Plane | Part-6 | Ex-5A

In the following article, we are going to discuss/solve few Short Answer Type Questions of S.N.Dey Mathematics-Class 12 of the chapter Plane. In the previous article , we have solved few Short Answer type Questions.
Plane | Part-6 | Ex-5A
Plane | Part-6 | Ex-5A

10. If the points ~(2-x, 2, 2), ~(2, 2-y, 2)~ and ~(2, 2, 2-z)~ are coplanar then prove that ~\frac 2x+\frac 2y+\frac 2z=1.


If the points ~(2-x,2,2),~(2,2-y,2)~ and ~(2,2,2-z)~ are coplanar then

\begin{vmatrix} 2-x& 2 & 2 \\ 2& 2-y & 2 \\ 2& 2 &2-z  \\ \end{vmatrix}=0 \\ \text{or,}~~\begin{vmatrix} -x & 0 &z  \\ 0& -y &z  \\ 2& 2 &2-z  \\ \end{vmatrix}=0~~[\text{By}~~R_1' \rightarrow R_1-R_3,~R_2' \rightarrow R_2-R_3] \\ \text{or,}~~ -x\begin{vmatrix} -y& z \\ 2& 2-z \\ \end{vmatrix}+z\begin{vmatrix} 0 & -y \\ 2& 2 \\ \end{vmatrix} =0  \\ \text{or,}~~-x[-y(2-z)-2z]+z[0+2y]=0 \\ \text{or,}~~ xy(2-z)+2zx+2yz=0 \\ \text{or,}~~ 2xy+2yz+2zx=xyz \\ \text{or,}~~\frac{2xy+2yz+2zx}{xyz}=1 \\ \therefore~~\frac 2x+\frac 2y+\frac 2z=1~~\text{(proved)}

11. Let ~\vec{n}~ be a vector of magnitude ~2\sqrt{3}~ such that it makes equal acute angles with the coordinate axes. Find the vector and cartesian forms of the equation of a plane passing through ~(1, -1, 2)~ and normal to ~\vec{n}.


Suppose that the vector ~\vec{n}~ makes an angle ~\theta~(\leq 90^{\circ})~ with the coordinate axes.

\therefore~ \vec{n}=2\sqrt{3}(\hat{i}\cos\theta+\hat{j}\cos\theta+\hat{k}\cos\theta)

 \text{Again,}~\cos^2\theta+\cos^2\theta+\cos^2\theta=1 \\ \text{or,}~~ 3\cos^2\theta=1 \Rightarrow \cos\theta=\frac{1}{\sqrt{3}}~~(\because~\theta \leq 90^{\circ})

\therefore~~\vec{n}=2\sqrt{3} \left(\frac{\hat{i}}{\sqrt{3}}+\frac{\hat{j}}{\sqrt{3}}+\frac{\hat{k}}{\sqrt{3}}\right) \\ \text{or,}~~\vec{n}=2\hat{i}+2\hat{j}+2\hat{k}.

So, the equation of the plane passing through ~(1,-1,2)~ and normal to ~\vec{n}~ is given by 

~\vec{n} \cdot [\vec{r}-(\hat{i}-\hat{j}+2\hat{k})]=0 \\ \text{or,}~~ (2\hat{i}+2\hat{j}+2\hat{k}) \cdot [\vec{r}-(\hat{i}-\hat{j}+2\hat{k})]=0 \\ \text{or,}~~ (2\hat{i}+2\hat{j}+2\hat{k})\cdot \vec{r}-(2\hat{i}+2\hat{j}+2\hat{k}) \cdot (\hat{i}-\hat{j}+2\hat{k})=0 \\ \text{or,}~~ 2\vec{r} \cdot (\hat{i}+\hat{j}+\hat{k})-(2-2+4)=0 \\ \text{or,}~~2\vec{r} \cdot (\hat{i}+\hat{j}+\hat{k})-4=0 \\ \text{or,}~~ \vec{r} \cdot (\hat{i}+\hat{j}+\hat{k})=2 \rightarrow(1)

So, the equation ~(1)~ represents the vector equation of the plane. Now, putting ~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~ in (1) we get,

~(x\hat{i}+y\hat{j}+z\hat{k}) \cdot (\hat{i}+\hat{j}+\hat{k})=2 \Rightarrow~~ x+y+z=2.

