Question

The equation of the plane through the point (2, 1, 3) and perpendicular to the vector \(4\hat{i}+5\hat{j}+6\hat{k}\) is

• A. 4x+5y+6z=28
• B. 2x+y+3z=17
• C. 4x+5y+6z=33
• D. 8x+5y+18z=21
• E. 4x+5y+6z=31

Answer 1

Answer: (E)

Step 1: General Equation of a Plane  
The equation of a plane passing through a point \((x_0, y_0, z_0)\) and having normal vector \( \overrightarrow{N} = A\hat{i} + B\hat{j} + C\hat{k} \) is given by: \[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]

Step 2: Given Data 
The point is \( (2,1,3) \) and the normal vector is \( 4\hat{i} + 5\hat{j} + 6\hat{k} \), so \( A = 4 \), \( B = 5 \), and \( C = 6 \).

Step 3: Substitute in Plane Equation 
\[ 4(x - 2) + 5(y - 1) + 6(z - 3) = 0 \] Expanding: \[ 4x - 8 + 5y - 5 + 6z - 18 = 0 \] \[ 4x + 5y + 6z - 31 = 0 \]

Final Answer: \( 4x + 5y + 6z = 31 \).