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  • Provide Solution For R D Sharma Maths Class 12 Chapter 28 The Plane Exercise 28 Point 12 Question 4 Maths Textbook Solution

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.12 Question 4 Maths Textbook Solution.

Answers (1)

Answer: Therefore, the distance is 13 units.

Hint: Use distance formula \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}

Given:

\begin{aligned} &\vec{r}=2 \hat{\imath}-4 \hat{\jmath}+2 \hat{k}+\lambda(3 \hat{\imath}+4 \hat{\jmath}+2 \hat{k}) \text { and } \\ &\vec{r} \cdot(\hat{\imath}-2 \hat{\jmath}+\hat{k})=0 \text { and point }(2,12,5) \end{aligned}

Solution:

\begin{aligned} &\text { General point on the line is }\\ &(2+3 \lambda,-4+4 \lambda, 2+2 \lambda)\\ &\text { The equation of the plane is } \vec{r} \cdot(\hat{\imath}-2 \hat{\jmath}+\hat{k})=0 \end{aligned}

The point of intersection of the line and the plane will be ,
Substituting general point of the line in the equation of plane and finding the perpendicular value of \lambda

[(2+3 \lambda) \hat{\imath}+(-4+4 \lambda) \hat{\jmath}+(2+2 \lambda) \hat{k}](\hat{\imath}-2 \hat{\jmath}+\hat{k})=0

[(2+3 \lambda) 1+(-4+4 \lambda)(-2)+(2+2 \lambda) 1]=0

12-3 \lambda=0 \Rightarrow \lambda=4\therefore The \, point \: of \: intersection\: is (14,12,10)

Distance \: of \: this\: point \: from (2,12,5) is

=\sqrt{(14-2)^{2}+(12-12)^{2}+(10-5)^{2}}

=\sqrt{12^{2}+5^{2}}

=\sqrt{144+25}









 

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