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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 8 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 25 Scalar Triple Product  Exercise 25.1 Question 8 Maths Textbook Solution.

Answers (1)

Answer :-  \left [ \overrightarrow{AB}\: \overrightarrow{AC}\: \overrightarrow{AD} \right ]\neq 0

Hint :- To show vectors are coplanar, their scalar triple product must be zero.

Given:-  points whose position vectors are,

\begin{aligned} &\vec{A}=6 \hat{\imath}-7 \hat{\jmath} \\ &\vec{B}=16 \hat{\imath}-19 \hat{\jmath}-4 \hat{k} \\ &\vec{C}=3 \hat{\imath}-6 \hat{k} \\ &\vec{D}=2 \hat{\imath}-5 \hat{\jmath}+10 \hat{k} \end{aligned}

\begin{aligned} &\overrightarrow{A B}=(16 \hat{\imath}-19 \hat{\jmath}-4 \hat{k})-(6 \hat{\imath}-7 \hat{\jmath}) \\ &=(16-6) \hat{\imath}+(-19+7) \hat{\jmath}+(-4-0) \hat{k} \\ &=10 \hat{\imath}-12 \hat{\jmath}-4 \hat{k} \\ \end{aligned}

\begin{aligned} &\overrightarrow{A C}=(3 \hat{\imath}-6 \hat{k})-(6 \hat{\imath}-7 \hat{\jmath}) \\ &=(3-6) \hat{\imath}+(0+7) \hat{\jmath}+(-6-0) \hat{k} \\ &=-3 \hat{\imath}+7 \hat{\jmath}-6 \hat{k} \\ \end{aligned}

\begin{aligned} &\overrightarrow{A D}=(2 \hat{\imath}-5 \hat{j}+10 \hat{k})-(6 \hat{\imath}-7 \hat{\jmath}) \\ &=(2-6) \hat{\imath}+(-5+7) \hat{\jmath}+(10-0) \hat{k} \\ &=-4 \hat{\imath}+2 \hat{\jmath}+10 \hat{k} \end{aligned}

Now, we have to prove. \left [ \overrightarrow{AB}\: \overrightarrow{AC}\: \overrightarrow{AD} \right ]\neq 0

\begin{aligned} &\therefore[\overrightarrow{A B} \overrightarrow{A C} \overrightarrow{A D}]=\left|\begin{array}{ccc} 10 & -12 & -4 \\ -3 & 7 & -6 \\ -4 & 2 & 10 \end{array}\right| \\ \end{aligned}

\begin{aligned} &=10(70+12)+12(-30-24)-4(-6+28) \\ &=820-648+(-88) \\ &=84 \neq 0 \end{aligned}

Thus, the vectors are not coplanar.

 

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