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Please solve RD Sharma class 12 chapter The Plane exercise 28.3  question 1 maths textbook solution

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Answer:

\hat{r}(4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k})=3

Hint:

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

Given:

Plane passing through the point having the position vector 2 \hat{\imath}-\hat{\jmath}+\hat{k} and perpendicular to vector 4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}

Solution:

\begin{aligned} &\vec{a}=2 \hat{\imath}-\hat{\jmath}+\hat{k} \\ &\vec{n}=4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k} \end{aligned}

Vector equation of the plane passing through point a and normal to n is ,

\begin{aligned} &(\vec{r}-\vec{a}) \cdot \vec{n}=0 \\ &\Rightarrow \vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n} \end{aligned}

\begin{aligned} &\Rightarrow \vec{r} \cdot(4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k})=(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \cdot(4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \\ &=\hat{r}(4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k})=(8-2-3)=3 \\ &=\hat{r}(4 \hat{\imath}+2 \hat{\jmath}-3 \hat{k})=3 \end{aligned}

 

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