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Provide solution for RD Sharma maths class 12 chapter Vector or Cross Product exercise 24.1 question 2 sub question (i)

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Answer         : \sqrt{26}

Hint               : To solve this equation we use determinate formula then magnitude formula

Given             : \vec{a}=3 \hat{\imath}+4 \hat{\jmath}

                        \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k}

                       Find value |\vec{a} \times \vec{b}|

Solution        :

\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{array}\right| \\\\ &=\hat{\imath}(4 \times 1-0 \times 1)-\hat{\jmath}(3 \times 1-0 \times 1)+\hat{k}(3 \times 1+4 \times 1) \\\\ &=4 \hat{\imath}-3 \hat{\jmath}-\hat{k} \end{aligned}

\begin{aligned} &|\vec{a} \times \vec{b}|=\sqrt{4^{2}+3^{2}+1^{2}} \\\\ &=\sqrt{16+9+1} \\\\ &=\sqrt{26} \end{aligned}

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