\therefore~~ x+y+z=2~~ represents the cartesian equation of the plane.

12. Through the point ~P(\alpha,\beta,\gamma)~ a plane is drawn perpendicular to ~OP~ where ~O~ is the origin. Let this plane meet the coordinate axes at ~L,~M~ and ~N.~ Show that the area of the triangle ~LMN=\frac{r^5}{(2\alpha \beta \gamma)}~ where ~|\vec{OP}|=r.


The direction ratios of ~\vec{OP}~ are ~ \alpha-0,~\beta-0,~\gamma-0~ or, ~\alpha,\beta,\gamma.

\therefore~|\vec{OP}|=r \Rightarrow \sqrt{\alpha^2+\beta^2+\gamma^2}=r \rightarrow(1)

Again, ~OP~ is normal to the plane . The equation of the plane through the point ~(\alpha,\beta,\gamma)~ and perpendicular to the plane is 

~\alpha(x-\alpha)+\beta(y-\beta)+\gamma(z-\gamma)=0 \\ \text{or,}~~ \alpha x-\alpha^2+\beta y-\beta^2+\gamma z-\gamma^2=0 \\ \text{or,}~~ \alpha x+\beta y+\gamma z=\alpha^2+\beta^2+\gamma^2 \\ \text{or,}~~ \alpha x+\beta y+\gamma z=r^2~[\text{By (1)}] \\ \therefore~~ \frac{x}{r^2/\alpha}+\frac{y}{r^2/\beta}+\frac{z}{r^2/\gamma}=1 \\ \therefore~~ L \equiv \left(\frac{r^2}{\alpha},0,0\right),~~ M \equiv \left(0,\frac{r^2}{\beta},0\right),~N \equiv \left(0,0,\frac{r^2}{\gamma}\right).

So, the direction ratios of ~OL, ~OM~ and ~ON~ are ~~\frac{r^2}{\alpha},0,0;~~0,\frac{r^2}{\beta},0~~\text{and}~~0,0,\frac{r^2}{\gamma}~ respectively.

The Volume of ~OLMN 

\frac 16\begin{vmatrix}  \frac{r^2}{\alpha}&0  &0  \\ 0& \frac{r^2}{\beta} &0  \\ 0& 0 &\frac{r^2}{\gamma}  \\ \end{vmatrix}=\frac 16\cdot \frac{r^2}{\alpha} \cdot \frac{r^2}{\beta} \cdot \frac{r^2}{\gamma}=\frac{r^6}{6\alpha \beta \gamma}

Again, the volume of  ~~OLMN=\frac 13 \times OP \times \Delta LMN

\therefore~ \frac{r^6}{6\alpha \beta \gamma}=\frac 13 \times r \times \Delta LMN \\ \text{or,}~~ \Delta LMN=\frac{r^5}{2\alpha \beta \gamma}~~(\text{proved})

13. A variable plane which is at a constant distance ~3p~ from the origin ~O~ cuts the axes at ~L,~M~ and ~N.~ Show that the locus of the point of intersection of the planes through ~L,~M,~N~ drawn parallel to the coordinate axes parallel to the coordinate planes is ~9(x^{-2}+y^{-2}+z^{-2})=p^{-2}.


Let ~P(\alpha,\beta,\gamma)~ be the point of intersection of three planes passing through the points ~L,~M~ and ~N.

The equation of planes passing through ~P~ and parallel to ~xy,~yz~ and ~zx~ planes are ~z=\gamma,~x=\alpha~ and ~y=\beta~ respectively and these planes meet z-axis, x-axis and y-axis at ~N(0,0,\gamma),~L(\alpha,0,0)~ and ~M(0,\beta,0)~ respectively.

The equation of the plane through the points ~L,M,N~ can be written as 

~\frac{x}{\alpha}+\frac y\beta +\frac z\gamma=1 \longrightarrow(1)

Now, the distance of the plane (1) from the origin is ~3p~ unit.

