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Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 24.1 question 3 sub question (ii)

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Answer         : \frac{1}{11}(\hat{\imath}+\hat{\jmath}-3 \hat{k})

Hint               : To solve this equation we use \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}

Given            : \vec{a}=2 \hat{\imath}+\hat{\jmath}+\hat{k}
                        \vec{b}=\hat{\imath}+2 \hat{\jmath}+\hat{k}

Solution       : \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}

\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & 1 \\ 1 & 2 & 1 \end{array}\right|

            \begin{aligned} &=\hat{\imath}(1 \times 1-2 \times 1)-\hat{\jmath}(2 \times 1-1 \times 1)+\hat{k}(2 \times 2-1 \times 1) \\\\ &=\hat{\imath}(-1)-\hat{\jmath}(1)+\hat{k}(3) \\\\ &=-\hat{\imath}-\hat{\jmath}+3 \hat{k} \end{aligned}

|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+(-1)^{2}+3^{2}}

            \begin{aligned} &=\sqrt{11} \\\\ &\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{1}{\sqrt{11}}(-\hat{\imath}-\hat{\jmath}+3 \hat{k}) \end{aligned}

 

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