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  • Please solve RD Sharma class 12 chapter 28 The Plane exercise Fill in the blank question 17 maths textbook solution

Please solve RD Sharma class 12 chapter 28 The Plane exercise Fill in the blank question 17 maths textbook solution

Answers (1)

Answer:

 sin^{-1}\left [ \frac{9}{\sqrt{156}} \right ]

is the angle between the line and the plane

Hint:

 Find \theta using vector dot product

Given:

 \vec{r}=\left ( 5\hat{i}-\hat{j}-4\hat{k} \right )+\lambda \left ( 2\hat{i}-\hat{j}-\hat{k} \right ) \text { and the plane }\vec{r}. \left ( 3\hat{i}-4\hat{j}-\hat{k} \right )+5=0

Solution:

Line

\vec{r}=\left ( 5\hat{i}-\hat{j}-4\hat{k} \right )+\lambda \left ( 2\hat{i}-\hat{j}-\hat{k} \right )

Is parallel to the vector

\vec{b}= 2\hat{i}-\hat{j}+\hat{k}

Normal to the plane is

\vec{n}= 3\hat{i}-4\hat{j}-\hat{k}

Let \theta is the angle between line and plane

Then,

sin\theta =\frac{\left | \vec{b}.\vec{n} \right |}{\left | \vec{b} \right |\left | \vec{n} \right |}\Rightarrow \frac{\left | \left ( 2\hat{i}-\hat{j}+\hat{k} \right ).\left ( 3\hat{i}-4\hat{j}+\hat{k} \right ) \right |}{\sqrt{6}.\sqrt{26}}

=\frac{\left | 6+4-1 \right |}{\sqrt{156}}=\frac{9}{\sqrt{156}}

\theta = sin^{-1}\left [ \frac{9}{\sqrt{156}} \right ]

Posted by

Gurleen Kaur

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