Question

Fimd the distance of plint p whose position vector is 3j+6j+8k from y axis

Answer 1

Answer:

√73

Step-by-step explanation:

Hi,

Position vector of given point is P(3i + 6j + 8k)

Let F be the foot of the perpendicular of the point P onto y axis,

then the co-ordinates of point F are 6j

Hence distance of point P from y - axis is |PF|

But PF is given by position vector 3i + 8k

So length of vector PF

= |PF|

= √3² + 8²

= √73.

Hence, the point P is at a distance of √73 units from y-axis.

Hope, it helps !