Differentiation (Part-38) | S N De

 

1. Reduce the equation of the plane $~4x+3y-6z-12=0~$ to the intercept form and find its intercepts on the coordinate axes.

Solution.

The given equation can be written as 

$4x+3y-6z-12=0 \\ \text{or,}~~ 4x+3y-6z=12 \\ \text{or,}~~ \frac{4x}{12}+\frac{3y}{12}-\frac{6z}{12}=1 \\ \therefore~ \frac x3+\frac y4+\frac{z}{(-2)}=1 \longrightarrow(1)$

So, the equation (1) represents the intercept form of the plane and the lengths of the intercepts from $~x-$ axis, $~y-$ axis and $~z-$ axis are given by $~3,4~$ and $~-2.$

2. Find the normal vector to the plane $~2x-y+2z=5.$ Also, find a unit vector normal to the plane 

Solution.

The normal vector to the plane $~2x-y+2z=5~$ is given by $~2\hat{i}-\hat{j}+2\hat{k}.$

So, the unit vector normal to the plane is 

$\frac{2\hat{i}-\hat{j}+2\hat{k}}{|2\hat{i}-\hat{j}+2\hat{k}|}=\frac{~2\hat{i}-\hat{j}+2\hat{k}}{\sqrt{2^2+(-1)^2+2^2}}=\frac 13(2\hat{i}-\hat{j}+2\hat{k}).$

3. Find the vector equation of the following planes in scalar product form $~(\vec{r}\cdot \vec{n}=d) :$

$(i)~ \vec{r}=(2\hat{i}-\hat{k})+\lambda \hat{i}+\mu(-\hat{i}-2\hat{j}-\hat{k})$

Solution.

We know that the equation $~\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}~$ represents a plane passing through a point having position vector $~\vec{a}~$ and parallel to vectors $~\vec{b}~$ and $~\vec{c}~$ [where $~\lambda~$ and $~\mu~$ are scalars]. Here, $~\vec{a}=2\hat{i}-\hat{k},~\vec{b}=\hat{i}~$ and $~\vec{c}=\hat{i}-2\hat{j}-\hat{k}.$

The given plane is perpendicular to the vector 

$\vec{n}=\vec{b} \times \vec{c}=\begin{vmatrix} \hat{i} &  \hat{j}&\hat{k}  \\ 1& 0 &0  \\ 1& -2 & -1 \\ \end{vmatrix}\\ \text{or,}~~ \vec{n}=\hat{i}(0-0)-\hat{j}(-1-0)+\hat{k}(-2-0) \\ \text{or,}~~ \vec{n}=\hat{j}-2\hat{k}.$

So, the vector equation of the plane in scalar product form is 

$~\vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n} \\ \text{or,}~~ \vec{r} \cdot (\hat{j}-2\vec{k})=(2\hat{i}-\hat{k}) \cdot (\hat{j}-2\hat{k})=2 $ 

Hence, the required vector equation is given by  $~\vec{r} \cdot (\hat{i}-2\vec{k})=2.$

$(ii)~ \vec{r}=(1+s-t)\hat{i}+(2-s)\hat{j}+(3-2s+2t)\hat{k}$

Solution.

$\vec{r}=(1+s-t)\hat{i}+(2-s)\hat{j}+(3-2s+2t)\hat{k} \\ \therefore~\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+s(\hat{i}-\hat{j}-2\hat{k})+t(-\hat{i}+2\hat{k})$

The given plane is perpendicular to the vector

$\vec{n}=\begin{vmatrix} \hat{i} &  \hat{j}&\hat{k}  \\ 1& -1 &-2  \\ -1& 0 & 2 \\ \end{vmatrix}=\hat{i}(-2-0)-\hat{j}(2-2)+\hat{k}(0-1) \\ \therefore ~ \vec{n}=-2\hat{i}-\hat{k}.$

So, the general form $~(\vec{r} \cdot \vec{n}=d)~$ of the plane is given by 

$~\vec{r} \cdot (-2\hat{i}-\hat{k})=(\hat{i}+2\hat{j}+3\hat{k}) \cdot (-2\hat{i}-\hat{k}) \\ \text{or,}~~ \vec{r} \cdot (-2\hat{i}-\hat{k})=-5~(\text{ans.})$

$(iii)~ \vec{r}=\hat{i}-\hat{j}+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(4\hat{i}-2\hat{j}+3\hat{k})$

Solution.

