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Explain solution RD Sharma class 12 chapter 25 Scalar triple product exercise multiple choice question 16 maths

Answers (1)

Answer:

 \begin{aligned} &\lambda =-2,\lambda =1\pm \sqrt{3} \end{aligned}

Hint:

 If vectors are coplanar, their scalar triple product is zero.

\begin{aligned} &\therefore \text { For }\begin{vmatrix} \lambda &1 &2 \\ 1 &\lambda &-1 \\ 2 &-1 &\lambda \end{vmatrix}\\ &=\lambda (\lambda ^2-1)-1(\lambda +2)+2(-1-2\lambda )\\ &=\lambda ^3-\lambda -\lambda -2-2-4\lambda \\ &=\lambda ^3-6\lambda -4 \end{aligned}

Now

\begin{aligned} &\lambda ^3-6\lambda -4=0\\ &(\lambda +2)(\lambda ^2-2\lambda -2)=0\\ &\lambda =-2,\lambda =1\pm \sqrt{3} \end{aligned}

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Gurleen Kaur

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