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  • Provide solution for RD Sharma maths Class 12 Chapter 28 The Plane Exercise 28 .9 Question 1 textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 28 The Plane Exercise 28.9 Question 1 textbook solution.

Answers (1)

Answer : \frac{47}{13}

Hint : \text { Distance }=\frac{\left|A x_{1}+B y_{1}+C z_{1}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}

Given : Point  (2 \hat{i}-\hat{j}-4 \hat{k}) from the plane vector \vec{r} \cdot(3 \hat{i}-4 \hat{j}+12 \hat{k})=9

Solution :

For the given plane

Vector, \vec{r} \cdot(3 \hat{i}-4 \hat{j}+12 \hat{k})=9

Cartesian form is

3 x-4 y+12 z=9

We can write as,

3 x-4 y+12 z-9=0

The point given is (2 \hat{i}-\hat{j}-4 \hat{k})

Which can be written as (2,-1,-4)

We know that

\begin{aligned} &\text { Distance }=\frac{\left|A x_{1}+B y_{1}+C z_{1}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}} \\ &\text { Distance }=\frac{|(2 \times 3)+(-1 \times-4)+(-4 \times 12)+(-9)|}{\sqrt{3^{2}+(-4)^{2}+12^{2}}} \end{aligned}

So we get,

=\frac{|6+4-48-9|}{\sqrt{9+16+144}}

On further calculation

\begin{aligned} &=\frac{|-47|}{\sqrt{169}} \\ \end{aligned}

\begin{aligned} &=\frac{47}{13} \text { Units } \end{aligned}

Posted by

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