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  • Explain Solution R.D. Sharma Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 14 Maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 14 Maths Textbook Solution.

Answers (1)

Answer :-  x=6

Hint :- If given vectors  are coplanar, than their scalar triple product is always zero.

Given:-   \begin{aligned} &\mathrm{A}=4 \hat{\imath}+3 \hat{\jmath}+3 \hat{k} \\ \end{aligned}

\begin{aligned} &\mathrm{B}=5 \hat{\imath}+x \hat{\jmath}+7 \hat{k} \\ &\mathrm{C}=5 \hat{\imath}+3 \hat{\jmath} \\ &\mathrm{D}=7 \hat{\imath}+6 \hat{j}+\hat{k} \end{aligned}

We have to form any three vectors from above points & make their scalar triple product is zero to find the value of x.

\begin{aligned} &\therefore \overrightarrow{A B}=(5 \hat{\imath}+x \hat{\jmath}+7 \hat{k})-(4 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}) \\ &=(5-4) \hat{\imath}+(\mathrm{x}-3) \hat{\jmath}+(7-3) \hat{k} \\ &=\hat{\imath}+(\mathrm{x}-3) \hat{\jmath}+4 \hat{k} \\ \end{aligned}

\begin{aligned} &\overrightarrow{A C}=(5 \hat{\imath}+3 \hat{\jmath})-(4 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}) \\ &=(5-4) \hat{\imath}+(3-3) \hat{\jmath}+(0-3) \hat{k} \end{aligned}

=\hat{i}-3\hat{k}

\begin{aligned} &\overrightarrow{A D}=(7 \hat{\imath}+6 \hat{\jmath}+\hat{k})-(4 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}) \\ &=(7-4) \hat{\imath}+(6-3) \hat{\jmath}+(1-3) \hat{k} \\ &=3 \hat{\imath}+3 \hat{\jmath}-2 \hat{k} \end{aligned}

We know \left [ \overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD} \right ]=0

\therefore\left|\begin{array}{ccc} 1 & (x-3) & 4 \\ 1 & 0 & -3 \\ 3 & 3 & -2 \end{array}\right|

\begin{aligned} &=1(0+9)-(\mathrm{x}-3)(-2+9)+4(3-0) \\ &=9-7 \mathrm{x}+21+12 \\ &=42-7 \mathrm{x} \\ &\therefore 42-7 \mathrm{x}=0 \\ &\therefore 7 \mathrm{x}=42 \\ &\mathrm{x}=6 \end{aligned}

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