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=0

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The query is from Three Dimensional Geometry topic in division Three Dimensional Geometry of Mathematics – Class 12

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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.