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  • Explain Solution R.D.Sharma Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 13 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 13 Maths Textbook Solution.

Answers (1)

Answer :-  \lambda =5

Hint :- If the shown are coplanar, their scalar triple product is zero.

Given:- points

\begin{aligned} &\mathrm{A}=3 \hat{\imath}+2 \hat{\jmath}+\hat{k} \\ &\mathrm{~B}=4 \hat{\imath}+\lambda \hat{\jmath}+5 \hat{k} \\ &\mathrm{C}=4 \hat{\imath}+2 \hat{\jmath}-2 \hat{k} \\ &\mathrm{D}=6 \hat{\imath}+5 \hat{\jmath}-\hat{k} \end{aligned}

Vectors formed by these points are coplanar.

 So, we will form any three vectors & their scalar triple product will be zero & will get the value of \lambda.

\begin{aligned} &\therefore \overrightarrow{A B}=(4 \hat{\imath}+\lambda \hat{\jmath}+5 \hat{k})-(3 \hat{\imath}+2 \hat{\jmath}+\hat{k}) \\ &=(4-3) \hat{\imath}+(\lambda-2) \hat{\jmath}+(5-1) \hat{k} \\ &=\hat{\imath}+(\lambda-2) \hat{\jmath}+4 \hat{k} \\ \end{aligned}

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

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

\begin{aligned} &\overrightarrow{A D}=(6 \hat{\imath}+5 \hat{\jmath}-\hat{k})-(3 \hat{\imath}+2 \hat{\jmath}+\hat{k}) \\ &=(6-3) \hat{\imath}+(5-2) \hat{\jmath}+(-1-1) \hat{k} \\ &=3 \hat{\imath}+3 \hat{\jmath}-2 \hat{k} \end{aligned}

\begin{aligned} &{[\overrightarrow{A B} \overrightarrow{A C} \overrightarrow{A D}]=0} \\ &=\left|\begin{array}{ccc} 1 & (\lambda-2) & 4 \\ 1 & 0 & -3 \\ 3 & 3 & -2 \end{array}\right| \end{aligned}

\begin{aligned} &=1(0+9)-(\lambda-2)(-2+9)+4(3-0) \\ &=9-(\lambda-2)(7)+12 \\ &=9-7 \lambda+14+12 \\ \end{aligned}

\begin{aligned} &=35-7 \lambda \\ &\therefore 35-7 \lambda=0 \\ &\therefore 7\lambda=35 \\ &\lambda=5 \end{aligned}

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