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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise 25.1 Question 7 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 25 Scalar Triple Product  Exercise 25.1 Question 7 Maths Textbook Solution.

Answers (1)

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

Hint :- To prove vectors are coplanar scalar product of three vectors should be zero.

Given:-  points A=\hat{i}+4\hat{j}-3\hat{k}\

B=3\hat{i}+2\hat{j}-5\hat{k}

C=3\hat{i}+8\hat{j}-5\hat{k}

D=3\hat{i}+2\hat{j}+\hat{k}

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

Similarly, \begin{aligned} &\overrightarrow{A C}=(-3 \hat{\imath}+8 \hat{\jmath}-5 \hat{k})-(-\hat{\imath}+4 \hat{\jmath}-3 \hat{k}) \\ \end{aligned}

\begin{aligned} &=(-3+1) \hat{\imath}+(8-4) \hat{\jmath}+(-5+3) \hat{k} \\ &=-2 \hat{\imath}+4 \hat{\jmath}-2 \hat{k} \\ &\overrightarrow{A D}=(-3 \hat{\imath}+2 \hat{\jmath}+\hat{k})-(-\hat{\imath}+4 \hat{\jmath}-3 \hat{k}) \\ &=(-3+1) \hat{\imath}+(2-4) \hat{\jmath}+(1+3) \hat{k} \\ &=-2 \hat{\imath}-2 \hat{\jmath}+4 \hat{k} \end{aligned}

Now, if  \left [ \vec{AB}\: \: \vec{AC}\: \: \vec{AD} \right ]=0 Vectors are coplanar

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

\begin{aligned} &=4(16-4)+2(-8-4)-2(4+8) \\ &=4(12)+2(-12)-2(12) \\ &=48-24-24 \\ &=0 \end{aligned}

Thus given vectors are coplanar.

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