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  • Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 26

Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 26

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Answer: Required equation of plane is  x-y+z=0  and angle  \sin \theta=\frac{-1}{\sqrt{3}}

Hint: Use formula  \sin \theta=\frac{\vec{n} \cdot \vec{b}}{|\vec{n}||\vec{b}|}

Given: Point  (2,1,-1) \text { and } 2 x+y-z=3 \text { and } x+2 y+z=2

Solution:

Line of intersection  \mathrm{n}_{1} \times \mathrm{n}_{2}

                \vec{A} \vec{R} \cdot\left[n_{1} \times n_{2}\right]=0

Scalar triple product, \left[\begin{array}{lll} \overrightarrow{A R} & n_{1} & n_{2} \end{array}\right]

\overrightarrow{A R}= Position Vector of \overrightarrow{ R} – Position Vector of  \overrightarrow{ A}

        \begin{aligned} &=(x-2) \hat{i}+(y-1) \hat{j}+(z+1) \hat{k} \\ & \end{aligned}

        \left|\begin{array}{ccc} x-2 & y-1 & z+1 \\ 2 & 1 & -1 \\ 1 & 2 & 1 \end{array}\right|=0

        \begin{aligned} &(x-2)(1+2)-(y-1)(2+1)+(z+1)(4-1)=0 \\ & \end{aligned}

        x-2-y+1+z+1=0

        x-y+z=0  is required equation of plane

Now, the angle between the plane and the y-axis is  \sin \theta=\frac{\vec{n} \cdot \vec{b}}{|\vec{n}||\vec{b}|}

\begin{aligned} &\sin \theta=\frac{-1}{\sqrt{1+1+1} \sqrt{0+1+0}} \\ & \end{aligned}

\sin \theta=\frac{-1}{\sqrt{3}} \\                                      \left[\sin \theta=\frac{\vec{n} \cdot \vec{b}}{|\vec{n}||\vec{b}|}\right]

\theta=\sin ^{-1} \frac{-1}{\sqrt{3}}

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