Question

Find the point in the plane 2x+3y-z= 5 which is nearest to the origin.

Answer 1

x=1, y=1 and z=5
point=(1,1,5) is your answer...

Answer 2

The given problem can be solved by two methods,with calculus(A lengthy approach) and without calculus.

So, equation of given plane
$2x + 3y - z = 5$

here we know that Direction ratio of normal to the plane is (2,3,-1)

That is from this we can find the equation which passes through origin and meet the plane, let that equation has general point (2k,3k,-k)

So,it passes through plane
$2(2k) + 3(3k) - ( - k) = 5 \\ \\ 4k + 9k + k = 5 \\ \\ k = \frac{5}{14}$
Thus,coordinates of that point which is nearest to the plane

$x = 2k = 2 \times \frac{5}{14} = \frac{5}{7} \\ \\ y = 3k = 3 \times \frac{5}{14} = \frac{15}{14} \\ \\ z = - k = - \frac{5}{14}$
Coordinates of point which lies on the plane and nearest to the origin are
$\bigg( \frac{5}{7}, \: \frac{15}{14}, \: \frac{ - 5}{14} \bigg)$
Hope it helps you.