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  • Explain solution RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 24 maths

Explain solution RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 24 maths

Answers (1)

Answer:

cos^{-1}\frac{11}{21}

Hint:

 cos\theta =\frac{\left | \overrightarrow{n_1}\overrightarrow{n_2} \right |}{\left | \overrightarrow{n_1} \right |.\left | \overrightarrow{n_2} \right |}

Given:

 \overrightarrow{r}.(\widehat{i}-2\widehat{j}-2\widehat{k})=1 \text { and }\overrightarrow{r}.(3\widehat{i}-6\widehat{j}+2\widehat{k})=0

Solution:

\overrightarrow{r}.(\widehat{i}-2\widehat{j}-2\widehat{k})=1 \qquad \qquad \dots(i)

So the vector, normal to the plane is

\overrightarrow{n_1}=(\widehat{i}-2\widehat{j}-2\widehat{k})

\overrightarrow{r}.(3\widehat{i}-6\widehat{j}+2\widehat{k})=0 \qquad \qquad \dots (ii)

So the vector, normal to the plane is

\overrightarrow{n_2}=(3\widehat{i}-6\widehat{j}+2\widehat{k})

Angle between (i) & (ii)

\begin{aligned} &\Rightarrow cos\theta =\frac{\left | \overrightarrow{n_1}\overrightarrow{n_2} \right |}{\left | \overrightarrow{n_1} \right |.\left | \overrightarrow{n_2} \right |}\\ &\Rightarrow cos\theta = \frac{\left | (\widehat{i}-2\widehat{j}-2\widehat{k}).(3\widehat{i}-6\widehat{j}+2\widehat{k}) \right |}{\left | (\widehat{i}-2\widehat{j}-2\widehat{k}) \right |\left | (3\widehat{i}-6\widehat{j}+2\widehat{k}) \right |}\\ &\Rightarrow cos\theta =\frac{3+12-4}{\sqrt{1+4+4}.\sqrt{9+36+4}}\\ &\Rightarrow cos\theta =\frac{11}{3\times 7}\\ &\Rightarrow cos\theta =\frac{11}{21}\\ &\Rightarrow \theta =cos^{-1}\frac{11}{21} \end{aligned}

Posted by

Gurleen Kaur

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