Plane | Part-5 | Ex-5A Examhoop

In the following article, we are going to discuss/solve few Short Answer Type Questions of S.N.Dey Mathematics-Class 12 . In the previous article , we have completed solution of VSA type Questions.

1. Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point $~P(3, 2, 1)~$ from the plane $~2x-y+z+1=0.~$ Find also the image of the point in the plane.

Solution.

Let $~O(\alpha,\beta,\gamma)~$ be the image of the point $~P(3,2,1)~$ in the plane $~2x-y+z+1=0.$

PO is perpendicular to the plane and S is the midpoint of PO and the foot of the perpendicular.

Direction ratios of PS are $~2,-1,1.$

$\therefore~$ the equation of PS are

$~\frac{x-3}{2}=\frac{y-2}{-1}=\frac{z-1}{1}=\mu(\neq 0)$

The general point on line is $~S(2\mu+3,-\mu+2,\mu+1).$

If this point lies on plane, then

$~2(2\mu+3)-(-\mu+2)+1(\mu+1)+1=0 \\ \text{or,}~~ 6\mu+6=0 \Rightarrow \mu=-1$

So, the coordinates of S are $~\{2(-1)+3,-(-1)+2,-1+1\}=(1,3,0).$

As S is the midpoint of PO $~\left(\frac{3+\alpha}{2},\frac{2+\beta}{2},\frac{1+\gamma}{2}\right)=(1,3,0).$

By comparing both sides , we get

$~\frac{3+\alpha}{2}=1 \Rightarrow \alpha=2-3=-1,~~ \frac{2+\beta}{2}=3 \Rightarrow \beta=6-2=4,~~ \frac{1+\gamma}{2}=0 \Rightarrow \gamma=-1.$

So, the image of the point $~P(-1,4,-1).$

$\therefore~$ Perpendicular distance between two points is

$~d=\sqrt{(1-3)^2+(3-2)^2+(0-1)^2}~\text{units}=\sqrt{4+1+1}~\text{units}=\sqrt{6}~\text{units}$

Read More : Complete S N Dey Solution of Plane (Ex-5A) Chapter with PDF link

2. Show that, $\left(-\frac 12, 2, 0\right)~$ is the circum-centre of the triangle formed by the points $~(1, 2, 1), ~(-2, 2, -1)~$ and $~(1, 1, 0).$

Solution.

Let $~A \equiv (1,2,1),~ B \equiv (-2,2,-1),~C \equiv (1,1,0).$

$\therefore~ AB =\sqrt {(-2-1)^2+(2-2)^2+(-1-1)^2}=\sqrt{13},\\~~BC=\sqrt{(1+2)^2+(1-2)^2+(0+1)^2}=\sqrt{11},\\~~CA=\sqrt{(1-1)^2+(2-1)^2+(1-0)^2}=\sqrt{2}.$

$\therefore~ AB^2=BC^2+CA^2.$

$\therefore\Delta ABC~$ is a right angled triangle with hypotenuse $AB.$

So, circum centre of the triangle is the midpoint of AB.

$\therefore~$ the circum-centre of the triangle is

$~\left(\frac{1-2}{2},\frac{2+2}{2},\frac{1-1}{2}\right)=\left(-\frac 12,2,0\right).$

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3. Find the equations of the three planes which are parallel to the coordinate axes and which pass through the points $~(3, 1, 1)~$ and $~(1, -2, 3).$

Solution.

For Planes parallel to x-axis :

The direction ratios of x-axis are $~1,0,0.$ So, the equation of the plane parallel to the x-axis and passing through the points $~(3,1,1)~$ and $~(1,-2,3)~$ is

$\begin{vmatrix} x-3& y-1 &z-1 \\ 3-1& 1+2 &1-3 \\ 1&0 &0 \\ \end{vmatrix}=0 \\ \text{or,}~~ \begin{vmatrix} x-3& y-1 & z-1 \\ 2& 3 &-2 \\ 1&0 &0 \\ \end{vmatrix}=0 \\ \text{or,}~~ \begin{vmatrix} y-1& z-1 \\ 3& -2 \\ \end{vmatrix} =0 \\ \text{or,}~~ -2(y-1)-3(z-1)=0 \\ \text{or,}~~ -2y+2-3z+3=0 \\ \text{or,}~~ -2y-3z=-5 \Rightarrow 2y+3z=5.$

