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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28 .12 Question 8 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28 .12 Question 8 Maths Textbook Solution.

Answers (1)

Answer: Therefore, required point (7,10,0)

Hint :Use distance formula \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

Given:\frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2}

Solution:

If \: line \frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2} cuts \: the XY-plane

Then z=0

So, let \: coordinate\: of \: point \: be (x, y, 0)

\frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2}=k(s a y)

\Rightarrow x=3 k+1, y=7 k-4, z=2 k-4

Since\: z=0

\Rightarrow 2 k-4=0

\Rightarrow 2 k=4 \Rightarrow k=2

Now, x=7, y=10, z=0

Therefore, required point (7,10,0)

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