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  • Explain solution RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 8 maths

Explain solution RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 8 maths

Answers (1)

Answer:

 Option (b)

Hint:

 Use scalar product of vector

Given:

 \overrightarrow{r}=2\widehat{i}-2\widehat{j}+3\widehat{k}+\lambda (\widehat{i}+\widehat{j}+4\widehat{k})

from the plane

\overrightarrow{r}.(\widehat{i}+5\widehat{j}+\widehat{k})=5

Solution:

We have the straight line given as,

\overrightarrow{r}=2\widehat{i}-2\widehat{j}+3\widehat{k}+\lambda (\widehat{i}+\widehat{j}+4\widehat{k})

And the plane as,

\begin{aligned} &\overrightarrow{r}.(\widehat{i}+5\widehat{j}+\widehat{k})=5\\ &x-5y+2=5\\ &\text { i.e. }x-5y+2-5=0 \end{aligned}

the normal vector of the plane given as,

\overrightarrow{n}=(\widehat{i}-5\widehat{j}+\widehat{k})

if the straight line and the plane are parallel,

scalar product will be zero.

\begin{aligned} &(\widehat{i}+\widehat{j}+4\widehat{k}).(\widehat{i}-5\widehat{j}+\widehat{k})=1+[1\times (-5)]+(4\times 1)\\ &=1-5+4\\ &=0 \end{aligned}

Hence, the point (2, -2, 3) is on the straight line. Distance from point (2, -2, 3) to the plane will be equal to the distance of the line from the plane. We know, that the distance of a point (x0, y0, z0) from plane Ax + By + Cz + D = 0 ...(2) is

\begin{aligned} &=\frac{\left | Ax_0+By_0+Cz_0+D \right |}{\sqrt{A^2+B^2+C^2}} \end{aligned}

On comparing, equation (1), i.e., x - 5y + z - 5 = 0 with equation (2), we get,

A = 1, B = -5, C = 1, D = -5

So, the distance is,

\begin{aligned} &=\frac{\left [ (1\times 2)+(-5\times -2)+(1\times 3)+(-5) \right ]}{\sqrt{1^2+(-5)^2+1^2}}\\ &=\frac{\left | 2+10+3-5 \right |}{\sqrt{1+25+1}}\\ &=\frac{10}{\sqrt{27}}\\ &=\frac{10}{3\sqrt{3}} \end{aligned}

Posted by

Gurleen Kaur

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