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Let,  \vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}, \vec{a} \cdot \overrightarrow{\mathrm{c}}=7,2 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+43=0, \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} Then |\vec{a} \cdot \overrightarrow{\mathrm{b}}| | is equal to

Option: 1

8


Option: 2

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Option: 3

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Option: 4

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Answers (1)

best_answer

\overline{\mathrm{a}}=(1,2, \lambda) \quad \overline{\mathrm{b}}=(3,-5,-\lambda)

\begin{aligned} & \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=7, \quad \overline{\mathrm{b}} \cdot \overline{\mathrm{c}}=\frac{-43}{2} \\ & (\overline{\mathrm{a}}-\overline{\mathrm{b}}) \times \overline{\mathrm{c}}=0 \\ & \mathrm{c}=\mathrm{x}[-2,7,2 \lambda]=(-2 \mathrm{x}, 7 \mathrm{x}, 2 \lambda \mathrm{x}) \\ & \text { now, } \quad \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=-2 \mathrm{x}+14 \mathrm{x}+2 \lambda^2 \mathrm{x}=7 \\ & 2 \lambda^2 \mathrm{x}+12 \mathrm{x}=7 \end{aligned} ________________(i)

 \begin{array}{r} \overline{\mathrm{b}} \cdot \overline{\mathrm{c}}=-6 \mathrm{x}-35 \mathrm{x}-2 \lambda^2 \mathrm{x}=\frac{-43}{2} \\ -41 \mathrm{x}-2 \lambda^2 \mathrm{x}=\frac{-43}{2} \end{array} ___________________(ii)

by adding (1) + (2)

\begin{aligned} & -29 \mathrm{x}=\frac{-29}{2} \Rightarrow \mathrm{x}=\frac{1}{2} \\ & \therefore \lambda^2+6=7 \quad \Rightarrow \lambda^2=1 \\ & |\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}|=\left|3-10-\lambda^2\right|=|-8| \end{aligned}

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vishal kumar

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