Given equation of the plane is $5x+2y -3z=17.$ Then,
A. point (2, 2, -1) lies on the plane
B. The direction cosines of normal to plane are (2, 2,-1)
C. Coordinates of foot of perpendicular from origin are (5, 2, -3)
D. The plane does not pass through the origin
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
Given plane: $5x + 2y - 3z = 17$
A. Check if $(2,2,-1)$ lies on the plane:
$5*2 + 2*2 - 3*(-1) = 10 + 4 + 3 = 17$ ✅
So, A is True
B. Direction cosines of normal vector:
Normal vector $n = (5, 2, -3)$, not $(2,2,-1)$ ❌
B is False
C. Coordinates of foot of perpendicular from origin:
Foot of perpendicular: $\left(\frac{5*17}{38}, \frac{2*17}{38}, \frac{-3*17}{38}\right) = \left(\frac{85}{38}, \frac{34}{38}, \frac{-51}{38}\right)$ ❌
C is False
D. Plane does not pass through origin:
At origin $(0,0,0)$: $5*0 + 2*0 -3*0 = 0 \ne 17$ ✅
D is True
Answer: A and D