# Plane | Part-3 | Ex-5A

5. Find the vector equation of the following planes :

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

Solution.

$\vec{r}=\lambda\hat{i}-2\mu \hat{i}+3\hat{j}-\mu \hat{j}+2\lambda\hat{k}+\mu\hat{k} \\ \text{or,}~~ \vec{r}=3\hat{j}+\lambda(\hat{i}+2\hat{k})+\mu(-2\hat{i}-\hat{j}+\hat{k}) \\ \therefore~ \vec{n}=(\hat{i}+2\hat{k}) \times (-2\hat{i}-\hat{j}+\hat{k}) \\ \text{or,}~~ \vec{n}=\begin{vmatrix} \hat{i} & \hat{j}&\hat{k} \\ 1& 0 &2 \\ -2& -1 & 1 \\ \end{vmatrix}=\hat{i}(0+2)-\hat{j}(1+4)+\hat{k}(-1-0) \\ \text{or,}~~ \vec{n}=2\hat{i}-5\hat{j}-\hat{k}.$

Hence, the non-parametric form of the plane is

$~\vec{r} \cdot (2\hat{i}-5\hat{j}-\hat{k})=3\hat{j}\cdot (2\hat{i}-5\hat{j}-\hat{k}) \\ \therefore~~ \vec{r} \cdot (2\hat{i}-5\hat{j}-\hat{k})=-15.$

$(ii)~\vec{r}=(2\hat{i}+2\hat{j}-\hat{k})+\lambda(\hat{i}+2\hat{j}+3\hat{k})+\mu(5\hat{i}-2\hat{j}+7\hat{k})$

Solution.

$\vec{r}=(2\hat{i}+2\hat{j}-\hat{k})+\lambda(\hat{i}+2\hat{j}+3\hat{k})+\mu(5\hat{i}-2\hat{j}+7\hat{k}) \\ \therefore \vec{n}=(\hat{i}+2\hat{j}+3\hat{k}) \times (5\hat{i}-2\hat{j}+7\hat{k}) \\ \text{or,}~~ \vec{n}=\begin{vmatrix} \hat{i} & \hat{j}&\hat{k} \\ 1& 2 &3 \\ 5& -2 & 7 \\ \end{vmatrix}=\hat{i}(14+6)-\hat{j}(7-15)+\hat{k}(-2-10) \\ \text{or,}~~ \vec{n}=20\hat{i}+8\hat{j}-12\hat{k} \\ \therefore~~ \vec{n}=4(5\hat{i}+2\hat{j}-3\hat{k}).$

So, the non-parametric form of the plane is

$\vec{r} \cdot(20\hat{i}+8\hat{j}-12\hat{k})=4(5\hat{i}+2\hat{j}-3\hat{k})\cdot(2\hat{i}+2\hat{j}-\hat{k}) \\ \text{or,}~~ 4\vec{r} \cdot(5\hat{i}+2\hat{j}-3\hat{k})=4(10+4+3) \\ \text{or,}~~ \vec{r} \cdot (5 \hat{i}+2\hat{j}-3\hat{k})=17~~\text{(ans.)}$

#### S N Dey mathematics class 12 solutions -Plane-Ex-5A

6. Find the vector equations of the plane passing through the points $~3\hat{i}+4\hat{j}+2\hat{k},~2\hat{i}-2\hat{j}-\hat{k}~$ and $~7\hat{i}+6\hat{k}.$

Solution.

Let the position vectors of three points lying on the plane be denoted by $~A(3\hat{i}+4\hat{j}+2\hat{k}),~B(2\hat{i}-2\hat{j}-\hat{k})~$ and $~C(7\hat{i}+6\hat{k})~$ respectively.

So, $~\vec{AB} \cdot \vec{AC}~$ lies on the plane .

Therefore, $~\vec{AB} \times \vec{AC}~$ is normal vector to the plane.

$~\vec{AB}=(2\hat{i}-2\hat{j}-\hat{k})-(3\hat{i}+4\hat{j}+2\hat{k})=-\hat{i}-6\hat{j}-3\hat{k}, \\~~\vec{AC}=(7\hat{i}+6\hat{k})-(3\hat{i}+4\hat{j}+2\hat{k})=4\hat{i}-4\hat{j}+4\hat{k}.$

$\therefore~~\vec{AB} \times \vec{AC}\\=\begin{vmatrix} \hat{i} & \hat{j}&\hat{k} \\ -1& -6 &-3 \\ 4& -4 & 4 \\ \end{vmatrix}=\hat{i}(-24-12)-\hat{j}(-4+12)+\hat{k}(4+24) \\ \text{or,}~~ \vec{AB} \times \vec{AC}=-36\hat{i}-8\hat{j}+28\hat{k}=-4(9\hat{i}+2\hat{j}-7\hat{k})$

So, the equation of the plane passing through $~A~$ and normal to $~\vec{AB} \times \vec{AC}~$ is

$-4\vec{r} \cdot (9\vec{i}+2\vec{j}-7\vec{k})=-4(9\vec{i}+2\vec{j}-7\vec{k}) \cdot (3\vec{i}+4\vec{j}+2\vec{k}) \\ \text{or,}~~ \vec{r} \cdot (9\vec{i}+2\vec{j}-7\vec{k})=27+8-14=21~~\text{(ans.)}$

7. Find the equation of the plane passing through the following points :

$(i)~(2,3,4),~(4,-1,2)~$ and $~(-3,5,1).$

Solution.

