\displaystyle \textbf{Question 1: }~\text{Find the vector equation of a plane which is at a distance of }3 \\ \text{ units from the origin and has }\widehat{k}\text{ as the unit vector normal to it.}
\displaystyle \text{Answer:}
\displaystyle \text{Given that the normal vector, }\widehat{n}=\widehat{k}
\displaystyle \text{Now, }\widehat{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{\widehat{k}}{\left|\widehat{k}\right|}=\frac{\widehat{k}}{1}=\widehat{k}
\displaystyle \text{The equation of a plane in normal form is}
\displaystyle \overrightarrow{r}\cdot\widehat{n}=d\text{ (where }d\text{ is the distance of the plane from the origin)}
\displaystyle \text{Substituting }\widehat{n}=\widehat{k}\text{ and }d=3\text{ in the relation, we get}
\displaystyle \overrightarrow{r}\cdot\widehat{k}=3

\displaystyle \textbf{Question 2: }~\text{Find the vector equation of a plane which is at a distance of }5 \\ \text{ units from the origin and which is normal to the vector }\widehat{i}-2\widehat{j}-2\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{It is given that the normal vector, }\overrightarrow{n}=\widehat{i}-2\widehat{j}-2\widehat{k}
\displaystyle \text{Now, }\widehat{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{\widehat{i}-2\widehat{j}-2\widehat{k}}{\sqrt{1+4+4}}=\frac{\widehat{i}-2\widehat{j}-2\widehat{k}}{3}=\frac{1}{3}\widehat{i}-\frac{2}{3}\widehat{j}-\frac{2}{3}\widehat{k}
\displaystyle \text{The equation of a plane in normal form is}
\displaystyle \overrightarrow{r}\cdot\widehat{n}=d\text{ (where }d\text{ is the distance of the plane from the origin)}
\displaystyle \text{Substituting }\widehat{n}=\frac{1}{3}\widehat{i}-\frac{2}{3}\widehat{j}-\frac{2}{3}\widehat{k}\text{ and }d=5
\displaystyle \overrightarrow{r}\cdot\left(\frac{1}{3}\widehat{i}-\frac{2}{3}\widehat{j}-\frac{2}{3}\widehat{k}\right)=5

\displaystyle \textbf{Question 3: }~\text{Reduce the equation }2x-3y-6z=14\text{ to the normal form} \\ \text{and hence} \text{find the length of} \text{perpendicular from the origin to the plane. Also, find} \\ \text{the direction cosines of the normal to the plane.}
\displaystyle \text{Answer:}
\displaystyle \text{The given equation of the plane is}
\displaystyle 2x-3y-6z=14\quad (1)
\displaystyle \sqrt{2^{2}+(-3)^{2}+(-6)^{2}}=\sqrt{4+9+36}=\sqrt{49}=7
\displaystyle \text{Dividing (1) by }7\text{, we get}
\displaystyle \frac{2}{7}x-\frac{3}{7}y-\frac{6}{7}z=2\quad (2)
\displaystyle \text{The Cartesian equation of the normal form of a plane is}
\displaystyle lx+my+nz=p\quad (3)
\displaystyle \text{where }l,m,n\text{ are direction cosines of the normal to the plane and }p\text{ is the length of the} \\ \text{perpendicular from the origin to the plane.}
\displaystyle \text{Comparing (2) and (3), we get}
\displaystyle l=\frac{2}{7},\ m=\frac{-3}{7},\ n=\frac{-6}{7}
\displaystyle \text{length of the perpendicular from the origin to the plane : }p=2