$\vec{r}=\hat{i}-\hat{j}+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(4\hat{i}-2\hat{j}+3\hat{k})\\ \therefore ~ \vec{n}=(\hat{i}+\hat{j}+\hat{k})\times (4\hat{i}-2\hat{j}+3\hat{k}) \\ \text{or,}~~ \vec{n}=\begin{vmatrix}\hat{i} &  \hat{j}&\hat{k}  \\ 1& 1 &1  \\ 4& -2 & 3 \\ \end{vmatrix}=\hat{i}(3+2)-\hat{j}(3-4)+\hat{k}(-2-4)\\ \therefore~~ \vec{n}=5\hat{i}+\hat{j}-6\hat{k}$

So, the general form $~(\vec{r} \cdot \vec{n}=d)~$ of the plane is given by  

$~\vec{r} \cdot (5\hat{i}+\hat{j}-6\hat{k})=(\hat{i}-\hat{j}) \cdot (5\hat{i}+\hat{j}-6\hat{k})=5-1 \\ \text{or,}~~ \vec{r} \cdot (5\hat{i}+\hat{j}-6\hat{k})=4$

4. Find the cartesian form of the equation of the following planes :

$(i)~ \vec{r}=(\hat{i}-\hat{j})+s(-\hat{i}+\hat{j}+2\hat{k})+t(\hat{i}+2\hat{j}+\hat{k})$

Solution.

$\vec{r}=(\hat{i}-\hat{j})+s(-\hat{i}+\hat{j}+2\hat{k})+t(\hat{i}+2\hat{j}+\hat{k}) \\ \therefore~ \vec{n}=(-\hat{i}+\hat{j}+2\hat{k}) \times (\hat{i}+2\hat{j}+\hat{k}) \\ \text{or,}~~ \vec{n}=\begin{vmatrix} \hat{i} &  \hat{j}&\hat{k}  \\ -1& 1 &1  \\ 1& 2 & 1 \\ \end{vmatrix}=\hat{i}(1-4)-\hat{j}(-1-2)+\hat{k}(-2-1) \\ \text{or,}~~ \vec{n}=-3\hat{i}+3\hat{j}-3\hat{k}$

So, the general form $~(\vec{r} \cdot \vec{n}=d)~$ of the plane is given by  

$\vec{r} \cdot (-3\hat{i}+3\hat{j}-3\hat{k})=(\hat{i}-\hat{j})\cdot(-3\hat{i}+3\hat{j}-3\hat{k}) \\ \text{or,}~~ -3\vec{r}\cdot(\hat{i}-\hat{j}+\hat{k})=-3-3 \\ \therefore~ \vec{r} \cdot (\hat{i}-\hat{j}+\hat{k})=\frac{-6}{-3}=2~$

So, the cartesian form of the equation of the plane is given by 

$(x\hat{i}+y\hat{j}+z\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})=2 \\ \text{or,}~~ x-y+z=2.$

$(ii)~\vec{r}=(1+s+t)\hat{i}+(2-s+t)\hat{j}+(3-2s+2t)\hat{k}$

Solution.

$\vec{r}=\hat{i}+s\hat{i}+t\hat{i}+2\hat{j}-s\hat{j}+t\hat{j}+3\hat{k}-2s\hat{k}+2t\hat{k} \\ \text{or,}~~ \vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+s(\hat{i}-\hat{j}-2\hat{k})+t(\hat{i}+\hat{j}+2\hat{k}) \\ \therefore~ \vec{n}=(\hat{i}-\hat{j}-2\hat{k}) \times (\hat{i}+\hat{j}+2\hat{k}) \\ \text{or,}~~ \vec{n}=\begin{vmatrix} \hat{i} &  \hat{j}&\hat{k}  \\ 1& -1 &-2  \\ 1& 1 & 2 \\ \end{vmatrix}=\hat{i}(-2+2)-\hat{j}(2+2)+\hat{k}(1+1) \\ \text{or,}~~ \vec{n}=-4\hat{j}+2\hat{k}.$

So, the general form $~(\vec{r} \cdot \vec{n}=d)~$ of the plane is given by  

$\vec{r} \cdot (-4\hat{j}+2\hat{k})=(\hat{i}+2\hat{j}+3\hat{k}) \cdot (-4\hat{j}+2\hat{k}) \\ \text{or,}~~ -2 \vec{r} \cdot (2\hat{j}-\hat{k})=-8+6 \\ \text{or,}~~ \vec{r} \cdot (2\hat{j}-\hat{k})=\frac{-2}{-2}=1$

Hence, the cartesian form of the plane is given by 

$~(x\hat{i}+y\hat{j}+z\hat{k}) \cdot (2\hat{j}-\hat{k})=1 \\ \text{or,}~~ 2y-z=1~~\text{(ans.)}$