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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.3 question 6

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.3  question 6

Answers (1)

Answer:

\vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=2, \quad x+y+z=2

Hint:

(\vec{r}-\vec{a}) \cdot \vec{n}=0

Given:

Plane passing through (2,1,-1)

Solution:

|\vec{n}|=\sqrt{3}

Let \vec{n} vector make \alpha angle with x-axis, β angle with y-axis and \gamma angle with z-axis.

Since

\begin{aligned} &\alpha=\beta=\gamma \\ &\cos \alpha=\cos \beta=\cos \gamma \\ &l=m=n \end{aligned}

\begin{aligned} &\therefore l^{2}+m^{2}+n^{2}=1 \\ &\Rightarrow l^{2}+l^{2}+l^{2}=1 \\ &\Rightarrow 3 l^{2}=1 \\ &\Rightarrow l^{2}=\pm \frac{1}{\sqrt{3}} \end{aligned}

\therefore l=\frac{1}{\sqrt{3}}=m=n                   [ since its make acute angle ]

\begin{aligned} &\hat{n}=\frac{1}{\sqrt{3}} \hat{\imath}+\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k} \\ &\frac{\vec{n}}{|\hat{n}|}=\frac{1}{\sqrt{3}}(\hat{\imath}+\hat{\jmath}+\hat{k}) \end{aligned}

\begin{aligned} &\vec{n}=\frac{\sqrt{3}}{\sqrt{3}}(\hat{\imath}+\hat{\jmath}+\hat{k}) \ldots|\vec{n}|=\sqrt{3} \\ &\vec{n}=(\hat{\imath}+\hat{\jmath}+\hat{k}) \\ &(\vec{r}-\vec{a}) \cdot \vec{n}=0 \end{aligned}

\begin{aligned} &\Rightarrow \vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n} \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=(2 \hat{\imath}+\hat{\jmath}-\hat{k})(\hat{\imath}+\hat{\jmath}+\hat{k}) \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=2+1-1=2 \end{aligned}

For cartesian form,substitute \vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}

\begin{aligned} &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=2 \\ &\Rightarrow x+y+z=2 \end{aligned}

Posted by

infoexpert26

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