JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 14)
The distance of the point having position vector $$ - \widehat i + 2\widehat j + 6\widehat k$$
from the straight line passing through the point
(2, 3, – 4) and parallel to the vector, $$6\widehat i + 3\widehat j - 4\widehat k$$ is :
6
7
$$2\sqrt {13} $$
$$4\sqrt 3 $$
Explanation
_10th_April_Evening_Slot_en_14_2.png)
$$AD = \left| {{{\overrightarrow {AP} .\overrightarrow n } \over {\left| {\overrightarrow n } \right|}}} \right| = \sqrt {61} $$
$$ \Rightarrow PD = \sqrt {A{P^2} - A{D^2}} $$
$$ = \sqrt {110 - 61} $$
= 7
Comments (0)
