Find the distance between the point P(6,5,9) | vedclass

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(A) First,find the equation of the plane passing through $A(3,-1,2)$,$B(5,2,4)$,and $C(-1,-1,6)$.The vectors $\overline{AB} = (5-3)\hat{i} + (2-(-1))\

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$\frac{3 \sqrt{34}}{17}$

Vedclass · Answer

(A) First,find the equation of the plane passing through $A(3,-1,2)$,$B(5,2,4)$,and $C(-1,-1,6)$.The vectors $\overline{AB} = (5-3)\hat{i} + (2-(-1))\hat{j} + (4-2)\hat{k} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ and $\overline{AC} = (-1-3)\hat{i} + (-1-(-1))\hat{j} + (6-2)\hat{k} = -4\hat{i} + 0\hat{j} + 4\hat{k}$.The normal vector to the plane is $\vec{n} = \overline{AB} 	imes \overline{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 2 \ -4 & 0 & 4 \end{vmatrix} = \hat{i}(12-0) - \hat{j}(8 - (-8)) + \hat{k}(0 - (-12)) = 12\hat{i} - 16\hat{j} + 12\hat{k}$.Simplifying the normal vector by dividing by $4$,we get $\vec{n}' = 3\hat{i} - 4\hat{j} + 3\hat{k}$.The equation of the plane is $3(x-3) - 4(y+1) + 3(z-2) = 0$,which simplifies to $3x - 4y + 3z - 19 = 0$.The distance $d$ from point $P(6,5,9)$ to the plane $ax+by+cz+d=0$ is given by $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$.Substituting the values: $d = \frac{|3(6) - 4(5) + 3(9) - 19|}{\sqrt{3^2 + (-4)^2 + 3^2}} = \frac{|18 - 20 + 27 - 19|}{\sqrt{9 + 16 + 9}} = \frac{|6|}{\sqrt{34}} = \frac{6}{\sqrt{34}} = \frac{6\sqrt{34}}{34} = \frac{3\sqrt{34}}{17}$.

Find the distance between the point $P(6,5,9)$ and the plane determined by points $A(3,-1,2)$,$B(5,2,4)$,and $C(-1,-1,6)$.

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