Plane | Part-4 | Ex-5B Examhoop

In the previous article , we have solved few VSA type questions of Plane Chapter (Ex-2B) of S N De Mathematics(Chhaya). In the following article, we are going to discuss/solve Short Answer Type Questions of S.N.Dey Mathematics-Class 12 of the chapter Plane (Ex-5B).

1. Find the equation of the plane passing through the intersection of the planes $~x-2y+z = 1~$ and $~2x+y+z = 8~$ and parallel to the line with direction ratios proportional to $~1, 2, 1.~$ Find also the perpendicular distance of $~(1, 1, 1)~$ from this plane. [CBSE ’05]

Solution.

The given equations of the planes are $~x-2y+z=1 \longrightarrow(1),~~2x+y+z=8 \longrightarrow (2)$

The equation of the plane passing through the intersection of planes (1) and (2) is

$~x-2y+z-1+k(2x+y+z-8)=0 ~~(k \neq 0)\\ ~~~\text{or,}~~ (1+2k)x+ (k-2)y+(1+k)z=1+8k \longrightarrow(3)$

This plane (3) is parallel to the line with direction ratios proportional to $~1,2,1.$

$\therefore~~ (1+2k) \cdot 1+(k-2) \cdot 2+(1+k) \cdot 1=0 \\ \text{or,}~~ 1+2k+2k-4+1+k=0 \\ \text{or,}~~ 5k-2=0 \Rightarrow k=\frac 25.$

$\therefore~~$ The required equation of the plane is

$~\left(1+2 \times \frac 25\right)x+\left(\frac 25-2\right)y+\left(1+\frac 25\right)z=1+8 \times \frac 25 \\~~~ \text{or,}~~ (5+4)x+(2-10)y+(5+2)z=5+16 \\~~~ \therefore~ 9x-8y+7z=21.$

2. Find the equation of the plane parallel to the plane $~2x-2y-z-3 = 0~$ and situated at a distance of $~7~$ units from it.

Solution.

The equation of any plane parallel to the plane $~2x-2y-z=3~$ is $~2x-2y-z=k \longrightarrow(1)$

This plane (1) is situated at a distance $~7~$ units from the given plane .

$\therefore~~ \frac{|k-3|}{\sqrt{2^2+(-2)^2+(-1)^2}}=7 \\ ~~~\text{or,}~~ \frac{|k-3|}{\sqrt{4+4+1}}=7 \\ ~~~\text{or,}~~ \frac 13 |k-3|=7 \\~~~ \text{or,}~~ |k-3|=21 \\ ~~~\text{or,}~~ k-3= \pm 21 \\ ~~~\text{or,}~~ k=3 \pm 21 \\ ~~~\therefore~~ k=24,~~-18.$

So, the equation of the plane is $~~2x-2y-z=24~~$ and $~2x-2y-z=-18.$

3. Find the equation of the plane passing through the intersection of the planes $~2x+y-z=3~$ and $~5x-3y+4z+9=0~~$ and parallel to the line $~~\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}.$

Solution.

$~2x+y-z=3 \longrightarrow(1),~~ 5x-3y+4z+9=0 \longrightarrow(2)$

Equation of plane passing through the intersection of planes (1) and (2) is

$~2x+y-z-3 +k(5x-3y+4z+9)=0~~(k\neq 0) \\ \text{or,}~~ (2+5k)x+(1-3k)y+(4k-1)z+(9k-3)=0 \longrightarrow(3)$

The plane (3) is parallel to the straight line $~ \frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}.$

$\therefore~~ (2+5k) \times 2+(1-3k) \times 4+(4k-1) \times 5=0 \\ \text{or,}~~ 4+10k+4-12k+20k-5=0 \\ \text{or,}~~ 18k+3=0 \Rightarrow k=-\frac{3}{18}=-\frac 16.$

Now, we calculate the following values (from (3)).

$~2+5k=2+5 \times \left(-\frac 16\right)=2-\frac 56=\frac 76,\\~~1-3k=1-3\times \left(-\frac 16\right)=1+\frac 12=\frac 32,\\~~4k-1=4 \times \left(-\frac 16\right)-1=-\frac 23-1=-\frac 53,\\~~9k-3=9 \times \left(-\frac 16 \right)-3=-\frac 32-3=-\frac 92.$

So, the required equation of the plane is given by

$~\frac 76 \cdot x+\frac 32 \cdot y-\frac 53 \cdot z-\frac 92=0 \\ \text{or,}~~ 7x+9y-10z-27=0 \\ \text{or,}~~ 7x+9y-10z=27.$

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4. Find the distance of the point $~(1,-2,3)~$ from the plane $~x-y+z=5~$ measured along a line parallel to $~~\frac x2=\frac y3=\frac{z}{-6}.$

Solution.

