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  • Need solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 9.

Need solution for RD Sharma Maths Class 12 Chapter 25 Scalar Triple Product Exercise Very Short Answer Question, question 9.

Answers (1)

Answer:

10

Hint:

Use scalar triple product formula.

Given:

\left[\begin{array}{llll} 3 \vec{a}+7 \vec{b} & \vec{c} & \vec{d} \end{array}\right]=\lambda\left[\begin{array}{llll} \vec{a} & \vec{c} & \vec{d} \end{array}\right]+\mu\left[\begin{array}{lll} \vec{b} & \vec{c} & \vec{d} \end{array}\right]

Solution:

\left[\begin{array}{llll} 3 \vec{a}+7 \vec{b} & \vec{c} & \vec{d} \end{array}\right]=\lambda\left[\begin{array}{llll} \vec{a} & \vec{c} & \vec{d} \end{array}\right]+\mu\left[\begin{array}{lll} \vec{b} & \vec{c} & \vec{d} \end{array}\right]

L.H.S

\begin{array}{ll} =[3 \vec{a}+7 \vec{b} \quad \vec{c} \quad \vec{d}] \\ =\{(3 \vec{a}+7 \vec{b}) \times \vec{c}\} \cdot \vec{d} &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because[\vec{a} \quad \vec{b} \quad \vec{c}]=\vec{a} \times \vec{b} \cdot \vec{c}=\vec{a} \cdot(\vec{b} \times \vec{c})]} \\ =\{3(\vec{a} \times \vec{c})+7(\vec{b} \times \vec{c})\} \cdot \vec{d} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\text { By distributive property }]} \end{array}

\begin{aligned} &=3\{(\vec{a} \times \vec{c}) \cdot \vec{d}\}+7\{(\vec{b} \times \vec{c}) \cdot \vec{d}\} \\ &=3\left[\begin{array}{llll} \vec{a} & \vec{c} & \vec{d} \end{array}\right]+7[\vec{b} \quad \vec{c} \quad \vec{d}] \end{aligned} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because(\vec{a} \times \vec{b}) \cdot \vec{c}=\left[\begin{array}{lll} \vec{a} & \vec{b} & \vec{c} \end{array}\right]\right.

  Since, L.H.S = R.H.S

i.e.\left[\begin{array}{llll} 3 \vec{a}+7 \vec{b} & \vec{c} & \vec{d} \end{array}\right]=\lambda\left[\begin{array}{llll} \vec{a} & \vec{c} & \vec{d} \end{array}\right]+\mu\left[\begin{array}{lll} \vec{b} & \vec{c} & \vec{d} \end{array}\right]

Comparing and equating co efficient from both sides.

We get,

\begin{aligned} &\lambda=3 \text { and } \mu=7 \\ &\therefore \lambda+\mu=3+7=10 \end{aligned}

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