If A(3,-1,11),B(0,2,3),and C(4,8,11) are thre | vedclass

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(C) Given points are $A(3,-1,11)$,$B(0,2,3)$,and $C(4,8,11)$.First,find the equation of the line $BC$ passing through $B(0,2,3)$ and $C(4,8,11)$.The d

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$(2,5,7)$

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(C) Given points are $A(3,-1,11)$,$B(0,2,3)$,and $C(4,8,11)$.First,find the equation of the line $BC$ passing through $B(0,2,3)$ and $C(4,8,11)$.The direction ratios of line $BC$ are $(4-0, 8-2, 11-3) = (4, 6, 8)$.The equation of line $BC$ is $\frac{x-0}{4} = \frac{y-2}{6} = \frac{z-3}{8} = r$.Any point $P$ on the line $BC$ can be represented as $(4r, 6r+2, 8r+3)$.Since $AP$ is perpendicular to $BC$,the dot product of the direction ratios of $AP$ and $BC$ must be zero.The direction ratios of $AP$ are $(4r-3, 6r+2-(-1), 8r+3-11) = (4r-3, 6r+3, 8r-8)$.Since $AP \perp BC$,we have $4(4r-3) + 6(6r+3) + 8(8r-8) = 0$.$16r - 12 + 36r + 18 + 64r - 64 = 0$.$116r - 58 = 0 \Rightarrow r = \frac{58}{116} = \frac{1}{2}$.Substituting $r = \frac{1}{2}$ into the coordinates of $P$,we get $P = (4(\frac{1}{2}), 6(\frac{1}{2})+2, 8(\frac{1}{2})+3) = (2, 3+2, 4+3) = (2, 5, 7)$.Thus,the coordinates of the foot of the perpendicular are $(2, 5, 7)$.

If $A(3,-1,11)$,$B(0,2,3)$,and $C(4,8,11)$ are three points,then the coordinates of the foot of the perpendicular drawn from the point $A$ to the line joining the points $B$ and $C$ is

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