A line passing through the point A with position vector a= 4i+ 2j+2k is parallel to the vector b = 2i+3j+6k . Find the length of the perpendicular drawn on this line from a point P with position vector r= i + 2 j+ 3k

Join BYJU'S Learning Program

If P is a point on the line passing through the point A with position vector 2¯i+¯j−3¯¯¯k and parallel to ¯i+2¯j+¯¯¯k such that AP=2√6 then the position vector of P is

The correct option is D 4¯i+5¯j−¯¯¯kP⋅V of P=(2+t)^i+(1+2t)^j+(−3+t)^kP⋅V of A=(2,1,−3)∣∣∣−−→AP∣∣∣=√t2+4t2+t22 √6=t√6⇒t=2P=4^i+5^j−^k

If $P$ is a point on the line passing through the point $A$ with position vector $2 ¯ i + ¯ j - 3 ¯ ¯ ¯ k$ and parallel to $¯ i + 2 ¯ j + ¯ ¯ ¯ k$ such that $A P = 2 \sqrt{6}$ then the position vector of $P$ is

The correct option is D $4 ¯ i + 5 ¯ j - ¯ ¯ ¯ k$
$P \cdot V o f P = (2 + t)^i + (1 + 2 t)^j + (- 3 + t)^k$
$P \cdot V o f A = (2, 1, - 3)$
$∣ ∣ ∣ - - \to A P ∣ ∣ ∣ = \sqrt{t^{2} + 4 t^{2} + t^{2}}$
$2 \sqrt{6} = t \sqrt{6}$
$\Rightarrow t = 2$
$P = 4^i + 5^j -^k$