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Provide Solution For  R.D. Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 7 Maths Textbook Solution.

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Answer: \frac{x}{2}=\frac{y}{0}=\frac{z}{-1}

Hint: use simultaneous equation;

Given: \frac{x+2}{3}=\frac{y}{-2}=\frac{z-7}{6} \text { and } \frac{x+6}{1}=\frac{x+5}{-3}=\frac{z-1}{-1}

Solution:

          Let \begin{array}{r} \mathrm{L} 1=\frac{2+3}{3}=\frac{y}{-2}=\frac{z-7}{6} \end{array} and

                       \begin{array}{r} { L2 }=\frac{2+6}{1}=\frac{y+5}{-3}=\frac{z-1}{2} \end{array}

Equation of two lines .let the plane be  az+by+cz=0 -------(i)

Given that the required plane through the intersection of the lines L1 and L2 hence the normal to the plane is perpendicular to the line L1 and L2

                             \begin{array}{r} 3 x-3 y+6 z=0 \\ x-3 y+2 z=0 \\ \end{array}

Using cross multiplication, we get

                   \begin{array}{r} \frac{x}{-4+18}=\frac{y}{6-6}=\frac{z}{-9+8} \\ \end{array}

                     \begin{array}{r} \frac{x}{14}=\frac{y}{0}=\frac{z}{-1} \\ \frac{x}{2}=\frac{y}{0}=\frac{z}{-1} \end{array}

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