Question

The equation to the altitude of the triangle formed by $(1,1,1),(1,2,3),(2,-1,1)$ through $(1,1,1)$ is

• A. $r=(\overline{i}+\overline{j}+\overline{k})+t(\overline{i}-\overline{j}+2\overline{k})$
• B. $r=(\overline{i}+\overline{j}+\overline{k})+t(\overline{i}+3\overline{j}+2\overline{k})$
• C. $r=(\overline{i}+\overline{j}+\overline{k})+t(\overline{i}-3\overline{j}+2\overline{k})$
• D. $|\overline{r}|=5$

Answer 1

The correct option is A $r=(\overline{i}+\overline{j}+\overline{k})+t(\overline{i}-\overline{j}+2\overline{k})$

Let, $A(1,1,1),B(1,2,3),C(2,-1,-1)$ and $AP$ be the altitude, then

$D{C}^{\prime }s$ of $AP=(\lambda ,-3\lambda +1,-2\lambda +2)$

$D{C}^{\prime }s$ of $BC=(1,-3,-2)$

$\therefore P=(\lambda +1,-3\lambda +2,-2\lambda +3)$

$\because AP\perp BC$

$⟹\lambda +3(3\lambda -1)+2(2\lambda -2)=0$

$⟹\lambda +9\lambda -3+4\lambda -4=0$

$⟹14\lambda -7=0$

$⟹14\lambda =7$

$\therefore \lambda =\frac{7}{14}=\frac{1}{2}$

i.e., $P=(\frac{1}{2}+1,\frac{-3}{2}+2,\frac{-2}{2}+3)$

$P=(\frac{3}{2},\frac{1}{2},2)$

$AP=\frac{1}{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$

$r=i+j+k+t(i-j+2k)$