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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 6

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 6

Answers (1)

Answer:

 \frac{12}{13}x-\frac{3}{13}y+\frac{4}{13}z=5

Hint:

 You must know the rules of solving vector functions

Given:

The direction ratios of perpendicular from the origin to a plane are 12,-3,4 and the length of perpendicular is 5. Find the equation of plane. 

Solution:

We have

Direction ratio are 12,-3,4 of vector \vec{n}

\begin{aligned} &\vec{n}=12\hat{i}-3\hat{j}+4\hat{k}\\ &\left | \vec{n} \right |=\sqrt{(12)^{2}+(-3)^2+(4)^2}\\ &=\sqrt{144+9+16}=\sqrt{169}=13 \end{aligned}

Now,

\begin{aligned} &\hat{n}=\frac{\vec{n}}{\left | \vec{n} \right |}=\frac{12\hat{i}-3\hat{j}+4\hat{k}}{13}\\ &\frac{12}{13}\hat{i}-\frac{3}{13}\hat{j}+\frac{4}{13}\hat{k} \end{aligned}

Also we have,
length of the perpendicular from the origin d=5
∴Equation of plane in normal form is
\begin{aligned} &\vec{r}.\hat{n}=d \end{aligned}

[d= distance of plane from origin]

\begin{aligned} &\vec{r}.\left ( \frac{12}{13}\hat{i}-\frac{3}{13}\hat{j}+\frac{4}{13}\hat{k} \right )=5 \end{aligned}

Normal form is

\frac{12}{13}x-\frac{3}{13}y+\frac{4}{13}z=5

Posted by

Gurleen Kaur

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