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  • Explain Solution R.D.Sharma Class 12 Chapter 28 The Plane Exercise 28.12 Question 6 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 28 The Plane  Exercise 28.12 Question 6 Maths Textbook Solution.

Answers (1)

Answer: Therefore, the distance is 7 units.

Hint : Use distance formula \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}\left ( z_{2}-z_{1} \right )^{2}}

Given:P\left ( 3,4,4 \right ),A\left ( 3,-4,-5 \right ) &B\left ( 2,-3,1 \right )

and 2x+y+z=7

Solution:

The \: direction \: ratios \: of\: line \: joining\: A(3,-4,-5) \& B(2,-3,1) are

[(2-3),(-3+4),(1+5)]=(-1,1,6)

Now\: \: the \: \: equation \: of \: line \: passing \: through \: (3,-4,-5) \& B(2,-3,1) and\: having DR's (-1,1,6) is

\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}

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

Thus the general point on the line is given by \left ( 3-\lambda ,\lambda -4,6\lambda -5 \right )
Since the intersection of the line and the plane intersects each other,the line lies on the plane
Hence,
2 x+y+z=7

2(-\lambda+3)+(\lambda-4)+(6 \lambda-5)=7 \lambda=2

On \: substituting\: the \: value \: of\: \lambda,value of x, y \& z are

(3-2,2-4,6(2)-5) i.e (1,-2,7)

Distance\: \: between (3,4,4) and (1,-2,7)

=\sqrt{(3-1)^{2}+(4+2)^{2}+(4-7)^{2}}

=\sqrt{4+36+9}

=\sqrt{49}

= 7\: \: units
 

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