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 : |
A and B only B and C only C and D only A and D only |
A and D only |
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 |