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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 6 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 6 Maths Textbook Solution.

Answers (1)

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

Hint :- if any triads of vectors are not coplanar if & only if their scalar triple product \neq 0

Given:- four points A,B,C & D

So, let    \begin{aligned} &\vec{A}=6 \hat{\imath}-7 \hat{\jmath} \\ \end{aligned}

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

We have to show that any three vectors forming from four points have their scalar triple product\neq 0
\therefore \vec{AB}= Position of B-Position of A

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

Similarly,    \therefore \vec{AC}= Position of C -Position of A

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

Now,  \therefore \vec{AD}=  Position of D – Position of A
\begin{aligned} &=(2 \hat{\imath}-5 \hat{\jmath}+10 \hat{k})-(6 \hat{\imath}-7 \hat{\jmath}+0 \hat{k}) \\ &=(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 [ \vec{AB}.\vec{AC}.\vec{AD} \right ]\neq 0

\begin{aligned} &\therefore[\overrightarrow{A B} \cdot \overrightarrow{A C} \cdot \overrightarrow{A D}]=\left|\begin{array}{ccc} 10 & -12 & -4 \\ -3 & 7 & -6 \\ -4 & 2 & 10 \end{array}\right| \\ &=10(70+12)+12(-30-24)-4(-6+28) \\ &=820-648-88 \\ &=84 \neq 0 \end{aligned}

 

Thus, the vectors so formed are not coplanar

 

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