\therefore~ \frac{\left|\frac 0\alpha+\frac 0\beta+\frac 0\gamma -1\right|}{\sqrt{\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}}}=3p \\ \text{or,}~~ 3p \sqrt{\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}}=1 \\ \text{or,}~~ 9p^2\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)=1 \\ \text{or,}~~ 9(\alpha^{-2}+\beta^{-2}+\gamma^{-2})=p^{-2}.

Hence, the locus of the point ~P(\alpha,\beta,\gamma)~ is ~9(x^{-2}+y^{-2}+z^{-2})=p^{-2}~~(\text{proved})

14. A variable plane passes through the point ~(f,g,h)~ and meets the axes at ~L,M~ and ~N. If the planes through ~L,M,N~ and parallel to the axes meet at ~P~, then prove that the locus of ~P~ is ~\frac fx+\frac gy+\frac hz=1. 


Let the equation of the plane be ~\frac x\alpha+\frac y\beta+\frac z\gamma=1.~ If this plane passes through ~(f,g,h),~ then

~\frac f\alpha+\frac g\beta+ \frac h\gamma=1 \longrightarrow(1)

\therefore~ L \equiv (\alpha,0,0),~~M \equiv (0,\beta,0),~~N \equiv (0,0,\gamma).

The equation of the planes through ~L,M,N~ are ~ x=\alpha,~y=\beta,~ z=\gamma~ respectively.

So, using (1) we get, ~\frac fx+\frac hy+\frac hz=1\rightarrow(2)

So, from (2) we get the locus of P and hence the result follows. (proved)

15. A point moves on the plane ~\frac xa+\frac yb+\frac zc=1.~ The plane, drawn through perpendicular to ~OP~ meets the axes in ~L,M,N.~ If the planes through ~L,M,N,~ parallel to the coordinate planes meet in a point ~Q,~ then show that the locus of ~Q~ is given by the equation :



Let ~Q \equiv (\alpha,\beta,\gamma).~ So, ~L\equiv (\alpha,0,0),~~M \equiv (0,\beta,0)~ and ~N\equiv (0,0,\gamma).

The equation of any plane through the points ~L,M~ and ~N~ is given by  ~~\frac x\alpha+\frac y\beta+\frac z\gamma=1 \longrightarrow(1)

Let ~~P \equiv (x_1,y_1,z_1).~ So, the direction ratios of ~\vec{OP}~ are ~x_1-0,~y_1-0,~z_1-0~ i.e., ~x_1,y_1,z_1.

Now, ~\vec{OP}~ is normal to the plane ~(1).

\therefore~ \frac{x_1}{\frac 1\alpha}=\frac{y_1}{\frac 1\beta}=\frac{z_1}{\frac 1\gamma}=k~~(\neq 0,\text{say})

\therefore~~x_1=\frac k\alpha,~~y_1=\frac k\beta,~~z_1=\frac k\gamma \longrightarrow(2)

The point ~P~ lies on the plane ~\frac xa+\frac yb+\frac zc=1.

\therefore~~ \frac{x_1}{a}+\frac{y_1}{b}+\frac{z_1}{c}=1 \\ \text{or,}~~ \frac{k}{a\alpha}+\frac{k}{b\beta}+\frac{k}{c\gamma}=1~~[\text{By (2)}] \\ \therefore~~ \frac 1k=\frac{1}{a\alpha}+\frac{1}{b\beta}+\frac{1}{c\gamma} \longrightarrow(3)

Again, the point ~P~ lies on the plane  ~(1).

\therefore~~ \frac{x_1}{\alpha}+\frac{y_1}{\beta}+\frac{z_1}{\gamma}=1 \\ \text{or,}~~ \frac{k}{\alpha^2}+\frac{k}{\beta^2}+\frac{k}{\gamma^2}=1~~[\text{Using (2)}] \\ \text{or,}~~ \frac 1k=\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2} \longrightarrow(4)

Hence, from (3) and ~(4)~ we get,

~\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}=\frac{1}{a\alpha}+\frac{1}{b\beta}+\frac{1}{c\gamma} \\ \text{or,}~~~ \alpha^{-2}+\beta^{-2}+\gamma^{-2}=\frac{1}{a\alpha}+\frac{1}{b\beta}+\frac{1}{c\gamma}.

Hence, the locus of ~(\alpha,\beta,\gamma)~ is 


Leave a Comment