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  • The vector <img alt="\vec{a}=-\hat{\imath}+2 \hat{\jmath}+\hat{k}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cvec%7Ba%7D%3D-%5Chat%7B%5Cimath%7D&plus;2%20%5Chat%7B%5Cjmath%7D&p

The vector \vec{a}=-\hat{\imath}+2 \hat{\jmath}+\hat{k}  is rotated through a right angle, passing through the y-axis in its way and the resulting vector is \vec{b}.Then the projection of 3 \vec{a}+\sqrt{2} \vec{b} on \vec{c}=5 \hat{\imath}+4 \hat{\jmath}+3 \hat{k} is:

Option: 1

2 \sqrt{3}


Option: 2

1


Option: 3

3 \sqrt{2}


Option: 4

\sqrt{6}


Answers (1)

best_answer

\overline{\mathrm{b}}=\lambda \overline{\mathrm{a}}+\mu \hat{\mathrm{J}}
=\lambda(-\hat{i}+2 \hat{j}+\hat{\mathrm{k}})+\mu \hat{\mathrm{j}}
\bar{b}=-\lambda \hat{i}+(2 \lambda+\mu) \hat{j}+\lambda \hat{k}

|\overline{\mathrm{a}}|=|\overline{\mathrm{b}}|
|\overline{\mathrm{a}}|^{2}=|\overline{\mathrm{b}}|^{2} \quad \Rightarrow 6=\lambda^{2}+(2 \lambda+\mu)^{2}+\lambda^{2}\dots(1)

\because \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=0 \quad \Rightarrow \lambda+2(2 \lambda+\mu)+(1)(\lambda)=0
                             \Rightarrow 6 \lambda+2 \mu=0
                             \Rightarrow \mu=-3 \lambda \dots (2)

from (1) \& (2)
3 \lambda^{2}=6
\lambda^{2}=2               \Rightarrow \lambda= \pm \sqrt{2}
                            \Rightarrow \mu= \pm 3 \sqrt{2} 

\text{Projection of }\: 3 \bar{a}+2 \bar{b}$ on $\bar{c}$ is $=\frac{(3 \bar{a}+\sqrt{2} \bar{b}) \cdot \bar{c}}{|\bar{c}|}
                                                                =\frac{3 \bar{a} \cdot \bar{c}+\sqrt{2} \bar{b} \cdot \bar{c}}{|\bar{c}|}
                                                               =\frac{18+\sqrt{2}(-6 \sqrt{2})}{\sqrt{50}}
                                                               =\frac{6}{\sqrt{50}}=\frac{6}{5 \sqrt{2}}=\frac{3 \sqrt{2}}{5}

Case I: 
 (\overline{\mathrm{a}} . \overline{\mathrm{c}}=-5+8+3=6) \quad \frac{18+\sqrt{2}(-6 \sqrt{2})}{\sqrt{50}}
\lambda=\sqrt{2} \: \: \: \quad \overline{\mathrm{b}}=-\sqrt{2} \hat{i}+12 \sqrt{2}-3 \sqrt{2} \hat{j}+\sqrt{2} \hat{k}
                         \overline{\mathrm{b}}=-\sqrt{2} \hat{i}-\sqrt{2} \hat{j}+\sqrt{2} \hat{\mathrm{k}}
                         \bar{b} \cdot \bar{c}=-5 \sqrt{2}-4 \sqrt{2}+3 \sqrt{2}
                                   =-6 \sqrt{2}

Case II:
               \text { } \left.\quad \begin{array}{ll} \lambda =-\sqrt{2} \\ \mu=3 \sqrt{2} \end{array}\right]\quad \bar{b}=\sqrt{2} \hat{i}+(\sqrt{2}) \hat{\mathbf{j}}+(-\sqrt{2}) \hat{{k}}
             =\frac{18+\sqrt{2}(6 \sqrt{2})}{\sqrt{50}}
           =\frac{30}{\sqrt{50}}=\frac{30}{5 \sqrt{2}}=\frac{6}{\sqrt{2}}=3 \sqrt{2} \text { Ans. }


 

Posted by

shivangi.shekhar

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