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  • Let a unit vector  make angles <img alt="\alpha ,\beta, \ga

Let a unit vector \widehat{\mathrm{OP}} make angles \alpha ,\beta, \gamma with the positive directions of the co-ordinate axes OX, OY, OZ respectively, where \beta \in\left(0, \frac{\pi}{2}\right) . If \widehat{\mathrm{OP}}  is perpendicular to the plane through points (1,2,3), (2,3,4) and (1,5,7), then which one of the

following is true ? 

Option: 1

\alpha \in\left(0, \frac{\pi}{2}\right)$ and $\gamma \in\left(0, \frac{\pi}{2}\right)


Option: 2

\alpha \in\left(0, \frac{\pi}{2}\right)$ and $\gamma \in\left(\frac{\pi}{2}, \pi\right)


Option: 3

\alpha \in\left(\frac{\pi}{2}, \pi\right)$ and $\gamma \in\left(\frac{\pi}{2}, \pi\right)


Option: 4

\alpha \in\left(\frac{\pi}{2}, \pi\right)$ and $\gamma \in\left(0, \frac{\pi}{2}\right)


Answers (1)

best_answer

\because \overrightarrow{\mathrm{OP}} makes angle \alpha, \beta, \gamma with positive directions of the co-ordinate axes then \cos ^2 \alpha+\cos ^2 \beta+\cos ^2\gamma=1 . 

Point on planes are \mathrm{a}(1,2,3), \mathrm{b}(2,3,4) and c(1,5,7)

\begin{aligned} & \because \overrightarrow{a b}=\langle 1,1,1\rangle \\ & \overrightarrow{a c}=\langle 0,3,4\rangle \end{aligned}

normal vector of plane =\left[\begin{array}{lll} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ 0 & 3 & 4 \end{array}\right]

                                   \begin{aligned} & =\hat{\mathrm{i}}(1)-\hat{j}(4)+\hat{\mathrm{k}}(3) \\ & =\langle 1,-4,3\rangle \end{aligned}

direction cosine of normal is =\left\langle \pm \frac{1}{\sqrt{26}}, \pm \frac{4}{\sqrt{26}}, \pm \frac{3}{\sqrt{26}}\right\rangle

then direction cosine of op is \begin{aligned} & \left(-\frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}},-\frac{3}{\sqrt{26}}\right) \\ & \left(\because \beta \in\left(0, \frac{\pi}{2}\right)\right) \end{aligned}

Hence, \alpha \in\left(\frac{\pi}{2}, \pi\right) \text { and } \gamma \in\left(\frac{\pi}{2}, \pi\right)

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vinayak

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