Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD sharma maths class 12 chapter 24 vector or cross product Exercises Multiple choice questions question 13 maths textbook solution

Need solution for RD sharma maths class 12 chapter 24 vector or cross product Exercises Multiple choice questions question 13 maths textbook solution

Answers (1)

Answer: \left | \overrightarrow{a} \right |^{2}\left | \overrightarrow{b} \right |^{2}-\left ( \overrightarrow{a}.\overrightarrow{b} \right )^{2}

Given: \left ( \overrightarrow{a}\times \overrightarrow{b} \right )^{2}

Hint: Using \left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta

Explanation: We know,

                            \left | \overrightarrow{a}\times \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \theta

Squaring on both sides

              \begin{aligned} &(\vec{a} \times \vec{b})^{2}=(|\vec{a}||\vec{b}| \sin \theta)^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \\ &\Rightarrow(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right) \\ &\Rightarrow(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \end{aligned}………………….. (*)

              We know \left | \overrightarrow{a}. \overrightarrow{b} \right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta 

On squaring both sides

              \left ( \overrightarrow{a}.\overrightarrow{b} \right )^{2}=\left | \overrightarrow{a} \right |^{2}\left | \overrightarrow{b} \right |^{2}\cos \theta

              Put in (*)

              \begin{aligned} &\Rightarrow(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \\ &(\vec{a} \times \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} \end{aligned}

              

Posted by

Infoexpert

View full answer

Similar Questions

Latest Question