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  • Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise multiple choice question 15

Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise multiple choice question 15

Answers (1)

Answer:

 Option (a)

Hint:

 Use straight line & vector cross product.

Given:

 x - 1 = 2y - 5 = 2z and 3x = 4y -11 = 3z - 4

Solution:

The required plane is parallel to the lines

  x - 1 = 2y - 5 = 2z and 3x = 4y - 11 = 3z - 4

Equation of the lines can be written as,

\frac{x-1}{1}=\frac{2y-5}{1}=\frac{2z}{1}

or

\frac{x-1}{1}=\frac{y-\frac{5}{2}}{\frac{1}{2}}=\frac{z}{\frac{1}{2}}

and,

\frac{3x}{1}=\frac{4y-11}{1}=\frac{3z-4}{1}

or,

\frac{x}{\frac{1}{3}}=\frac{y-\frac{11}{4}}{\frac{1}{4}}=\frac{z-\frac{4}{3}}{\frac{1}{3}}=\mu , \text { Let }

So, we know the straight lines as

\begin{aligned} &(1+\lambda )\widehat{i}+\left ( \frac{5}{2}+\frac{1}{2}\lambda \right )\widehat{j}+\frac{1}{2}\widehat{k}=0 \text { or, }\\ &\widehat{i}+\frac{5}{2}\widehat{j}+\lambda \left ( \widehat{i}+\frac{1}{2}\widehat{j}+\frac{1}{2}\widehat{k} \right )=0 \end{aligned}

And,

\begin{aligned} &\frac{1}{3}\mu \widehat{i}+\left ( \frac{11}{4}+\frac{1}{4}\mu \right )\widehat{j}+\left ( \frac{4}{3}+\frac{1}{3}\mu \right )\widehat{k}=0, \text { or }\\ &\frac{11}{4}\widehat{j}+\frac{4}{3}\widehat{k}+\mu \left ( \frac{1}{3}\widehat{i}+\frac{1}{4}\widehat{j}+\frac{1}{3}\widehat{k} \right )=0 \end{aligned}

We have the normal vector of the plane as,

\begin{aligned} &\overrightarrow{n}=\left (\widehat{i}+\frac{1}{2}\widehat{j}+\frac{1}{2}\widehat{k} \right )\times \left ( \frac{1}{3}\widehat{i}+\frac{1}{4}\widehat{j}+\frac{1}{3}\widehat{k} \right )\\ &=\begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ \\ 1 &\frac{1}{2} &\frac{1}{2} \\ \\ \frac{1}{3} &\frac{1}{4} &\frac{1}{3} \end{vmatrix}\\ &=\left ( \left ( \frac{1}{2}\times \frac{1}{3} \right )-\left ( \frac{1}{2}\times \frac{1}{4} \right ) \right )\widehat{i}-\left ( \left ( 1\times \frac{1}{3} \right )-\left ( \frac{1}{2}\times \frac{1}{3} \right ) \right )\widehat{j}+\left ( \left ( 1\times \frac{1}{4} \right )-\left (\frac{1}{2}\times \frac{1}{3} \right ) \right )\widehat{k}\\ &=\left ( \frac{1}{6}-\frac{1}{8} \right )\widehat{i}-\left ( \frac{1}{3}-\frac{1}{6} \right )\widehat{j}+\left ( \frac{1}{4}-\frac{1}{6} \right )\widehat{k}\\ &=\left (\frac{1}{24}\widehat{i}-\frac{1}{6}\widehat{j}+\frac{1}{12}\widehat{k} \right ) \end{aligned}

So, the equation of plane is

\begin{aligned} &(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0\\ &\text { Where } \overrightarrow{a}=2\widehat{i}+3\widehat{j}+3\widehat{k},\\ &\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}\\ &\left ( x\widehat{i}+y\widehat{j}+z\widehat{k} \right ).\overrightarrow{n}=(2\widehat{i}+3\widehat{j}+3\widehat{k}).\overrightarrow{n}\\ &\left ( x\widehat{i}+y\widehat{j}+z\widehat{k} \right ).\left ( \frac{1}{24}\widehat{i}-\frac{1}{6}\widehat{j}+\frac{1}{12}\widehat{k} \right )=(2\widehat{i}+3\widehat{j}+3\widehat{k}).\left ( \frac{1}{24}\widehat{i}-\frac{1}{6}\widehat{j}+\frac{1}{12}\widehat{k} \right ) \end{aligned}

x - 4y + 2z + 4 = 0, (by solving further)

Posted by

Gurleen Kaur

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