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  • Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 2 sub question 4 maths textbook solution

Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 2 sub question 4 maths textbook solution

Answers (1)

Answer:

              \cos ^{-1}\left ( -\frac{\sqrt{5}}{\sqrt{58}} \right )

Hint:

              \cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1} 2}^{2} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2} 2}}

Given:

              2x-3y+4z=1 \hspace{0.2cm}And \hspace{0.2cm} -x+y=4

Solution:

Here,

\begin{aligned} &a_{1}=2, b_{1}=-3, c_{1}=4, \text { on comparing with } 2 x-3 y+4 z=1 \\ &a_{2}=-1, b_{2}=1, c_{2}=0, \text { on comparing with }-x+y=4 \\ &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &=\frac{(2)(-1)+(-3)(1)+(4)(0)}{\sqrt{2^{2}+(-3)^{2}+\left(4^{2}\right)} \cdot \sqrt{(-1)^{2}+(1)^{2}}} \\ &=\frac{-5}{\sqrt{29} \times \sqrt{2}} \\ &=-\frac{\sqrt{5}}{\sqrt{58}} \\ & \end{aligned}

\theta=\cos ^{-1}\left(-\frac{\sqrt{5}}{\sqrt{58}}\right)

 

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