The equation of any straight line passing through the point $~(1,-2,3)~$ and parallel to the straight line $~\frac x2=\frac y3=\frac{z}{-6}~$ is

$~\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6} \longrightarrow(1)$

Any point on this straight line (1) can be written as $~(2k+1,3k-2,-6k+3)~$ where $~k~$ is any non-zero real number.

If this point lies on the plane $~x-y+z=5,~$ then

$~(2k+1)-(3k-2)+(-6k+3)=5 \Rightarrow k=\frac 17.$

$\therefore~~$ The point is

$~(2k+1,3k-2,-6k+3)=\left(2 \times \frac 17+1, 3 \times \frac 17-2, -6 \times \frac 17+3\right)=\left(\frac 97,-\frac{11}{7},\frac{15}{7}\right).$

So, the required distance = The distance between $~(1,-2,3)~$ and $~\left(\frac 97,-\frac{11}{7},\frac{15}{7}\right)$

$=\sqrt{\left(\frac 97-1\right)^2+\left(-\frac{11}{7}+2\right)^2+\left(\frac{15}{7}-3\right)^2}\\~~~=\sqrt{\left(\frac{9-7}{7}\right)^2+\left(\frac{-11+14}{7}\right)^2+\left(\frac{15-21}{7}\right)^2}\\~~~=\sqrt{\left(\frac 27\right)^2+\left(\frac 37\right)^2+\left(-\frac 67\right)^2}\\~~~=\frac 17 \sqrt{2^2+3^2+(-6)^2}\\~~~=\frac 17 \sqrt{4+9+36}\\~~~=\frac 17 \times 7=1~~\text{unit}.$

5. Find the length and foot of the perpendicular from the point $~(1,1,2)~$ to the plane $~\vec{r} \cdot (\hat{i}-2\hat{j}+4\hat{k})+5=0.$

Solution.

The equation of the straight line passing through the point $~(1,1,2)~$ and perpendicular to the plane $~ \vec{r} \cdot (2\hat{i}-2\hat{j}+4\hat{k})+5=0$ is

$~\frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{4} \longrightarrow(1)$

The coordinates of any point on the straight line (1) can be written as $~(2k+1,-2k+1,4k+2),~$ where $~k~$ is any non-zero real number.

If this point lies on the given plane , then

$~(2k+1) \cdot 2+ (-2k+1) \cdot (-2)+(4k+2) \cdot 4+5=0 \\ \text{or,}~~ 4k+2+4k-2+16k+8+5=0 \\ \text{or,}~~ 24k+13=0 \Rightarrow k=-\frac{13}{24}.$

Now, the distance between the points $~(1,1,2)~$ and $~(2k+1,-2k+1,4k+2)~$ is

$\sqrt{(2k+1-1)^2+(-2k+1-1)^2+(4k+2-2)^2}\\=\sqrt{(2k)^2+(-2k)^2+(4k)^2}\\=\sqrt{k^2} \cdot \sqrt{2^2+(-2)^2+4^2}\\=\sqrt{\left(-\frac{13}{24}\right)^2} \cdot \sqrt{24}\\=\frac{13}{24} \times \sqrt{24}=\frac{13 \times 2\sqrt{6}}{24}=\frac{13}{12} \sqrt{6}~~\text{unit}$

The coordinates of the foot of the perpendicular is

$(2k+1,-2k+1,4k+2)\\=\left(2 \times \frac{-13}{24}+1, -2 \times \frac{-13}{24}+1,4 \times \frac{-13}{24}+2\right)\\=\left(1-\frac{13}{12},1+\frac{13}{12},2-\frac{13}{6}\right)\\=\left(-\frac{1}{12},\frac{25}{12},-\frac 16\right)~~\text{(ans.)}$

6. Find the equation of the plane that contains the line of intersection of the planes $~\vec{r} \cdot (\hat{i}+2\hat{j}+3\hat{k})-4=0~$ and $~\vec{r} \cdot (2\hat{i}+\hat{j}-\hat{k})+5=0~$ and which is perpendicular to the plane $~\vec{r} \cdot (5\hat{i}+3\hat{j}-6\hat{k})+8=0~$ .

