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  • Need solution for RD Sharma maths class 12 chapter 25 Scalar triple product exercise multiple choice question 3

Need solution for RD Sharma maths class 12 chapter 25 Scalar triple product exercise multiple choice question 3

Answers (1)

Answer:

 \pm 1

Hint:

 Use formula of scalar triple product

Given:

\overrightarrow{a},\overrightarrow{b}\: and\; \overrightarrow{c} be the non-coplanar and mutually perpendicular unit vectors.

\therefore \left |\overrightarrow{a} \right |=\left |\overrightarrow{b} \right |= \left |\overrightarrow{c} \right |=1

Now

\begin{aligned} &\left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]=\left | \overrightarrow{a}\times \overrightarrow{b} \right |.\overrightarrow{c}\\ &=\left | \overrightarrow{a}\times \overrightarrow{b} \right |.1\\ &\because \overrightarrow{c} \text { is unit vector }\\ &\therefore \left [ \overrightarrow{a}\overrightarrow{b}\overrightarrow{c} \right ]=\left | \overrightarrow{a}\times \overrightarrow{b} \right | \text { or } -\left | \overrightarrow{a}\times \overrightarrow{b} \right |\\ &\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |sin \: \theta \dots \dots \dots (1)\\ &-\left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |sin \: \theta \dots \dots \dots (2)\\ &=(1)(1)sin\frac{\pi }{2}\\ &=1 \dots \dots \dots From(1)\\ &=-(1)(1)sin\frac{\pi }{2}\\ &=-1 \dots \dots \dots From(2) \end{aligned}

Thus value could be

\begin{aligned} &\pm 1 \end{aligned}

Posted by

Gurleen Kaur

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