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Provide solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 8.

Answers (1)

Answer:

144

Hint:

Use scalar triple product formula.

Given:

 |\vec{a}|=3 \text { and }|\vec{b}|=4 \text { and }\left[\begin{array}{lll} \overrightarrow{\boldsymbol{a}} & \vec{b} & \vec{a} \times \vec{b}]+(\vec{a} \cdot \vec{b})^{2} \end{array}\right.

Solution:

\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{a} \times \vec{b} \end{array}\right]+(\vec{a} \cdot \vec{b})^{2}

\begin{aligned} &=\{(\vec{a}) \times(\vec{b}) \cdot(\vec{a} \times \vec{b})\}+(\vec{a} \cdot \vec{b})^{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c}] \end{array}=(\vec{a} \times \vec{b}) \cdot \vec{c}\right]\right.\\ &=(\vec{a} \times \vec{b}) \cdot(\vec{a} \times \vec{b})+(\vec{a} \cdot \vec{b})^{2} \end{aligned}

\begin{aligned} &=|(\vec{a} \times \vec{b})|^{2}+(\vec{a} \cdot \vec{b})^{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \vec{a} \times \vec{a}=|\vec{a}|^{2}\right] \\ &=(|\vec{a}||\vec{b}| \sin \theta)^{2}+(|\vec{\alpha}||\vec{b}| \cos \theta)^{2} & {\left[\begin{array}{l} \because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \\ \vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \end{array}\right]} \end{aligned}

\begin{aligned} &=(|\vec{a}||\vec{b}|)^{2} \sin ^{2} \theta+(|\vec{a}||\vec{b}|)^{2} \cos ^{2} \theta \\ &=(|\vec{a}||\vec{b}|)^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \\ &=(|\vec{a}||\vec{b}|)^{2} \cdot 1 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \sin ^{2} A+\cos ^{2} A=1\right] \\ &=(3.4)^{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because|\vec{a}|=3,|\vec{b}|=4] \\ &=(12)^{2}=144 \end{aligned}

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