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  • Need Solution for R.D. Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 9 Maths Textbook Solution.

Need Solution for R.D. Sharma Maths Class 12 Chapter 25 Scalar Triple Product  Exercise 25.1 Question 9 Maths Textbook Solution.

Answers (1)

Answer :-  \lambda =1

Hint :- When vectors are coplanar their scalar triple product is zero.

Given:  four points,

\begin{aligned} &\mathrm{A}=-\hat{\jmath}-\hat{k} \\ &\mathrm{~B}=4 \hat{\imath}+5 \hat{\jmath}+\lambda \hat{k} \\ &\mathrm{C}=3 \hat{\imath}+9 \hat{\jmath}+4 \hat{k} \\ &\mathrm{D}=-4 \hat{\imath}+4 \hat{\jmath}+4 \hat{k} \\ \end{aligned}

\begin{aligned} &\therefore \overrightarrow{A B}=(4 \hat{\imath}+5 \hat{\jmath}+\lambda \hat{k})-(-\hat{\jmath}-\hat{k}) \\ &=(4-0) \hat{\imath}+(5+1) \hat{\jmath}+(\lambda+1) \hat{k} \\ &=4 \hat{\imath}+6 \hat{\jmath}+(\lambda+1) \hat{k} \end{aligned}

\begin{aligned} &\overrightarrow{A C}=(3 \hat{\imath}+9 \hat{\jmath}+4 \hat{k})-(-\hat{\jmath}-\hat{k}) \\ &=(3-0) \hat{\imath}+(9+1) \hat{\jmath}+(4+1) \hat{k} \\ &=3 \hat{\imath}+10 \hat{\jmath}+5 \hat{k} \\ \end{aligned}

\begin{aligned} &\overrightarrow{A D}=(-4 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})-(-\hat{\jmath}-\hat{k}) \\ &=(-4-0) \hat{\imath}+(4+1) \hat{\jmath}+(4+1) \hat{k} \\ &=-4 \hat{\imath}+5 \hat{\jmath}+5 \hat{k} \end{aligned}

Now, As vectors are given coplanar then,  \left [ \overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD} \right ]=0

\begin{aligned} &\therefore[\overrightarrow{A B} \overrightarrow{A C} \overrightarrow{A D}]=\left|\begin{array}{ccc} 4 & 6 & (\lambda+1) \\ 3 & 10 & 5 \\ -4 & 5 & 5 \end{array}\right| \\ \end{aligned}

\begin{aligned} &=4(50-25)-6(15+20)+(\lambda+1)(15+40) \\ &=100-210+55 \lambda+55 \\ &=55 \lambda+(-55) \\ &=55 \lambda-55 \\ \end{aligned}

As vectors are coplanar

\begin{aligned} &\therefore 55 \lambda-55=0 \\ &\therefore \lambda=1 \end{aligned}

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