Solution.

$\vec{r} \cdot (\hat{i}+2\hat{j}+3\hat{k})-4=0 \longrightarrow (1),~~\vec{r} \cdot (2\hat{i}+\hat{j}-\hat{k})+5=0 \longrightarrow(2)$

The equation of the plane that contains the line of intersection of planes (1) and (2) can be written as

$~[\vec{r} \cdot (\hat{i}+2\hat{j}+3\hat{k})-4]+\lambda [\vec{r} \cdot (2\hat{i}+\hat{j}-\hat{k})+5]=0 \\ \text{or,}~~ \vec{r} \cdot [(1+2\lambda)\hat{i}+(2+\lambda)\hat{j}+(3-\lambda)\hat{k}]+(5\lambda-4)=0 \longrightarrow(3)$

The plane (3) is perpendicular to the plane $~\vec{r} \cdot (5\hat{i}+3\hat{j}-6\hat{k})+8=0.$

$\therefore~~ (1+2\lambda) \cdot 5+(2+\lambda) \cdot 3+(3-\lambda) \cdot (-6)=0 \\ \text{or,}~~ 5+10\lambda+6+3\lambda-18+6\lambda=0 \\ \text{or,}~~ 19\lambda-7=0 \Rightarrow \lambda=\frac{7}{19}.$

Now, we calculate the following rules (from (3)).

$~ 1+2\lambda=1 +2 \times \frac{7}{19}=1+\frac{14}{19}=\frac{33}{19},\\~~2+\lambda=2+\frac{7}{19}=\frac{38+7}{19}=\frac{45}{19},\\~~ 3-\lambda=3-\frac{7}{19}=\frac{57-7}{19}=\frac{50}{17},\\~~ 5\lambda-4=5 \times \frac{7}{19}-4=\frac{35-76}{19}=-\frac{41}{19}.$

After substituting the aforesaid values in (3) we get the required equation of the plane as follows .

$\vec{r} \cdot \left[\frac{33}{19}\hat{i}+\frac{45}{19}\hat{j}+\frac{50}{19}\hat{k}\right]-\frac{41}{19}=0 \\ \text{or,}~~ \vec{r} \cdot (33\hat{i}+45\hat{j}+50\hat{k})-41=0 \longrightarrow(4)$

Putting $~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k},~$ we get from (4),

$~(x\hat{i}+y\hat{j}+z\hat{k}) \cdot (33 \hat{i}+45\hat{j}+50\hat{k})=41 \\ \text{or,}~~ 33x+45y+50z=41.$

7. Find the distance between the point $~P(6,5,9)~$ and the plane passing through $~~A(3,-1,2),~ B(5,2,4)~$ and $~~C(-1,-1,6).$

Solution.

The equation of the plane passing through the points $~A(3,-1,2),~B(5,2,4)~$ and $~C(-1,-1,6)~$ is

$\begin{vmatrix} x-3& y+1 & z-2 \\ 5-3& 2+1 &4-2 \\ -1-3&-1+1 &6-2 \\ \end{vmatrix}=0 \\ \text{or,}~~ \begin{vmatrix} x-3 & y+1 &z-2 \\ 2& 3 & 2 \\ -4& 0 &4 \\ \end{vmatrix}=0 \\ \text{or,}~~ -4\begin{vmatrix} y+1& z-2 \\ 3&2 \\ \end{vmatrix} +4\begin{vmatrix} x-3& y+1 \\ 2& 3 \\ \end{vmatrix} =0 \\ \text{or,}~~ -4(2y+2-3z+6)+4(3x-9-2y-2)=0 \\ \text{or,}~~ -(2y-3z+8)+(3x-2y-11)=0 \\ \text{or,}~~ 3x-2y-11-(2y-3z+8)=0 \\ \text{or,}~~ 3x-4y+3z-19=0 \longrightarrow(1)$

$\therefore~~$ Distance of the point $~P(6,5,9)~$ from the plane (1) is

$=\frac{|3 \times 6-4 \times 5+ 3\times 9-19|}{\sqrt{3^2+(-4)^2+3^2}}=\frac{|6|}{\sqrt{34}}=\frac{6}{\sqrt{34}}~~\text{unit.}$

8. Find the equation of the plane through the points $~(5,2,4),~(-1,-1,6)~$ and $~(3,-1,2)~$ and find the distance of this plane from the point $~(6,5,9).$ \

Hints : Follow Question No. 7

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9. Find the equation of the plane through the points $~(3,2,2),~(1,0,-1)~$ and parallel to the line $~\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-2}{3}.~$ Convert to vector form.

