Get Answers to all your Questions

header-bg qa

Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 24.1 question 9 sub question (ii)

Answers (1)

Answer          : \frac{\sqrt{6}}{2} \text { sq.units }

Hint                : To solve this we use area of parallelogram formula

Given             : 2 \hat{\imath}+\hat{k} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}

Solution        : Area of parallelogram   =\frac{1}{2}\left(d_{1} \times d_{2}\right)

\begin{gathered} d_{1}=2 \hat{\imath}+\hat{k} \\\\ d_{2}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{gathered}

\begin{aligned} &d_{1} \times d_{2}\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 0 & 1 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(0 \times 1-1 \times 1)-\hat{\jmath}(2 \times 1-1 \times 1)+\hat{k}(2) \\\\ &=-\hat{\imath}-\hat{\jmath}+2 \hat{k} \end{aligned}

\begin{aligned} &\frac{1}{2}\left|d_{1} \times d_{2}\right|=\frac{1}{2} \sqrt{(-1)^{2}+(-1)^{2}+2^{2}} \\\\ &=\frac{1}{2} \sqrt{1+1+4} \\\\ &=\frac{\sqrt{6}}{2} \text { sq.units } \end{aligned}

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads