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  • Need solution for RD Sharma maths class 12 chapter Vector or Cross Product exercise 24.1 question 19

Need solution for RD Sharma maths class 12 chapter Vector or Cross Product exercise 24.1 question 19

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Answer        : \frac{1}{\sqrt{165}}(10 \hat{\imath}+7 \hat{\jmath}-6 \hat{k})

Hint             : To solve this equation , we use determination method

Given          : A=(3,-1,2)


                    \begin{aligned} &B=(1,-1,-3) \\\\ &C=(4,-3,1) \end{aligned}

Solution     : \overrightarrow{A B}=-2 \hat{\imath}-5 \hat{k}

                    \begin{aligned} &\overrightarrow{A C}=\hat{\imath}-2 \hat{\jmath}-\hat{k} \\\\ &\vec{C}=x \overrightarrow{A B}+y \overrightarrow{A C} \end{aligned}

If \vec{a} is perpendicular to \vec{c}

\vec{d} is perpendicular to \overrightarrow{AB} ,\vec{d} perpendicular to \overrightarrow{AC}

\begin{aligned} &\vec{d}=\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -2 & 0 & -5 \\ 1 & -2 & -1 \end{array}\right| \\\\ &=-10 \hat{\imath}-7 \hat{\jmath}+4 \hat{k} \\\\ &\hat{d}=\frac{\vec{d}}{|\vec{d}|}=\frac{-10 \hat{\imath}-7 \hat{\jmath}+4 \hat{k}}{\sqrt{165}} \end{aligned}

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