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  • Provide solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 5.

Provide solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 5.

Answers (1)

Answer:

Hint:

Use scalar triple product formula.

Given:

Given vectors are  .

Solution:

Let

\left.\begin{array}{c} \vec{\alpha}=\hat{\imath}+\hat{\jmath} \\ \vec{\beta}=\hat{\imath}+2 \hat{\jmath} \\ \vec{\gamma}=\hat{\imath}+\hat{\jmath}+\pi \hat{k} \end{array}\right\}                        Eq.(i)

We know that the volume of a parallel piped whose three adjacent edge are

\vec{\alpha} \; \; \vec{\beta}\: \text{and} \: \vec{\gamma}  is equal to modulus of scalar triple product i.e. \left |\left [\vec{\alpha} \; \; \vec{\beta}\: \: \vec{\gamma} \right ] \right |

So, first we will find \left[\begin{array}{lll} \vec{\alpha} & \vec{\beta} & \vec{\gamma} \end{array}\right] \text { then }\left|\left[\begin{array}{lll} \vec{\alpha} & \vec{\beta} & \vec{\gamma} \end{array}\right]\right|

\begin{aligned} &\therefore\left[\begin{array}{lll} \vec{\alpha} & \vec{\beta} & \vec{\gamma} \end{array}\right]=\vec{\alpha} \cdot(\vec{\beta} \times \vec{\gamma}) \\ &=(\hat{\imath}+\hat{\jmath}) \cdot\{(\hat{\imath}+2 \hat{\jmath}) \times(\hat{\imath}+\hat{\jmath}+\pi \hat{k})\} \end{aligned}                                                        From (1)

\begin{aligned} &=(\hat{\imath}+\hat{\jmath}) \cdot\left|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & 0 \\ 1 & 1 & \pi \end{array}\right| \\ &=(\hat{\imath}+\hat{\jmath}) \cdot\{\hat{\imath}(2 \pi-0)-\hat{\jmath}(\pi-0)+\hat{k}(1-2)\} \\ &=(\hat{\imath}+\hat{\jmath}) \cdot\{2 \pi \hat{\imath}-\pi \hat{\jmath}-\hat{k}\} \end{aligned}

\begin{aligned} &=2 \pi(\hat{\imath} . \hat{\imath})-\pi(\hat{\imath} . \hat{\jmath})-(\hat{\imath} . \hat{k})+2 \pi(\hat{\jmath} \cdot \hat{\imath})-\pi(\hat{\jmath} \cdot \hat{\jmath})-(\hat{\jmath} \cdot \hat{k})\\ &=2 \pi-\pi=\pi \end{aligned}

Now, \left |\left [\vec{\alpha} \; \; \vec{\beta}\: \: \vec{\gamma} \right ] \right |=\left |\pi \right |=\pi cubic units

 

 

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