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  • Provide solution for RD Sharma maths class 12 chapter 28 The Plane exercise multiple choice question 14

Provide solution for RD Sharma maths class 12 chapter 28 The Plane exercise multiple choice question 14

Answers (1)

Answer:

 Option (c)

Hint:

 Calculate the perpendicular distance of the plane from the origin.

Given:

 \frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}

Solution:

Let, the equation of the plane be Ax + By + Cz + D = 0, as the plane is perpendicular to, so

We have,

A = 3, B = 0  and C = 4 as the plane passes through (1, 1, 1), we have,

\begin{aligned} &(A\times 1)+(B\times 1)+(C\times 1)+D=0\\ &A+B+C+D=0\\ &3+0+4+D=0\\ &D=-7 \end{aligned}

So, the equation of the plane becomes

3x + 4z - 7 = 0

Now, the perpendicular distance of the plane from the origin is

\begin{aligned} &\frac{\left | Ax_0+By_0+Cz_0+D \right |}{\sqrt{A^2+B^2+C^2}}=\frac{\left | (3\times 0)+(0\times 0)+(4\times 0)-7 \right |}{\sqrt{3^2+0^2+4^2}}\\ &\left [ \because (x_0, y_0,z_0)\approx (0,0,0) \right ]\\ &=\frac{\left | 0-7 \right |}{\sqrt{9+16}}\\ &=\frac{\left | -7 \right |}{\sqrt{25}}\\ &=\frac{7}{5} \end{aligned}

Posted by

Gurleen Kaur

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