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  • Please solve RD Sharma Class 12 Chapter 24 Vector or Cross Product Exercise Very Short Answer Question, question 8 Maths textbook solution.

Please solve RD Sharma Class 12 Chapter 24 Vector or Cross Product Exercise Very Short Answer Question, question 8 Maths textbook solution.

Answers (1)

ANSWER:

\theta=\frac{\pi}{4} \text { or }\left(\pi-\frac{\pi}{4}\right) \cong \frac{3 \pi}{4}

HINT:

 Just use the formula of cross product .

GIVEN:

\overrightarrow{a} and \overrightarrow{b} be  two vectors  having magnitude  3 and \frac{\sqrt{2}}{3}

SOLUTION:

i.e, \left | \overrightarrow{a} \right | =3

\left | \overrightarrow{b} \right |=\frac{\sqrt{2}}{3}

Moreover, \overrightarrow{a} \times \overrightarrow{b} is a unit vector

\begin{aligned} &\therefore \vec{a} \times \vec{b}=1 \\ &\therefore \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \operatorname{Sin} \theta \cdot \widehat{n} \end{aligned}

1=3\left(\frac{\sqrt{2}}{3}\right) \operatorname{\sin} \theta \cdot(1)

\sin \theta = \frac{1}{\sqrt{2}}

\theta=\frac{\pi}{4} \text { or }\left(\pi-\frac{\pi}{4}\right) \cong \frac{3 \pi}{4}

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