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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 9 Sub Question 1 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 9 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer: -22x+19y+5z=31

Hint: use vector cross product.

Given: \begin{aligned} &\frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2} \text { and } \\ &\frac{x-1}{x}=\frac{y-2}{1}=\frac{z-3}{5} \end{aligned}

Solution: the direction ratio of the line \frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2} \text { is } \pi=(-3,-2 k, 2)

The direction ratio of the line \frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5} \text { is } \pi2=( k,1, 5)

Since the line \frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2} and   \frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}are perpendicular so      

                \begin{aligned} &\pi_{1} \cdot \pi_{2}=0 \\ &(-3,-2 k, 2) \cdot(k, 1,5)=0 \\ &-3 k-2 k+10=0 \\ &-5 k=-10 \\ &k=2 \end{aligned}

The equation of the line are \frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2}and \frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}

The equation of the plane containing the line perpendicular lines \frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2}and \frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}

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