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  • Show that the points hat{i}-hat{j}+3hat{k} and 3( hat{i}+hat{j}+hat{k}) are equidistant from the plane r.( 5hat{i}+2hat{j}-7hat{k})+9=0and lies on the opposite of it.

Show that the points \hat{i}-\hat{j}+3\hat{k}  and 3\left ( \hat{i}+\hat{j}+\hat{k} \right ) are equidistant from the plane r.\left ( 5\hat{i}+2\hat{j}-7\hat{k} \right )+9=0and lies on the opposite of it.

Answers (1)

Given two points,

\\\vec{A}=\hat{i}-\hat{j}+3\hat{k}\\ \vec{B}=3\left ( \hat{i}+\hat{j}+\hat{k} \right )=3\hat{i}+3\hat{j}+\hat{k}\\ \vec{r}.\left ( 5\hat{i}+2\hat{j}-7\hat{k} \right )+9=0

Also,

\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\\\\

Where,

Therefore,

\left (x\hat{i}+y\hat{j}+z\hat{k} \right ).\left (5\hat{i}+2\hat{j}-7\hat{k} \right )+9=0\\ \\ \Rightarrow 5x + 2y - 7z + 9 = 0

We must show that the points A and B are equidistant from the plane

5x + 2y - 7z + 9 = 0

We also need to show that the points lie on the opposite side of the plane.

Normal of the plane is, \vec{N=} 5{i} + 2\hat{j} - 7\hat{k}

We know, the perpendicular distance of the position vector of a point

\vec{A} = l\hat{i} + m\hat{j} + n\hat{k} \Rightarrow A(l, m, n) to the plane, p: ax + by + cz + d =  0 is given as:

D=\left | \frac{p(l,m,n)}{\left |\vec{N} \right |} \right |

Where  \left |\vec{N} \right |=Normal \: vector\: of\: the\: plane

\vec{N} =a\vec{i}+b\vec{j}+c\vec{k}

Thus, the perpendicular distance of the point \vec{A} =\vec{i}-\vec{j}+3\vec{k}=A(1,-1,3)  to the plane 5x + 2y - 7z + 9 = 0 having normal  \vec{N} =5\vec{i}+2\vec{j}-7\vec{k} is given by,

\left | D_{1} \right |=\left |\frac{5(1)+2(-1)-7(3)+9}{|5\hat{i}+2\hat{j}-7\hat{k}|} \right |\\ \Rightarrow \left | D_{1} \right |=\left |\frac{5-2-21+9}{\sqrt{5^{2}+2^{2}+(-7)^{2}}} \right |\\ \Rightarrow \left | D_{1} \right |=\left |\frac{-9}{\sqrt{25+4+49}} \right |\\ \Rightarrow \left | D_{1} \right |=\left |\frac{9}{\sqrt{78}} \right |\\

Hence, the perpendicular distance of the point \vec{B}=3\hat{i}+3\hat{j}+3\hat{k}=B(3,3,3) to the plane 5x + 2y - 7z + 9 = 0 having normal  \vec{N}=5\hat{i}+2\hat{j}-7\hat{k}

\left | D_{2} \right |=\left |\frac{5(3)+2(3)-7(3)+9}{|5\hat{i}+2\hat{j}-7\hat{k}|} \right |\\ \Rightarrow \left | D_{2} \right |=\left |\frac{15+6-21+9}{\sqrt{5^{2}+2^{2}+(-7)^{2}}} \right |\\ \Rightarrow \left | D_{2} \right |=\left |\frac{9}{\sqrt{25+4+49}} \right |\\ \Rightarrow \left | D_{2} \right |=\left |\frac{9}{\sqrt{78}} \right |\\

Therefore, |D1| = |D2|

However, D1 and D2 have different signs.

Therefore, the points A and B will lie on opposite sides of the plane.

Hence, we have successfully shown that the points are equidistant from the plane and lie on opposite sides of the plane.

Posted by

infoexpert24

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