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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.1 question 4 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.1 question 4 maths

Answers (1)

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Answer:

Hence,P externally divides the line segment AB in the ratio 2:1.

Hint:

Use determinant then solve the equation of plane.

Given:

The equation of the plane three points L(2,2,1), M(3,0,1) \text { and } \mathrm{N}(4,-1,0) .

Solution:

The equation of the plane passing through three points can be given by:

L(2,2,1), M(3,0,1) \text { and } \mathrm{N}(4,-1,0) .

\left|\begin{array}{lll} x-2 & y-2 & z-1 \\ x-3 & y-0 & z-1 \\ x-4 & y+1 & z-0 \end{array}\right|=0

Operating R_{2} \rightarrow R_{2}-R_{1}, \text { and } R_{3} \rightarrow R_{3}-R_{1}

\left|\begin{array}{ccc} x-2 & y-2 & z-1 \\ -1 & 2 & 0 \\ -2 & 3 & 1 \end{array}\right|=0

Solving the above determinant, we get

\begin{aligned} &(x-2)[2-0]-(y-2)[-1-0]+(z-1)[-3+4]=0 \\\\ &\Rightarrow(2 x-4)+(y-2)+(z-1)=0 \\\\ &\Rightarrow 2 x+y+z-7=0 \end{aligned}

Therefore, the equation of the plane is 2 x+y+z-7=0

Now, the equation of the line passing through two given points A(3,-4,-5) and B(2,-3,1) is

\begin{aligned} &\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}=\lambda \\\\ &\Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda \end{aligned}

\begin{aligned} &\Rightarrow \frac{x-3}{-1}=\lambda, \frac{y+4}{1}=\lambda, \frac{z+5}{6}=\lambda \\\\ &\Rightarrow x=(-\lambda+3), y=(\lambda-4), z=(6 \lambda-5) \end{aligned}

At the point of intersection, these points satisfy the equation of the plane

2 x+y+z-7=0

Putting the value of x, y and z in the equation of the plane, we get the value of \lambda.

\begin{aligned} &2(-\lambda+3)+(\lambda+4)+(6 \lambda-5)=0 \\\\ &\Rightarrow-2 \lambda+6+\lambda-4+6 \lambda-5-7=0 \\\\ &\Rightarrow 5 \lambda=10 \\\\ &\Rightarrow \lambda=2 \end{aligned}

Thus, the point of intersection is P(1,-2,7).

Now,

Let P divides the line AB in the ratio m:n

By the section formula, we have

Here,A(3,-4,-5) \text { and } \mathrm{B}(2,-3,1) \text { and } P(1,-2,7)

\begin{aligned} &\Rightarrow \frac{2 m+3 n}{m+n}=1 \\\\ &\Rightarrow m+n=2 m+3 n \\\\ &\Rightarrow m+2 n=0 \end{aligned}

\begin{aligned} &\Rightarrow m=-2 n \\\\ &\Rightarrow \frac{m}{n}=\frac{-2}{1} \end{aligned}

Hence, P externally divides the line segment AB in the ratio 2:1.

Posted by

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