\displaystyle \textbf{Question 4: }~\text{Reduce the equation }\overrightarrow{r}\cdot(\widehat{i}-2\widehat{j}+2\widehat{k})+6=0\text{ to normal form and} \\ \text{hence find the length of perpendicular} \text{from the origin to the plane.}
\displaystyle \text{Answer:}
\displaystyle \text{The given equation of the plane is}
\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}-2\widehat{j}+2\widehat{k}\right)+6=0
\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}-2\widehat{j}+2\widehat{k}\right)=-6\text{ or }\overrightarrow{r}\cdot\overrightarrow{n}=-6\text{, where }\overrightarrow{n}=\widehat{i}-2\widehat{j}+2\widehat{k}
\displaystyle \left|\overrightarrow{n}\right|=\sqrt{1+4+4}=3
\displaystyle \text{For reducing the given equation to normal form, we need to divide it by }\left|\overrightarrow{n}\right|.\text{ Then, we get}
\displaystyle \frac{\overrightarrow{r}\cdot\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{-6}{\left|\overrightarrow{n}\right|}
\displaystyle \overrightarrow{r}\cdot\left(\frac{\widehat{i}-2\widehat{j}+2\widehat{k}}{3}\right)=\frac{-6}{3}
\displaystyle \overrightarrow{r}\cdot\left(\frac{1}{3}\widehat{i}-\frac{2}{3}\widehat{j}+\frac{2}{3}\widehat{k}\right)=-2
\displaystyle \text{Dividing both sides by }-1\text{, we get}
\displaystyle \overrightarrow{r}\cdot\left(-\frac{1}{3}\widehat{i}+\frac{2}{3}\widehat{j}-\frac{2}{3}\widehat{k}\right)=2\quad (1)
\displaystyle \text{The equation of the plane in normal form is}
\displaystyle \overrightarrow{r}\cdot\widehat{n}=d\quad (2)
\displaystyle \text{(where }d\text{ is the distance of the plane from the origin)}
\displaystyle \text{Comparing (1) and (2),}
\displaystyle \text{length of the perpendicular from the origin to the plane }=d=2\text{ units}

\displaystyle \textbf{Question 5: }~\text{Write the normal form of the equation of the plane }2x-3y+6z+14=0.
\displaystyle \text{Answer:}
\displaystyle \text{The given equation of the plane is}
\displaystyle 2x-3y+6z+14=0
\displaystyle 2x-3y+6z=-14\quad (1)
\displaystyle \sqrt{2^{2}+(-3)^{2}+6^{2}}=\sqrt{4+9+36}=\sqrt{49}=7
\displaystyle \text{Dividing (1) by }7\text{, we get}
\displaystyle \frac{2}{7}x-\frac{3}{7}y+\frac{6}{7}z=-2
\displaystyle \text{Multiplying both sides by }-1\text{, we get}
\displaystyle -\frac{2}{7}x+\frac{3}{7}y-\frac{6}{7}z=2
\displaystyle \text{This is the normal form of the given equation of the plane.}

\displaystyle \textbf{Question 6: }~\text{The direction ratios of the perpendicular from the origin to a plane are } \\ 12,-3,4 \text{ and the length of the perpendicular is }5.\text{ Find the equation of the plane.}
\displaystyle \text{Answer:}
\displaystyle \text{It is given that the direction ratios of the normal vector }\overrightarrow{n}\text{ are }12,-3,4.
\displaystyle \text{So, }\overrightarrow{n}=12\widehat{i}-3\widehat{j}+4\widehat{k}
\displaystyle \left|\overrightarrow{n}\right|=\sqrt{12^{2}+(-3)^{2}+4^{2}}=\sqrt{144+9+16}=\sqrt{169}=13
\displaystyle \text{Now, }\widehat{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{12\widehat{i}-3\widehat{j}+4\widehat{k}}{13}=\frac{12}{13}\widehat{i}-\frac{3}{13}\widehat{j}+\frac{4}{13}\widehat{k}
\displaystyle \text{Length of the perpendicular from the origin to the plane, }d=5
\displaystyle \text{Equation of the plane in normal form is}
\displaystyle \overrightarrow{r}\cdot\widehat{n}=d
\displaystyle \overrightarrow{r}\cdot\left(\frac{12}{13}\widehat{i}-\frac{3}{13}\widehat{j}+\frac{4}{13}\widehat{k}\right)=5

