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  • Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 13

Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 13

Answers (1)

Answer: \sin ^{-1}\left(\frac{\sqrt{7}}{\sqrt{52}}\right)

Hint:  Use formula \sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}

Given: Line is \frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2} and plane is  3 x+4 y+z+5=0

Solution: We know that line

            \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}   is parallel to plane a_{2} x+b_{2} y+c_{2} z+d_{2}=0

            If  a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0

and the angle between them is given by

            \sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}                        …………….. (1)

Now, given equation by line is

            \frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2}

So, a_{1}=3, b_{1}=-1, c_{1}=2

Equation of plane is  3 x+4 y+z+5=0

So, \begin{aligned} a_{2}=& 3, b_{2}=4, c_{2}=1 \& d_{2}=-5 \\ \end{aligned}

            \sin \theta=\frac{3 \times 3+(-1) \times 4+2 \times 1}{\sqrt{3^{2}+(-1)^{2}+2^{2}} \sqrt{3^{2}+4^{2}+1^{2}}} \\

            \sin \theta =\frac{9-4+2}{\sqrt{9+1+4} \sqrt{9+16+1}}

                      \begin{aligned} &=\frac{7}{\sqrt{14} \sqrt{26}} \times \frac{\sqrt{7}}{\sqrt{7}} \\ & \end{aligned}

                      =\frac{7 \sqrt{7}}{7 \sqrt{52}}=\frac{\sqrt{7}}{\sqrt{52}}

           \sin \theta=\frac{\sqrt{7}}{\sqrt{52}} \Rightarrow \theta=\sin ^{-1}\left(\frac{\sqrt{7}}{\sqrt{52}}\right)

The angle between the plane and the line is  \theta=\sin ^{-1}\left(\frac{\sqrt{7}}{\sqrt{52}}\right)

 

Posted by

infoexpert27

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