Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Three particles are moving along the vectors <img alt="\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \quad \overrightarrow{\mathrm{B}}=\hat{\mathrm{j}}+\hat{\mathrm{k

Three particles P, Q$ and $R are moving along the vectors \overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \quad \overrightarrow{\mathrm{B}}=\hat{\mathrm{j}}+\hat{\mathrm{k}} \quad$ and $\quad \overrightarrow{\mathrm{C}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}} respectively. They strike on a point and start to move in different directions. Now particle P is moving normal to the plane which contains vector \overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}. Similarly particle \mathrm{Q} is moving normal to the plane which contains vector \overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{C}}. The angle between the direction of motion of P$ and $Q$ is $\cos ^{-1}\left(\frac{1}{\sqrt{x}}\right). Then the value of x is___________.
 

Answers (1)

best_answer

\begin{aligned} \hat{n}_{\hat{1}} &=\frac{\bar{A} \times \bar{B}}{|\bar{A} \times \bar{B}|}=\frac{(\hat{i}+\hat{\jmath}) \times(\hat{\jmath}+\hat{k})}{|(\hat{i}+\hat{\jmath}) \times(\hat{\jmath}+\hat{k})|} \\ &=\frac{\hat{k}+(-\hat{\jmath})+0+\hat{i}}{|\hat{k}+(-\hat{\jmath})+\hat{i}|} \\ \hat{n}_{1} &=\frac{\hat{i}+(-\hat{\jmath})+\hat{k}}{\sqrt{3}} \\ \end{aligned}

\hat{n_{1}}\rightarrowUnit vector normal to the plane containing \bar{A}\&\bar{B}( along the motion of particle P)

\begin{aligned} \hat{n}_{2} &=\frac{(\bar{A} \times \bar{C})}{|\bar{A} \times \bar{C}|}=\frac{(\hat{i}+\hat{\jmath}) \times(-\hat{i}+\hat{\jmath})}{|(\hat{i}+\hat{\jmath}) \times(-\hat{i}+\hat{\jmath})|} \\ &=\frac{0+\hat{k}+\hat{k}+0}{2} \\ \hat{n}_{2} &=\frac{2 \hat{k}}{ 2}=\hat{k} \end{aligned}

\hat{n}_{2} \rightarrow Unit vector perpendicular to the plane containing vector \bar{A}\&\bar{C} (along the direction of motion of particle Q.)

\begin{aligned} \hat{n}_{1} \cdot \hat{n}_{2} &=1 \times 1 \cos \theta \\ \cos \theta &=\left[\frac{\hat{i}+(-\hat{j})+\hat{k}}{\sqrt{3}}\right] . \hat {k}\end{aligned}

             \begin{aligned} =\frac{1}{\sqrt{3}} \\ \end{aligned}

\therefore x=3

Posted by

vishal kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question