For Planes parallel to y-axis :

The direction ratios of y-axis are $~0,1,0.$ So, the equation of the plane parallel to the y-axis and passing through the points $~(3,1,1)~$ and $~(1,-2,3)~$ is

$\begin{vmatrix} x-3 & y-1 & z-1 \\ 3-1& 1+2 & 1-3 \\ 0& 1 &0 \\ \end{vmatrix}=0 \\ \text{or,}~~\begin{vmatrix} x-3 &y-1 &z-1 \\ 2 &3 &-2 \\ 0& 1 &0 \\ \end{vmatrix}=0 \\ \text{or,}~~- \begin{vmatrix} x-3& z-1 \\ 2& -2 \\ \end{vmatrix} =0 \\ \text{or,}~~-[-2(x-3)-2(z-1)]=0 \\ \text{or,}~~ 2(x-3)+2(z-1)=0 \\ \text{or,}~~ x-3+z-1=0 \Rightarrow x+z=4.$

For Planes parallel to z-axis :

The direction ratios of z-axis are $~0,0,1.$ So, the equation of the plane parallel to the z-axis and passing through the points $~(3,1,1)~$ and $~(1,-2,3)~$ is

$\begin{vmatrix} x-3 & y-1 & z-1 \\ 3-1& 1+2 & 1-3 \\ 0& 0 &1 \\ \end{vmatrix}=0 \\ \text{or,}~~\begin{vmatrix} x-3 &y-1 &z-1 \\ 2 &3 &-2 \\ 0& 0 &1 \\ \end{vmatrix}=0 \\ \text{or,}~~ \begin{vmatrix} x-3& y-1 \\ 2& 3 \\ \end{vmatrix} =0 \\ \text{or,}~~ 3(x-3)-2(y-1)=0 \\ \text{or,}~~ 3x-9-2y+2=0 \\ \text{or,}~~ 3x-2y=7.$

4. Let $~(2, 3, -1)~$ be the coordinates of the foot of the perpendicular drawn from the origin to a plane. Find equation of that plane.

Solution.

Since $~(2,3,-1)~$ is the coordinates of the foot of the perpendicular drawn from the origin to the plane, the direction ratios of the normal to the plane are $~ 2-0,~3-0,~ -1-0~$ or, $~2,3,-1.$

Let $~2x+3y-z+d=0~$ be the equation of the plane which passes through the point $~(2,3,-1).$

$\therefore~ 2\times 2+3 \times 3-(-1)+d=0 \\ \text{or,}~~ 4+9+1+d=0 \Rightarrow d=-14.$

So, the equation of the plane is given by $~2x+3y-z=14.$

5. Show that the equation of the plane passing through the point (1, 2, 3) and parallel to the plane 3x+4y-5z=3 is given by 3x+4y-5z=-4.

Solution.

The equation of the plane parallel to $~3x+4y-5z=3~$ can be written as $~3x+4y-5z=d\longrightarrow(1)$

Since the plane (1) passes through the point $~(1,2,3)~$ ,

$~3 \times 1+4 \times 2-5 \times 3=d \\ \text{or,}~~ 3+8-15=d \\ \text{or,}~~ d=11-15=-4.$

So, the equation of the plane is given by $~3x+4y-5z=-4.$

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6. Find the equation of the plane passing through the points $~(1, 1, 2)~$ and $~(2, 4, 3)~$ and perpendicular to the plane $~x-3y+7z=6,~$ convert the equation to vector form.

Solution.

The equation of any plane passing through the point $~(1,1,2)~$ is

$~a(x-1)+b(y-1)+c(z-2)=0 \longrightarrow(1),~~$ where $~a,b,c~$ are constants.

