The orthocentre of the triangle formed by the | vedclass

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(C) Let the vertices be $A(2,1,5)$,$B(3,2,3)$,and $C(4,0,4)$.First,find the equation of the altitude from $A$ to $BC$. The direction ratios of $BC$ ar

Vedclass · Accepted Answer

$(3,1,4)$

Vedclass · Answer

(C) Let the vertices be $A(2,1,5)$,$B(3,2,3)$,and $C(4,0,4)$.First,find the equation of the altitude from $A$ to $BC$. The direction ratios of $BC$ are $(4-3, 0-2, 4-3) = (1, -2, 1)$.Let $P$ be the foot of the perpendicular from $A$ to $BC$. $P$ lies on $BC$,so $P = (3+k, 2-2k, 3+k)$ for some $k$.The vector $AP = (3+k-2, 2-2k-1, 3+k-5) = (k+1, 1-2k, k-2)$.Since $AP \perp BC$,the dot product $(k+1)(1) + (1-2k)(-2) + (k-2)(1) = 0$.$k+1 - 2 + 4k + k - 2 = 0 \Rightarrow 6k - 3 = 0 \Rightarrow k = 1/2$.Thus,$P = (3.5, 1, 3.5)$. The vector $AP = (1.5, 0, -1.5)$,which is parallel to $(1, 0, -1)$.The equation of altitude $AP$ is $\frac{x-2}{1} = \frac{y-1}{0} = \frac{z-5}{-1}$.Next,find the equation of the altitude from $B$ to $AC$. The direction ratios of $AC$ are $(4-2, 0-1, 4-5) = (2, -1, -1)$.Let $Q$ be the foot of the perpendicular from $B$ to $AC$. $Q$ lies on $AC$,so $Q = (2+2m, 1-m, 5-m)$ for some $m$.The vector $BQ = (2+2m-3, 1-m-2, 5-m-3) = (2m-1, -m-1, 2-m)$.Since $BQ \perp AC$,$(2m-1)(2) + (-m-1)(-1) + (2-m)(-1) = 0$.$4m - 2 + m + 1 - 2 + m = 0 \Rightarrow 6m - 3 = 0 \Rightarrow m = 1/2$.Thus,$Q = (3, 0.5, 4.5)$. The vector $BQ = (0, -1.5, 1.5)$,which is parallel to $(0, -1, 1)$.The equation of altitude $BQ$ is $\frac{x-3}{0} = \frac{y-2}{-1} = \frac{z-3}{1}$.Solving the two altitude equations: $y=1$ from $AP$,and from $BQ$,$\frac{y-2}{-1} = \frac{z-3}{1} \Rightarrow 1-2 = -(z-3) \Rightarrow -1 = -z+3 \Rightarrow z=4$. Since $x=3$ from $BQ$,the orthocentre is $(3,1,4)$.

The orthocentre of the triangle formed by the points $(2,1,5)$,$(3,2,3)$,and $(4,0,4)$ is

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