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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.7 question 1 subquestion (iii)

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.7 question 1 subquestion (iii)

Answers (1)

Answer:

 Required vector equation is

\vec{r}.(-\hat{i}+\hat{j}+\hat{k})=0

Hint:

 Using scalar form

\vec{n}=\vec{b}.\vec{n}

Given:

 \vec{r}=(\hat{i}+\hat{j})+\lambda (\hat{i}+2\hat{j}-\hat{k})+\mu (-\hat{i}+2\hat{j}-2\hat{k})

Solution:

We know that the equation

\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}

represents a plane passing through a point whose vector is \vec{a}and parallel to the vector \vec{b} and \vec{c}

Here

\begin{aligned} &\vec{a}=\hat{i}+\hat{j}+0\hat{k}\\ &\vec{b}=\hat{i}+2\hat{j}-\hat{k}\\ &\vec{c}=-\hat{i}+2\hat{j}-2\hat{k} \end{aligned}

Normal vector

\begin{aligned} &\vec{n}=\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &2 &-1 \\ -1 &1 &-2 \end{vmatrix}\\ &=-3\hat{i}+3\hat{j}+3\hat{k} \end{aligned}

The vector equation of the plane in scalar product form is

\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n} \end{aligned}

\begin{aligned} &\Rightarrow \vec{r}\: (-3\hat{i}+3\hat{j}+3\hat{k})= (\hat{i}+\hat{j}+0\hat{k}). (-3\hat{i}+3\hat{j}+3\hat{k})\\ &\Rightarrow \vec{r}\: (-3\hat{i}+3\hat{j}+3\hat{k})=-3+3\\ &\Rightarrow \vec{r}\: [3(-\hat{i}+\hat{j}+\hat{k})]=0\\ &\Rightarrow \vec{r}\: (-\hat{i}+\hat{j}+\hat{k})=0 \end{aligned}

 

 

 

Posted by

Gurleen Kaur

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