Find the distance of the point (2, 12, 5) from the point of intersection of the line vector(r) = 2i - 4j + 2k + λ(3i + 4j + 2k)

Question

Find the distance of the point (2, 12, 5) from the point of intersection of the line

\(\vec{r}=2\hat{i}-4\hat{j}+2\hat{k}+\lambda(3\hat{i}+4\hat{j}+2\hat{k})\) and the plane \(\vec{r}.(\hat{i}-2\hat{j}+\hat{k})=0.\)

Answer 1

Given line and plane are

\(\vec{r}=2\hat{i}-4\hat{j}+2\hat{k}+\lambda(3\hat{i}+4\hat{j}+2\hat{k})\, .....(i)\)

\(\vec{r}.(\hat{i}-2\hat{j}+\hat{k})=0\, ......(ii)\)

For intersection point Q, we solve equations (i) and (ii) by putting the value of \(\vec{r}\) from (i) in (ii)

Hence position vector of intersecting point is \(14\hat{i}+12\hat{i}+10\hat{k}.\)

Co-ordinate of intersecting point, Q \(\equiv\) (14, 12, 10)

Required distance