\displaystyle \textbf{Question 7: }~\text{Find a unit normal vector to the plane }x+2y+3z-6=0.
\displaystyle \text{Answer:}
\displaystyle \text{The given equation of the plane is}
\displaystyle x+2y+3z-6=0
\displaystyle x+2y+3z=6
\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)=6\text{ or }\overrightarrow{r}\cdot\overrightarrow{n}=6
\displaystyle \text{where }\overrightarrow{n}=\widehat{i}+2\widehat{j}+3\widehat{k}\quad (1)
\displaystyle \left|\overrightarrow{n}\right|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14}
\displaystyle \text{Unit vector to the plane, }\widehat{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{\widehat{i}+2\widehat{j}+3\widehat{k}}{\sqrt{14}}=\frac{1}{\sqrt{14}}\widehat{i}+\frac{2}{\sqrt{14}}\widehat{j}+\frac{3}{\sqrt{14}}\widehat{k}

\displaystyle \textbf{Question 8: }~\text{Find the equation of a plane which is at a distance of }3\sqrt{3} \text{ units from} \\ \text{the origin and the normal to which is equally inclined with the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\alpha,\beta\text{ and }\gamma\text{ be the angles made by }\overrightarrow{n}\text{ with }x,y\text{ and }z\text{-axes respectively.}
\displaystyle \text{It is given that}
\displaystyle \alpha=\beta=\gamma
\displaystyle \cos\alpha=\cos\beta=\cos\gamma
\displaystyle l=m=n\text{, where }l,m,n\text{ are direction cosines of }\overrightarrow{n}
\displaystyle \text{But }l^{2}+m^{2}+n^{2}=1
\displaystyle l^{2}+l^{2}+l^{2}=1
\displaystyle 3l^{2}=1
\displaystyle l^{2}=\frac{1}{3}
\displaystyle l=\frac{1}{\sqrt{3}}
\displaystyle \text{So, }l=m=n=\frac{1}{\sqrt{3}}
\displaystyle \text{It is given that the length of the perpendicular of the plane from the origin, }p=3\sqrt{3}
\displaystyle \text{The normal form of the plane is }lx+my+nz=p
\displaystyle \frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+\frac{1}{\sqrt{3}}z=3\sqrt{3}
\displaystyle x+y+z=9

\displaystyle \textbf{Question 9: }~\text{Find the equation of the plane passing through the point }(1,2,1) \\ \text{ and perpendicular to the line joining the points }(1,4,2)\text{ and }(2,3,5). \\ \text{ Find also the perpendicular distance} \text{of the origin from this plane.}   
\displaystyle \text{Answer:}
\displaystyle \text{The normal is passing through the points }A(1,4,2)\text{ and }B(2,3,5).
\displaystyle \text{So, }\overrightarrow{n}=\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\left(2\widehat{i}+3\widehat{j}+5\widehat{k}\right)-\left(\widehat{i}+4\widehat{j}+2\widehat{k}\right)=\widehat{i}-\widehat{j}+3\widehat{k}
\displaystyle \text{We know that the vector equation of the plane passing through a point }(1,2,1)\text{ (} \\ \overrightarrow{a}\text{) and normal to }\overrightarrow{n}\text{ is}
\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}
\displaystyle \text{Substituting }\overrightarrow{a}=\widehat{i}+2\widehat{j}+\widehat{k}\text{ and }\overrightarrow{n}=\widehat{i}-\widehat{j}+3\widehat{k}\text{, we get}
\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}-\widehat{j}+3\widehat{k}\right)=\left(\widehat{i}+2\widehat{j}+\widehat{k}\right)\cdot\left(\widehat{i}-\widehat{j}+3\widehat{k}\right)
\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}-\widehat{j}+3\widehat{k}\right)=1-2+3
\displaystyle \overrightarrow{r}\cdot\left(\widehat{i}-\widehat{j}+3\widehat{k}\right)=2\quad (1)
\displaystyle \text{To find the perpendicular distance of this plane from the origin, we have to reduce} \\ \text{this to normal form.}
\displaystyle \overrightarrow{n}=\widehat{i}-\widehat{j}+3\widehat{k},\ \left|\overrightarrow{n}\right|=\sqrt{1+1+9}=\sqrt{11}
\displaystyle \text{Dividing (1) by }\sqrt{11}\text{, we get}
\displaystyle \overrightarrow{r}\cdot\left(\frac{1}{\sqrt{11}}\widehat{i}-\frac{1}{\sqrt{11}}\widehat{j}+\frac{3}{\sqrt{11}}\widehat{k}\right)=\frac{2}{\sqrt{11}}
\displaystyle \text{So, the perpendicular distance of plane (1) from the origin }=\frac{2}{\sqrt{11}}

