Let ( x -2/3)=( y +1/-2)=( z +3/-1) lie on the plane px - qy + z =5, for some p , q ∈ R. The shortest distance of the plane from the origin is:

Question

Let ( x -2/3)=( y +1/-2)=( z +3/-1) lie on the plane px - qy + z =5, for some p , q ∈ R. The shortest distance of the plane from the origin is: (A) √(3/109) (B) √(5/142) (C) √(5/71) (D) √(1/142)

Tardigrade Admin · Accepted Answer

(2, -1,-3) satisfy the given plane. So 2 p + q =8 Also given line is perpendicular to normal plane so 3 p+2 q-1=0 ⇒ p =15, q =-22 Eq. of plane 15 x-22 y+z-5=0 its distance from origin =(6/√710)=√(5/142)

Q. Let $\frac{x - 2}{3} = \frac{y + 1}{- 2} = \frac{z + 3}{- 1}$ lie on the plane $p x - q y + z = 5$ , for some $p, q \in R$ . The shortest distance of the plane from the origin is:

$\frac{3}{109}$

$\frac{5}{142}$

$\frac{5}{71}$

$\frac{1}{142}$

$(2, - 1, - 3)$ satisfy the given plane.
So $2 p + q = 8$
Also given line is perpendicular to normal plane so
$3 p + 2 q - 1 = 0$
$\Rightarrow p = 15, q = - 22$
Eq. of plane $15 x - 22 y + z - 5 = 0$
its distance from origin $= \frac{6}{710} = \frac{5}{142}$

Q. Let 3x−2​=−2y+1​=−1z+3​ lie on the plane px−qy+z=5, for some p,q∈R. The shortest distance of the plane from the origin is:

1093​​

1425​​

715​​

1421​​