The equation of the plane which passes through three given points $~(x_1,y_1,z_1),~(x_2,y_2,z_2)~$ and $~(x_3,y_3,z_3)~$ is

$\begin{vmatrix} x-x_1 & y-y_1&z-z_1 \\ x_2-x_1& y_2-y_1 & z_2-z_1 \\ x_3-x_1& y_3-y_1 & z_3-z_1 \\ \end{vmatrix}=0 \rightarrow(1)$

Using $(1)$ we get the required equation of the plane

$\begin{vmatrix} x-2& y-3 & z-4 \\ 4-2& -1-3 & 2-4 \\ -3-2& 5-3 & 1-4 \\ \end{vmatrix}=0 \\ \text{or,}~~ \begin{vmatrix} x-2& y-3 & z-4 \\ 2& -4 &-2 \\ -5& 2 & -3 \\ \end{vmatrix}=0 \\ \text{or,}~~ (x-2)(12+4)-(y-3)(-6-10)+(z-4)(4-20)=0 \\ \text{or,}~~ 16(x-2)+16(y-3)-16(z-4)=0 \\ \text{or,}~~ 16[(x-2)+(y-3)-(z-4)]=0 \\ \text{or,}~~ x-3+y-3-z+4=0 \\ \therefore ~ x+y-z=1~~\text{(ans.)}$

$(ii)~(3,3,0),~(1,1,1)~$ and $~(0,-1,0).$

Solution.

The required equation of the plane is given by

$\begin{vmatrix} x-3& y-3 & z-0 \\ 1-3& 1-3 & 1-0 \\ 0-3& -1-3 & 0-0 \\ \end{vmatrix}=0 \\ \text{or,}~~ \begin{vmatrix} x-3& y-3 & z \\ -2& -2 &1 \\-3& -4 &0 \\ \end{vmatrix}=0 \\ \text{or,}~~ (x-3)(0+4)-(y-3)(0+3)+z(8-6)=0 \\ \text{or,}~~ 4(x-3)-3(y-3)+2z=0 \\ \therefore~~ 4x-3y+2z=3~~\text{(ans.)}$

8. Show that the following points are coplanar :

$(i)~ (3,9,4),~(4,5,1),~(-4,4,4)~$ and $~(0,-1,-1)$

Solution.

The equation of the plane passing through the points $~(3,9,4),~(4,5,1),~(-4,4,4)~$ is given by

$\begin{vmatrix} x-3& y-9 & z-4 \\ 4-3& 5-9 & 1-4 \\ -4-3 &4-9 &4-4 \\ \end{vmatrix}=0 \\ \text{or,}~~ \begin{vmatrix} x-3& y-9 &z-4 \\ 1& -4 &-3 \\ -7& -5 &0 \\ \end{vmatrix}=0 \rightarrow(1)$

Now, the point $~(0,-1,-1)~$ lies on the plane  $(1)$ if the point satisfies the equation $(1).$

So, putting $~x=0,~y=-1,~z=-1~$ on the L.H.S. of the equation $(1),~$ we get,

$\begin{vmatrix} 0-3& -1-9 & -1-4 \\ 1& -4 &-3 \\ -7& -5 &0 \\ \end{vmatrix}=\begin{vmatrix} -3& -10 & -5 \\ 1& -4 &-3 \\ -7&-5 & 0 \\ \end{vmatrix}\\~~~=-3(0-15)-(-10)(0-21)-5(-5-28)=45-210+165=0$

Clearly, the point $~(0,-1,-1)~$ satisfies the equation $~(1)~$ and so the point lies on the plane.

Hence, the four points are coplanar.

$(ii)~(-1,-5,-3),~(1,1,-1),~(0,4,3)~$ and $~(-2,-2,1)$

Solution.

The equation of the plane containing three points $~(-1,-5,-3),~(1,1,-1),~(0,4,3)~$ is given by

$\begin{vmatrix} x-(-1) &y-(-5) &z-(-3) \\ 1-(-1)& 1-(-5) &-1-(-3) \\ 0-(-1)&4-(-5) &3-(-3) \\ \end{vmatrix}=0 \rightarrow(1)$

Now, if $~(-2,-2,1)~$ lies on the plane $(1)$ , the given point $~(-2,-2,1)~$ must satisfy the equation $(1).$

Putting $~x=-2,~y=-2,~z=1~$ on the L.H.S.  of the equation $(1)$ we get,

$\begin{vmatrix} -2+1&-2+5 & 1+3 \\ 2& 6 &2 \\ 1& 9 &6 \\ \end{vmatrix}=\begin{vmatrix}-1&3 & 4\\ 2& 6 &2 \\ 1& 9 &6 \\ \end{vmatrix}\\~~=-1(36-18)-3(12-2)+4(18-6)=-18-30+48=0.$

Clearly, the point $~(-2,-2,1)~$ satisfies the equation $~(1)~$ and so the point lies on the plane.

Hence, the four points are coplanar.

9. Find the equation of the plane which makes equal (non zero) intercepts on the axes and passes through the point $~(2,3,-1).$

Solution.

We know that the equation of the plane intercepts on the coordinate axes area  $~a,~b~$ and $~c~$ is given by

$\frac xa+\frac yb +\frac zc=1$

Given that the plane makes equal intercepts on the co-ordinate axes and so, $~a=b=c.$

$\therefore~$ the equation of the plane can be written as

$~\frac xa+\frac ya+\frac za=1 \Rightarrow x+y+z=1 \rightarrow(1)$

Since the plane $(1)$ passes through the point $~(2,3,-1)~$, we get from $(1),~ 2+3-1=a \Rightarrow a=4.$

Hence, the equation of the plane is $~x+y+z=4.$

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