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  • Need solution for RD Sharma maths class 12 chapter Vector or Cross Product exercise 24.1 question 23

Need solution for RD Sharma maths class 12 chapter Vector or Cross Product exercise 24.1 question 23

Answers (1)

Answer:

 L.H.S = R.H.S

Hint:

To solve this we use formula

Given:

|\vec{a} \times \vec{b}|^{2}=\left|\begin{array}{ll} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} \end{array}\right|

Solution:          

RHS \\\\\begin{aligned} &\left|\begin{array}{ll} \left|\vec{a}^{2}\right| & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} \end{array}\right| \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{a}) \end{aligned}                                              

=|\vec{a}|^{2}|\vec{b}|^{2}-(\vec{a} \cdot \vec{b})^{2} \quad[\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}]

\begin{aligned} &=|\vec{a}|^{2}|\vec{b}|^{2}-(|\vec{a}||\vec{b}| \cos \theta)^{2} \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta \\\\ &=|\vec{a}|^{2}|\vec{b}|^{2}\left(1-\cos ^{2} \theta\right) \end{aligned}

\begin{aligned} &=|\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta \\\\ &=|\vec{a} \times \vec{b}|^{2} \quad[\because \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta] \end{aligned}         

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