Question

Find the image of the point $(2,-1,5)$ in the line $\overrightarrow{r}=(11\hat{i}-2\hat{j}-\widehat{8k})+\lambda (10\hat{i}-4\hat{j}-\widehat{11k})$.

Answer 1

We have,

Equation of line is $\overrightarrow{r}=(11\overrightarrow{i}-2\overrightarrow{j}-8\overrightarrow{k})+\lambda (10\overrightarrow{i}-4\overrightarrow{j}-11\overrightarrow{k})$

Image of point is $\overrightarrow{c}=(2,-1,5)$.

Comparing that equation of line is $\overrightarrow{r}=\overrightarrow{p}+\lambda \overrightarrow{q}$.

We know that,

Foot of perpendicular of the perpendicular distance

$=\frac{|(\overrightarrow{c}-\overrightarrow{p})\times \overrightarrow{q}|}{\left|\overrightarrow{q}\right|}$

Now,

$(\overrightarrow{c}-\overrightarrow{p})\times \overrightarrow{q}=$$\left|\begin{array}{ccc}i& j& k\\ -9& 1& 13\\ 10& -4& -11\end{array}\right|$

$=41\hat{i}+31\hat{j}+26\hat{k}$

Then, $Dis\tan ce=\frac{|41\hat{i}+31\hat{j}+26\hat{k}|}{|10\hat{i}-4\hat{j}-11\hat{k}|}=\frac{\sqrt{3318}}{\sqrt{237}}=\sqrt{14}$

Hence, this is the answer.