The orthocentre of the triangle formed by the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The shortest distance between lines $\overline{r}=(2 \hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ and $\overline{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+\hat{j}-5 \hat{k})$ is

The shortest distance between the lines $\frac{x-1}{0}=\frac{y+1}{-1}=\frac{z}{1}$ and $x+y+z+1=0, 2x-y+z+3=0$ is

If the lines $\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}$ and $\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}$ are perpendicular,then $\lambda=$

If the lines $\frac{x - 1}{c} = \frac{y + 2}{-2} = \frac{z - 3}{4}$ and $\frac{x - 5}{1} = \frac{y - 3}{1} = \frac{z + 1}{c}$ are parallel,then $c = ....$

The equation of the line passing through the point $(2, 3, -4)$ and perpendicular to the $XOZ$ plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module