Plane | Part-3 | Ex-5A Examhoop

*S N Dey mathematics class 12 solutions -Plane-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,

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