 0 \implies a_{1}b_{1} = -a_{2}b_{2}$. Dividing by $a_{1}b_{1}$,we get $1 = -\frac{a_{2}b_{2}}{a_{1}b_{1}}$. Rearranging gives $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$. Wait,checking option $A$: $\frac{b_{2}}{a_{1}} = -\frac{a_{2}}{b_{1}}$ is equivalent to $b_{1}b_{2} = -a_{1}a_{2}$,which is not the condition. The condition is $a_{1}b_{1} + a_{2}b_{2} = 0$. Thus,$a_{1}b_{1} = -a_{2}b_{2}$. This can be written as $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$. None of the options perfectly match this standard form,but if we rearrange $a_{1}b_{1} = -a_{2}b_{2}$ as $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$,there might be a typo in the options. However,if we look at $\frac{a_{1}}{b_{2}} = -\frac{a_{2}}{b_{1}}$,this implies $a_{1}b_{1} = -a_{2}b_{2}$,which is exactly the condition for perpendicularity. Therefore,option $D$ is correct.

If $\overrightarrow{A} = a_{1} \hat{\imath} + a_{2} \hat{\jmath}$ and $\overrightarrow{B} = b_{1} \hat{\imath} + b_{2} \hat{\jmath}$ are perpendicular to each other,then:

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