माना कि overrightarrow{ A }=(hat{i}+hat{j}) एवं overrightarrow{ B }=(2 hat{i}-hat{

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3-1.Vectors

hard

माना कि $\overrightarrow{ A }=(\hat{i}+\hat{j})$ एवं $\overrightarrow{ B }=(2 \hat{i}-\hat{j})$ है। एक समतल वेक्टर $\vec{C}$ इस प्रकार है कि $\overrightarrow{ A } \cdot \overrightarrow{ C }=\overrightarrow{ B } \cdot \overrightarrow{ C }=\overrightarrow{ A } \cdot \overrightarrow{ B }$, तो $\overrightarrow{ C }$ का परिमाण होगा

$\sqrt {\frac{5}{9}} $

$\sqrt {\frac{10}{9}} $

$\sqrt {\frac{20}{9}} $

$\sqrt {\frac{9}{12}} $

(JEE MAIN-2018)

Solution

$\begin{array}{l}
If\,\vec C = a\hat i + b\hat j\,then\,\vec A.\vec C = \vec A.\vec B\\
a + b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)\\
\vec B.\vec C = \vec A.\vec B\\
2a – b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {ii} \right)\\
Solving\,equation\,\left( i \right)\,and\,\left( {ii} \right)\,\,we\,get\\
a = \frac{1}{3},\,b = \frac{2}{3}\\
Magnitude\,of\,coplanar\,vector,\\
\left| {\vec C} \right| = \sqrt {\frac{1}{9} + \frac{4}{9}} = \sqrt {\frac{5}{9}}
\end{array}$