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Question

(A) The equation of the plane containing the two lines is given by the determinant form: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ a_1 & b_1 & c_1 \

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$\frac{11}{\sqrt{6}}$ units

Vedclass · Answer

(A) The equation of the plane containing the two lines is given by the determinant form: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$ Using the point $(1, 4, -4)$ from the second line and the direction vectors $(3, 5, 7)$ and $(1, 4, 7)$: $\begin{vmatrix} x-1 & y-4 & z+4 \ 3 & 5 & 7 \ 1 & 4 & 7 \end{vmatrix} = 0$ Expanding the determinant: $(x-1)(35-28) - (y-4)(21-7) + (z+4)(12-5) = 0$ $7(x-1) - 14(y-4) + 7(z+4) = 0$ Dividing by $7$: $(x-1) - 2(y-4) + (z+4) = 0$ $x - 2y + z - 1 + 8 + 4 = 0$ $x - 2y + z + 11 = 0$ The perpendicular distance from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. $d = \frac{|11|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{11}{\sqrt{1+4+1}} = \frac{11}{\sqrt{6}}$ units.

The perpendicular distance from the origin to the plane containing the two lines $\frac{x+2}{3}=\frac{y-2}{5}=\frac{z+5}{7}$ and $\frac{x-1}{1}=\frac{y-4}{4}=\frac{z+4}{7}$ is:

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