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  • Explain solution RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 12 maths

Explain solution RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 12 maths

Answers (1)

Answer:

 Option (c)

Hint:

 The plane, must satisfy both the equation.

Given:

 \frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}

Solution:

Let, the point of intersection of the line

\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}

And the plane x + y + z = 17 be (x0, y0, z0) So, (x0, y0, z0) satisfy the both equations,

\begin{aligned} &\frac{x_0-3}{1}=\frac{y_0-4}{2}=\frac{z_0-5}{2}=k, \text { (Let) }\\ &\text { i.e. }x_0=k+3\\ &y_0=2k+4\\ &z_0=2k+5 \end{aligned}

Put these values in the equation of plane x + y + z = 17

\begin{aligned} &(k+3)+(2k+4)+(2k+5)=17\\ &5k+12=17\\ &k=1\\ &\therefore x_0=1+3\\ &=4\\ &y_0=2\times 1+4\\ &=6\\ &z_0=2\times 1+5\\ &=7\\ \end{aligned}

Hence, the distance between point (4, 6, 7) and (3, 4, 5) is,

\begin{aligned} &=\sqrt{(4-3)^2+(6-4)^2+(7-5)^2}\\ &=\sqrt{1^2+2^2+2^2}\\ &=\sqrt{1+4+4}\\ &=3 \end{aligned}

Posted by

Gurleen Kaur

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