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Please solve RD Sharma class 12 chapter The Plane exercise 28.5 question 1 maths textbook solution

Answers (1)

Answer:

 The required vector equation of plane is

\vec{r}.(3\hat{i}-4\hat{k})+1=0

Hint:

 Using

\vec{r}.\vec{n}=\vec{a}.\vec{n}

Given:

 Vector equation of the plane passing through the points (1,1,1), (1,-1,1) and (-7,-3,-5)

Solution:

Let A (1,1,1), (1,-1,1) and (-7,-3,-5) be the coordinates.

The required plane passes through the point A (1, 1, 1) whose vector is

\vec{a}=\hat{i}+\hat{j}+\hat{k}

and is normal to the vector \vec{n} given by

\vec{n}=\vec{AB}\times \vec{AC} clearly.

\begin{aligned} &\vec{AB}=\vec{OB}-\vec{OA}=(\hat{i}-\hat{j}+\hat{k})-(\hat{i}+\hat{j}+\hat{k})\\ &=0\hat{i}-2\hat{j}+0\hat{k}\\ &\vec{AC}=\vec{OC}-\vec{OA}=(-7\hat{i}-3\hat{j}-5\hat{k})-(\hat{i}+\hat{j}+\hat{k})\\ &=-8\hat{i}-4\hat{j}+6\hat{k}\\ &\vec{n}=\vec{AB}-\vec{AC}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 0 &-2 &0 \\ -8 &-4 &-6 \end{vmatrix}=12\hat{i}+0\hat{j}-16\hat{k} \end{aligned}

The vector equation of required plane is: -

\begin{aligned} &\vec{r}.(12\hat{i}+0\hat{j}-16\hat{k})=(\hat{i}+\hat{j}+\hat{k}).(12\hat{i}+0\hat{j}-16\hat{k})\\ &\vec{r}[4(3\hat{i}-4\hat{k})]=12+0-16\\ &\vec{r}[4(3\hat{i}-4\hat{k})]=-4\\ &\vec{r}[(3\hat{i}-4\hat{k})]=-1\\ &\vec{r}[(3\hat{i}-4\hat{k})]+1=0 \end{aligned}

Posted by

Gurleen Kaur

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