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Let \mathrm{}\vec{a}=\hat{i}-\hat{j}+2 \hat{k} and let \mathrm{}\vec{b} be a vector such that \mathrm{} \vec{a} \times \vec{b}=2 \hat{i}-\hat{k} and \mathrm{}\vec{a} \cdot \vec{b}=3. Then the projection of \mathrm{}\vec{b} on the vector \mathrm{}\vec{a}-\vec{b} is :
 

Option: 1

\frac{2}{\sqrt{21}}


Option: 2

2 \sqrt{\frac{3}{7}}


Option: 3

\frac{2}{3} \sqrt{\frac{7}{3}}


Option: 4

\frac{2}{3}


Answers (1)

best_answer

\mathrm{|\overrightarrow{a} \times \overrightarrow{b}|=\sqrt{4+1}=\sqrt{5}}\\

\mathrm{\text { Also } \quad|\overrightarrow{a}|^{2}|\overrightarrow{b}|^{2}=|\overrightarrow{a} \cdot \overrightarrow{b}|^{2}+|\overrightarrow{a} \times \overrightarrow{b}|^{2}}\\

\mathrm{\Rightarrow 6|\overrightarrow{b}|^{2}=9+5}\\

\mathrm{\Rightarrow|\overrightarrow{b}|^{2}=\frac{7}{3}}

Required projection   \mathrm{=\frac{\overrightarrow{b} \cdot(\overrightarrow{a}-\overrightarrow{b})}{|\overrightarrow{a}-\overrightarrow{b}|}}             ........(i)

\mathrm{Now |\overrightarrow{a}-\overrightarrow{b}|^{2} =|\overrightarrow{a}|^{2}+|\overrightarrow{b}|^{2}-2 \overrightarrow{a} \cdot \overrightarrow{b} \\ =6+\frac{7}{3}-2 \cdot 3} \\

\mathrm{|\overrightarrow{a}-\overrightarrow{b}| =\sqrt{\frac{7}{3}} }

Using (i), projection   \mathrm{=\frac{\overrightarrow{a} \cdot \overrightarrow{b}-|\overrightarrow{b}|^{2}}{\sqrt{\frac{7}{3}}} }

                                \mathrm{=\frac{3-\frac{7}{3}}{\sqrt{\frac{7}{3}}}=\frac{2}{\sqrt{21}} }

Hence correct option is 1

Posted by

seema garhwal

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