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Provide solution for RD Sharma maths class 12 chapter 28 The Plane exercise very short answer type question 22

Answers (1)

Answer:

 \overrightarrow{r}.(2\widehat{i}-3\widehat{j}+6\widehat{k})=35

Hint:

 we will use equation of plane as

\overrightarrow{r}.\widehat{n}=\overrightarrow{d}

Given:

 Normal vector of plane is

2\widehat{i}-3\widehat{j}+6\widehat{k}

Solution:

\overrightarrow{n}=2\widehat{i}-3\widehat{j}+6\widehat{k}

and distance from the origin =5 units

We know that

\overrightarrow{r}.\widehat{n}=\overrightarrow{d}

\begin{aligned} &\Rightarrow \overrightarrow{r}.\frac{(2\widehat{i}-3\widehat{j}+6\widehat{k})}{\sqrt{4+9+36}}=5\\ &\Rightarrow \overrightarrow{r}.(2\widehat{i}-3\widehat{j}+6\widehat{k})=5\times \sqrt{49}\\ &\Rightarrow \overrightarrow{r}.(2\widehat{i}-3\widehat{j}+6\widehat{k})=5\times 7\\ &\Rightarrow \overrightarrow{r}.(2\widehat{i}-3\widehat{j}+6\widehat{k})=35 \end{aligned}

Posted by

Gurleen Kaur

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