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  • Let <img alt="\overrightarrow{\mathrm{a}}=\mathrm{a}_{1} \hat{i}+\mathrm{a}_{2} \hat{j}+\mathrm{a}_{3} \hat{k},\,\, \mathrm{a}_{i}>0, i=1,2,3" src="https://entrancecorner.oncodecogs.com/gif

Let \overrightarrow{\mathrm{a}}=\mathrm{a}_{1} \hat{i}+\mathrm{a}_{2} \hat{j}+\mathrm{a}_{3} \hat{k},\,\, \mathrm{a}_{i}>0, i=1,2,3 be a vector which makes equal angles with the coordinate axes OX, OY and OZ. Also, let the projection of \overrightarrow{\mathrm{a}} on the vector 3 \hat{i}+4 \hat{j} \text { be } 7. Let \overrightarrow{\mathrm{b}} be a vector obtained by rotating \overrightarrow{\mathrm{a}} with 90^{\circ}. If \vec{a}, \vec{b} and x-axis are coplanar, then projection of a vector \overrightarrow{\mathrm{b}} on 3 \hat{i}+4 \hat{j} is equal to:

Option: 1

\sqrt{7}


Option: 2

\sqrt{2}


Option: 3

2


Option: 4

7


Answers (1)

\vec{a}= a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}
\vec{a} =\lambda\left(\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\right)=\frac{\lambda}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})

Now projection of \vec{a} on given vector = 7
\frac{\lambda}{\sqrt{3}}\, \frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(3\hat{i}+4\hat{j} )}{5}=7
\lambda=5 \sqrt{3}

\vec{a}=5(\hat{i}+\hat{j}+\hat{k})
now \, \vec{b}=5\alpha(\hat{i}+\hat{j}+\hat{k})+\beta(\hat{i})
\vec{a} \cdot \vec{b}=0
\Rightarrow 25 \alpha(3)+5 \beta=0
\Rightarrow 15 \alpha+\beta=0 \Rightarrow \beta=-15 \alpha
\vec{b}=5 \alpha(-2 \hat{i}+\hat{j}+\hat{k})

Also |\vec{b}|=5 \sqrt{3}
\Rightarrow \alpha=\pm \frac{1}{\sqrt{2}}
\vec{b}=\pm \frac{5}{\sqrt{2}}(-2 \hat{i}+\hat{J}+\hat{k})

Projection\, of\, \; \vec{b} \; on \; 3\hat{i} +4 \hat{j} \; is
\frac{\vec{b} \cdot(3 \hat{i}+4 \hat{j})}{5}=\pm \frac{5}{\sqrt{2}}\left(\frac{-6+4}{5}\right)=\pm \sqrt{2}

Posted by

Ramraj Saini

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