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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.3 question 15

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.3  question 15

Answers (1)

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Answer:

The answer of given question is 24 \hat{\imath}-6 \hat{\jmath}+8 \hat{k}

Hint:

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

Given:

12x-3y+4z=1

Solution:

The vector equation of the plane 12x-3y+4z=1 can be written as

\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(12 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})=1 \\ &\Rightarrow \vec{r} \cdot(12 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})=1 \end{aligned}

The normal to this plane is

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

The unit vector becomes

\begin{aligned} &\vec{n}=\frac{12 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}}{13} \\ &=\frac{12}{13} \hat{\imath}-\frac{3}{13} \hat{\jmath}+\frac{4}{13} \hat{k} \end{aligned}

Now a vector normal to the plane with the magnitude 26 will be

\begin{aligned} &=26 \vec{n} \\ &\Rightarrow 26\left(\frac{12}{13} \hat{\imath}-\frac{3}{13} \hat{\jmath}+\frac{4}{13} \hat{k}\right) \\ &\Rightarrow 24 \hat{\imath}-6 \hat{\jmath}+8 \hat{k} \end{aligned}

A vector of magnitude 26 units normal to the plane is 24 \hat{\imath}-6 \hat{\jmath}+8 \hat{k}

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