The coordinates of the foot of the perpendicu | vedclass

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(C) First,find the normal vector $\vec{n}$ to the plane containing the two lines. The direction vectors of the lines are $\vec{v_1} = (6, 7, 8)$ and $

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$(0, 0, 0)$

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(C) First,find the normal vector $\vec{n}$ to the plane containing the two lines. The direction vectors of the lines are $\vec{v_1} = (6, 7, 8)$ and $\vec{v_2} = (3, 5, 7)$.$\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 6 & 7 & 8 \ 3 & 5 & 7 \end{vmatrix} = \hat{i}(49-40) - \hat{j}(42-24) + \hat{k}(30-21) = (9, -18, 9)$.We can simplify the normal vector to $\vec{n} = (1, -2, 1)$.The plane passes through the point $(-1, 1, 3)$ (from the first line). The equation of the plane is $1(x+1) - 2(y-1) + 1(z-3) = 0$,which simplifies to $x - 2y + z = 0$.Let the foot of the perpendicular from $P(1, -2, 1)$ to the plane be $F(x, y, z)$. The line passing through $P$ and perpendicular to the plane has direction ratios $(1, -2, 1)$.Thus,$\frac{x-1}{1} = \frac{y+2}{-2} = \frac{z-1}{1} = k$.$x = k+1, y = -2k-2, z = k+1$.Since $F$ lies on the plane $x - 2y + z = 0$,we have $(k+1) - 2(-2k-2) + (k+1) = 0$.$k+1 + 4k + 4 + k+1 = 0 \Rightarrow 6k + 6 = 0 \Rightarrow k = -1$.Substituting $k = -1$,we get $x = 0, y = 0, z = 0$.The coordinates of the foot of the perpendicular are $(0, 0, 0)$.

The coordinates of the foot of the perpendicular from the point $(1, -2, 1)$ on the plane containing the lines $\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}$ and $\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}$ is

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