Since the plane (1) passes through the point $~(2,4,3),$

$a(2-1)+b(4-1)+c(3-2)=0 \\ \text{or,}~~ a+3b+c=0 \longrightarrow(2)$

Since the plane (1) is perpendicular to the plane $~x-3y+7z=6,~$ so

$a-3b+7c=0 \longrightarrow(3)$

From $~(2)~$ and $~(3)~$ , we get by cross-multiplication,

$~\frac{a}{21+3}=\frac{b}{1-7}=\frac{c}{-3-3} \\ \text{or,}~~ \frac{a}{24}=\frac{b}{-6}=\frac{c}{-6} \\ \text{or,}~~ \frac a4=\frac{b}{-1}=\frac{c}{-1}=k(\neq 0,\text{say}) \\ \therefore~~ a=4k,~b=-k,~c=-k.$

Hence, by (1) we get the cartesian equation of plane as follows :

$~4k(x-1)-k(y-1)-k(z-2)=0 \\ \text{or,}~~ 4(x-1)-1(y-1)-(z-2)=0 \\ \text{or,}~~ 4x-4-y+1-z+2=0 \\ \text{or,}~~ 4x-y-z=1.$

Hence, the vector equation of the plane is

$~(x\hat{i}+y\hat{j}+z\hat{k}) \cdot (4\hat{i}-\hat{j}-\hat{k})=1 \\ \text{or,}~~ \vec{r} \cdot (4\hat{i}-\hat{j}-\hat{k})=1.$

7. Prove that the equation of the plane which passes through the points $~(1, 2, 3)~$ and $~(3, 2, -1)~$ and which is perpendicular to the plane $~3x+2y+6z+4=0~$ is $~2x-6y+z+7=0.$

Solution.

Equation of any plane passing through the point $~(1,2,3)~$ is $~a(x-1)+b(y-2)+c(z-3)=0 \longrightarrow(1),~$ where $a,b,c$ are constants.

If the plane (1) passes through the point $~(3,2,-1),$

$~a(3-1)+b(2-2)+c(-1-3)=0 \Rightarrow 2a+0 \cdot b-4c=0 \longrightarrow(2)$

Again, the plane (1) is perpendicular to the given plane $~3x+2y+6z=-4,$

$\therefore~ 3a+2b+6c=0 \longrightarrow(3)$

From (2) and (3), we get by cross-multiplication,

$\frac{a}{0+8}=\frac{b}{-12-12}=\frac{c}{4-0} \\ \text{or,}~~ \frac a8=\frac{b}{-24}=\frac c4 \\ \text{or,}~~ \frac a2=\frac{b}{-6}=\frac c1=k~(\neq 0,\text{say}) \\ \therefore~ a=2k,~b=-6k,~c=k.$

Hence, the equation of the plane is

$2k(x-1)-6k(y-2)+k(z-3)=0 \\ \text{or,}~~ 2(x-1)-6(y-2)+(z-3)=0 \\ \text{or,}~~ 2x-2-6y+12+z-3=0 \\ \therefore~~ 2x-6y+z+7=0~~\text{(proved)}$

8. Prove that the equation of the plane which passes through the point $~(-1, 3, 2)~$ and is perpendicular to the planes $~x+2y+22=5~$ and $~3x+3y+2z-8~$ is $~2x-4y+3z+8=0.$

Solution.

The equation of any plane passing through the point $~(-1,3,2)~$ is

$a(x+1)+b(y-3)+c(z-2)=0\longrightarrow(1),~~$ where $~a,~b,~c~$ are constants.

Since the plane (1) is perpendicular to the planes $~x+2y+2z=5~$ and $~3x+3y+2z=8,~$ so

$~a+2b+2c=5 \rightarrow (2),~~~3a+3b+2c=8 \rightarrow(3)$

Now, from (2) and (3) we get by cross-multiplication,

$~\frac{a}{4-6}=\frac{b}{6-2}=\frac{c}{3-6} \\ \text{or,}~~ \frac{a}{-2}=\frac{b}{4}=\frac{c}{-3}=k(\neq 0,\text{(say)} \\ \therefore~~ a=-2k,~~b=4k,~c=-3k.$

So, the equation of the plane is

$-2k(x+1)+4k(y-3)-3k(z-2)=0 \\ \text{or,}~~ -2(x+1)+4(y-3)-3(z-2)=0 \\ \text{or,}~~ -2x-2+4y-12-3z+6=0 \\ \text{or,}~~ -2x+4y-3z-8=0 \\ \therefore~ 2x-4y+3z+8=0~~\text{(proved)}$