\displaystyle \textbf{Question 10: }~\text{Find the vector equation of the plane which is at a distance of }\frac{6}{\sqrt{29}} \\ \text{ from the origin and its normal vector from the origin is }2\widehat{i}-3\widehat{j}+4\widehat{k}. \\ \text{ Also, find its cartesian form. }
\displaystyle \text{Answer:}
\displaystyle \text{Given, normal vector, }\overrightarrow{n}=2\widehat{i}-3\widehat{j}+4\widehat{k}
\displaystyle \text{Now, }\widehat{n}=\frac{\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{2\widehat{i}-3\widehat{j}+4\widehat{k}}{\sqrt{4+9+16}}=\frac{2\widehat{i}-3\widehat{j}+4\widehat{k}}{\sqrt{29}}=\frac{2}{\sqrt{29}}\widehat{i}-\frac{3}{\sqrt{29}}\widehat{j}+\frac{4}{\sqrt{29}}\widehat{k}
\displaystyle \text{The equation of the plane in normal form is}
\displaystyle \overrightarrow{r}\cdot\widehat{n}=d\text{ (where }d\text{ is the distance of the plane from the origin)}
\displaystyle \text{Substituting }\widehat{n}=\frac{2}{\sqrt{29}}\widehat{i}-\frac{3}{\sqrt{29}}\widehat{j}+\frac{4}{\sqrt{29}}\widehat{k}\text{ and }d=\frac{6}{\sqrt{29}}\text{ here, we get}
\displaystyle \overrightarrow{r}\cdot\left(\frac{2}{\sqrt{29}}\widehat{i}-\frac{3}{\sqrt{29}}\widehat{j}+\frac{4}{\sqrt{29}}\widehat{k}\right)=\frac{6}{\sqrt{29}}\quad (1)
\displaystyle \text{Cartesian form}
\displaystyle \text{For Cartesian form, substituting }\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}\text{ in (1), we get}
\displaystyle \left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right)\cdot\left(\frac{2}{\sqrt{29}}\widehat{i}-\frac{3}{\sqrt{29}}\widehat{j}+\frac{4}{\sqrt{29}}\widehat{k}\right)=\frac{6}{\sqrt{29}}
\displaystyle \frac{2x-3y+4z}{\sqrt{29}}=\frac{6}{\sqrt{29}}
\displaystyle 2x-3y+4z=6

\displaystyle \textbf{Question 11: }~\text{Find the distance of the plane }2x-3y+4z-6=0\text{ from the origin. }
\displaystyle \text{Answer:}
\displaystyle \text{The given equation of the plane is}
\displaystyle 2x-3y+4z=6\quad (1)
\displaystyle \sqrt{2^{2}+(-3)^{2}+4^{2}}=\sqrt{4+9+16}=\sqrt{29}
\displaystyle \text{Dividing (1) by }\sqrt{29}\text{, we get}
\displaystyle \frac{2}{\sqrt{29}}x-\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z=\frac{6}{\sqrt{29}}\text{, which is the normal form of plane (1).}
\displaystyle \text{So, the length of the perpendicular from the origin to the plane }=\frac{6}{\sqrt{29}}

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