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  • Please Solve R.D. Sharma class 12 Chapter 28 The Plane Exercise 28.13 Question 10 Sub Question 1 Maths Textbook Solution.

Please Solve R.D. Sharma class 12 Chapter 28 The Plane  Exercise 28.13 Question 10 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer:\theta =\sin ^{-1}\left ( \frac{1}{\sqrt{3}\sqrt{29}} \right )

Hint: use vector dot product

Since \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}

Solution :any point on the line \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=k is of the form \left ( 3k+2,4k-1,2k+2 \right )

if the point p\left ( 2k+2,4k-1,2k+2 \right ) lies in the plane

                    \begin{aligned} &x-y+z-5=0 \\ &(3 k+z)-(4 k-1)+(2 k+2)-5=0 \\ &3 k+2-4 k+1+2 k-5=0 \\ &k=0 \\ \end{aligned}

Thus, the coordinates of the point of intersection of the line and the planes are

                    \begin{aligned} &8(0)+2,4(0)-1,2(0)+2 \\ &p(2,-1,2) \\ \end{aligned}

Let 0 be the angle between the line and the plane thus

                   \begin{aligned} &\sin \theta=\frac{\mathrm{xl}+\mathrm{ym}+\mathrm{zn}}{\sqrt{\left(x^{2}+y^{2}+z^{2}\right)} \sqrt{l^{2}++n^{2}}} \end{aligned}

Where l, m, and n are the direction ratio of the line and x, y and z are the direction ratios of the normal to the plane

                    

 

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