Find the Cartesian equation of the plane passing through the point (3,2,-3) and the normal to the plane is \(4\hat{i}-2\hat{j}+5\hat{k}\)?(a) 4x-2y+5z+7=0(b) 3x-2y-3z+1=0(c) 4x-y+5z+7=0(d) 4x-2y-z+7=0I got this question in an international level competition.The query is from Three Dimensional Geometry topic in division Three Dimensional Geometry of Mathematics – Class 12

Find the Cartesian equation of the plane passing through the point (3,

1.	Find the Cartesian equation of the plane passing through the point (3,2,-3) and the normal to the plane is \(4\hat{i}-2\hat{j}+5\hat{k}\)?(a) 4x-2y+5z+7=0(b) 3x-2y-3z+1=0(c) 4x-y+5z+7=0(d) 4x-2y-z+7=0I got this question in an international level competition.The query is from Three Dimensional Geometry topic in division Three Dimensional Geometry of Mathematics – Class 12
Answer» The CORRECT option is (a) 4x-2y+5z+7=0 Explanation: The POSITION VECTOR of the point (3,2,-3) is \(\vec{a}=3\hat{i}+2\hat{j}-3\hat{k}\) and the normal vector \(\vec{N}\) perpendicular to the plane is \(\vec{N}=4\hat{i}-2\hat{j}+5\hat{k}\) Therefore, the vector equation of the plane is given by \((\vec{r}-\vec{a}).\vec{N}\)=0 Hence, \((\vec{r}-(3\hat{i}+2\hat{j}-3\hat{k})).(4\hat{i}-2\hat{j}+5\hat{k})\)=0 4(x-3)-2(y-2)+5(z+3)=0 4x-2y+5z+7=0.

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