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  • Please solve RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 17 maths textbook solution

Please solve RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 17 maths textbook solution

Answers (1)

Answer:

 45°

Hint:

 cos(90-\theta )=\frac{\left | \overrightarrow{n}.\overrightarrow{b} \right |}{\left | \overrightarrow{n} \right |.\left | \overrightarrow{b} \right |}

Given:

 \frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2} \text { and } x+y+4=0

Solution:

x+y+4=0 \qquad \qquad \qquad \dots(i)

So the vector normal to the plane (i) is

\overrightarrow{n}=\overrightarrow{i}+\overrightarrow{j}

Cartesian equation of plane is

\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2} \qquad \qquad \qquad \dots(ii)

\overrightarrow{b}=2\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k}

Since

cos(90-\theta )=\frac{\left | \overrightarrow{n}.\overrightarrow{b} \right |}{\left | \overrightarrow{n} \right |.\left | \overrightarrow{b} \right |}

\begin{aligned} &cos(90-\theta )=\frac{2+1}{\sqrt{1+1}.\sqrt{4+1+4}}\\ &cos(90-\theta )=\frac{3}{\sqrt{2}.3}\\ &cos(90-\theta )=\frac{1}{\sqrt{2}}\\ &\Rightarrow sin\theta =\frac{1}{\sqrt{2}}\\ &\Rightarrow \theta =45^{\circ} \end{aligned}

 

Posted by

Gurleen Kaur

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