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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 7

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 7

Answers (1)

Answer:

 \frac{1}{\sqrt{14}}\hat{i}+\frac{2}{\sqrt{14}}\hat{j}+\frac{3}{\sqrt{14}}\hat{k}

Hint:

 You must know the rules of solving vector functions

Given:

 Find a unit normal vector to the plane x + 2y + 3z - 6 = 0

Solution:

We have

\begin{aligned} &x + 2y + 3z - 6 = 0\\ &x + 2y + 3z =6 \end{aligned}

Equation of plane in normal form is

\begin{aligned} &\vec{r}.\hat{n}=d\\ &\therefore \vec{r}.(\hat{i}+2\hat{j}+3\hat{k})=6\\ &\vec{r}.\hat{n}=d\\ &\text { So, }\vec{n}=\hat{i}+2\hat{j}+3\hat{k}\\ &\left | \vec{n} \right |=\sqrt{(1)^2+(2)^2+(3)^2}\\ &=\sqrt{1+4+9}\\ &=\sqrt{14} \end{aligned}

Unit vector to the plane ,

\begin{aligned} &\hat{n}=\frac{\vec{n}}{\left | \vec{n} \right |}\\ &=\frac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{14}}\\ &\hat{n}=\frac{1}{\sqrt{14}}\hat{i}+\frac{2}{\sqrt{14}}\hat{j}+\frac{3}{\sqrt{14}}\hat{k} \end{aligned}

Posted by

Gurleen Kaur

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