Solution.

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

$~a(x-3)+b(y-2)+c(z-2)=0 \longrightarrow (1)$

The plane (1) is passing through the point $~(1,0,-1).$

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

Since the plane is parallel to the straight line $~~\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-2}{3},$

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

Eliminating $~a,~b,~c~$ from (1), (2), (3) we get,

$\begin{vmatrix} x-3& y-2 & z-2 \\ 2& 2 & 3 \\ 1& -2 & 3 \\ \end{vmatrix}=0 \\ \text{or,}~~ (x-3) \begin{vmatrix} 2& 3 \\ -2& 3 \\ \end{vmatrix} -(y-2) \begin{vmatrix} 2 &3 \\ 1& 3 \\ \end{vmatrix}+(z-2)\begin{vmatrix} 2& 2 \\ 1& -2 \\ \end{vmatrix} =0 \\ \text{or,}~~ (x-3)(6+6)-(y+2)(6-3)+(z-2)(-4-2)=0 \\ \text{or,}~~ 12(x-3)-3(y-2)-6(z-2)=0 \\ \text{or,}~~ 4(x-3)-(y-2)-2(z-2)=0 \\ \text{or,}~~ 4x-12-y+2-2z+4=0 \\ \text{or,}~~ 4x-y-2z=6 \longrightarrow(4)$

$\therefore~~$ The vector equation of (4) is

$\vec{r} \cdot (4\hat{i}-\hat{j}-2\hat{k})=6,~~$ where $~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}.$

10. Show that the lines $~\frac{x-a+d}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-a-d}{\alpha+\delta}~$ and $~\frac{x-b+c}{\beta-\gamma}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\gamma}~$ are coplanar.

Solution.

$~\frac{x-a+d}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-a-d}{\alpha+\delta} \\ \text{or,}~~ \frac{z-(a-d)}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-(a+d)}{\alpha+\delta} \longrightarrow(1)$

The straight line (1) passes through the point $~A(a-d,a,a+d)~$ and is parallel to $~~\vec{p}=(\alpha-\delta)\hat{i}+\alpha\hat{j}+(\alpha+\delta)\hat{k}.$

$\frac{a-b+c}{\beta-\delta}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\gamma} \\ \text{or,}~~ \frac{x-(b-c)}{\beta-\gamma}=\frac{y-b}{\beta}=\frac{z-(b+c)}{\beta+\gamma} \longrightarrow(2)$

Again, the straight line (2) passes through the point $~B(b-c,b,b+c)~$ and is parallel to $~\vec{q}=(\beta-\gamma)\hat{i}+\beta \hat{j}+(\beta+\gamma)\hat{k}.$

Clearly, given two straight lines are coplanar whenever $~\vec{AB},~\vec{p}~$ and $~\vec{q}~$ are coplanar.

$\therefore~~ \begin{vmatrix} (b-c)-(a-d) & b-a &(b+c)-(a+d) \\ \alpha-\delta& \alpha &\alpha+\delta \\ \beta-\gamma &\beta &\beta+\gamma \\ \end{vmatrix}\\= \begin{vmatrix} d-c & b-a &c-d \\ -\delta& \alpha &\delta \\ -\gamma& \beta &\gamma \\ \end{vmatrix},~~[C_1' \rightarrow C_1-C_2,~~C_3' \rightarrow C_3-C_2]\\= \begin{vmatrix} 0& b-a & c-d \\ 0& \alpha & \delta \\ 0& \beta & \gamma \\ \end{vmatrix}=0 \\ \therefore~~[\vec{AB}~~ \vec{p}~~\vec{q}]=0.$

$\therefore~~ ~\vec{AB},~\vec{p},~\vec{q}~$ are coplanar and hence the result follows.

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In the previous article , we have solved few VSA type questions of Plane Chapter (Ex-2B) of S N De Mathematics(Chhaya). In the following article, we are going to discuss/solve Short Answer Type Questions of S.N.Dey Mathematics-Class 12 of the chapter Plane (Ex-5B).

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