The distance of the point (1,3,-7) from the p | vedclass

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(A) The normal vector $\vec{n}$ is perpendicular to the direction vectors of the two lines,$\vec{v_1} = (1, -2, 3)$ and $\vec{v_2} = (2, -1, -1)$.$\ve

Vedclass · Accepted Answer

$\frac{10}{\sqrt{83}}$ units.

Vedclass · Answer

(A) The normal vector $\vec{n}$ is perpendicular to the direction vectors of the two lines,$\vec{v_1} = (1, -2, 3)$ and $\vec{v_2} = (2, -1, -1)$.$\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 2 & -1 & -1 \end{vmatrix} = \hat{i}(2+3) - \hat{j}(-1-6) + \hat{k}(-1+4) = 5\hat{i} + 7\hat{j} + 3\hat{k}$.The equation of the plane passing through $(1, -1, -1)$ with normal vector $(5, 7, 3)$ is $5(x-1) + 7(y+1) + 3(z+1) = 0$,which simplifies to $5x + 7y + 3z + 5 = 0$.The distance $d$ of the point $(1, 3, -7)$ from 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: $d = \frac{|5(1) + 7(3) + 3(-7) + 5|}{\sqrt{5^2 + 7^2 + 3^2}} = \frac{|5 + 21 - 21 + 5|}{\sqrt{25 + 49 + 9}} = \frac{10}{\sqrt{83}}$ units.

The distance of the point $(1,3,-7)$ from the plane passing through the point $(1,-1,-1)$ having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$ is

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