Question

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

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

Answer 1

The correct option is A $\overline{r}=(\overline{i}+\overline{j}+\overline{k})+t(\overline{i}-3\overline{j}-2\overline{k})$

$\text{Solve:}$

$\overrightarrow{BD}=(a-1)\hat{i}+(b-2)\hat{j}+(c-3)\hat{k}$

$\text{Points}{B}_{1}D\text{and}c\text{are three collinear}$

$\text{point}$

$\Rightarrow \lambda \overrightarrow{BC}=\left(\overrightarrow{BD}\right)$

$\Rightarrow (a-1)=\lambda \Rightarrow a=\lambda +1$

$(b-2)=-3\lambda \Rightarrow b=2-3\lambda$

$c-3=-2\lambda \Rightarrow c=3-2\lambda$

$\overrightarrow{AD}$ is perpendicular to $\overrightarrow{BC}$

$\Rightarrow \left(\overrightarrow{AD}\right)\cdot \left(\overrightarrow{BC}\right)=0$

$\overrightarrow{AD}=\lambda \hat{i}+(1-3\lambda )\hat{j}+(2-2\lambda )\hat{k}$

$\Rightarrow \overrightarrow{AD}\cdot \overrightarrow{BC}=\lambda -3+9\lambda -4+4\lambda =0$

$\Rightarrow 14\lambda =7$

$\Rightarrow \lambda =\frac{1}{2}$

$\text{SO,}a=\lambda +1=\frac{3}{2}$

$b=2-3\lambda =\frac{1}{2}\text{and}c=3-2\lambda$

So, equation of its altitude i.e. $\overrightarrow{AD}$

$\overrightarrow{AD}=\hat{i}+\hat{ȷ}+k+t[(a-1)\hat{i}+(b-1)\hat{ȷ}+(c-1)\hat{k}]$

$\Rightarrow \overrightarrow{AD}=\hat{i}+\hat{j}+\hat{k}+t(\frac{1}{2}\hat{i}+(-\frac{1}{2})\hat{ȷ}+\hat{k})$

$\Rightarrow \overrightarrow{AD}=\hat{i}+\hat{ȷ}+\hat{k}+t(\hat{ı}-\hat{ȷ}+2\hat{k})$

where$t=\frac{t}{2}$