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Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 24.1 question 8 sub question (ii)

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Answer         : 3 \sqrt{3} \text { sq. units }

Hint               : To solve this , we use area of parallelogram.

Given            :2 \hat{\imath}+\hat{\jmath}+3 \hat{k} ; \hat{\imath}-\hat{\jmath}

Solution       : Area of parallelogram=|\vec{a} \times \vec{b}|

\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right| \\\\ &=\hat{i}(1 \times 0-3(-1))-\hat{j}(2 \times 0-3 \times 1)+\hat{k}(2 \times 1-(-1) \times 1) \\\\ &=3 \hat{\imath}+3 \hat{j}+3 \hat{k} \end{aligned}

Area of parallelogram

\begin{aligned} &=\sqrt{3^{2}+3^{2}+3^{2}} \\\\ &=\sqrt{9+9+9} \\\\ &=\sqrt{27} \\\\ &=3 \sqrt{3} \text { sq. units } \end{aligned}

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