Skip to content
Related Articles

Related Articles

Improve Article

Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.12

  • Last Updated : 28 Mar, 2021
Geek Week

Question 1(i): Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses line through yz-plane.

Solution:

We know, the equation of line through the points (x1,y1,z1) and (x2,y2,z2) is \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}

Therefore, equation of line joining (5, 1, 6) and (3, 4, 1) is \frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}

\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}

Let, \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda, where \lambda is a constant.



⇒ x=-2\lambda+5, y=3\lambda+1, z=-5\lambda+6

Coordinates of any point on the line is in the form of (-2\lambda+5,3\lambda+1,-5\lambda+6)

Since, the line crosses the yz-plane, the point (-2\lambda+5,3\lambda+1,-5\lambda+6) must satisfy the equation of plane x=0,

-2\lambda+5=0 ⇒ \lambda=\frac{5}{2}    

Therefore, coordinates of points is given by, putting \lambda=\frac{5}{2} we get,

⇒ (0,\frac{17}{2},\frac{13}{2})

(ii) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses line through zx-plane.

Solution:

We know, the equation of line through the points (x1,y1,z1) and (x2,y2,z2) is \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}



Therefore, equation of line joining (5, 1, 6) and (3, 4, 1) is \frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}

 \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}

Let, \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda ,where \lambda is a constant.

⇒ x=-2\lambda+5, y=3\lambda+1, z=-5\lambda+6

Coordinates of any point on the line is in the form of (-2\lambda+5,3\lambda+1,-5\lambda+6)

Since, the line crosses the zx-plane, the point (-2\lambda+5,3\lambda+1,-5\lambda+6) must satisfy the equation of plane y=0,

⇒ 3\lambda+1=0 ⇒ \lambda=\frac{-1}{3}

Therefore, the coordinates of point is given by, putting \lambda=\frac{-1}{3} we get,

⇒ (\frac{17}{3},0,\frac{23}{3})

Question 2: Find the coordinates of point where the line through (3, -4, -5) and (2, -3, 1) crosses the plane 2x+y+z=7.

Solution:



We know, the equation of line through the points (x1,y1,z1) and (x2,y2,z2) is \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}

Therefore, line joining the points (3, -4, -5) and (2, -3, 1) is \frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}

⇒ \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}

Let \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda where \lambda    is constant.

⇒ x=-\lambda+3,y=\lambda-4,z=6\lambda-5

The coordinates of any point on the line is given by (-\lambda+3,\lambda-4,6\lambda-5)

The line crosses the plane, therefore, point must satisfy the plane equation.

2(-\lambda+3)+\lambda-4+6\lambda-5=7

⇒ \lambda=2

Therefore, The coordinates of point are given by, putting \lambda=2,



⇒ (-2+3, 2-4, 6(2)-5)  

(1, -2, 7)

Question 3: Find the distance of the point (-1, -5, -10) from the point of intersection of line

 \vec{r}=(2\hat{i}-\hat{j}+2\hat{k})+\lambda(3\hat{i}+4\hat{j}+2\hat{k}) and the plane \vec{r}.(\hat{i}-\hat{j}+\hat{k})=5.

Solution:

Given equation of line is \vec{r}=(2\hat{i}-\hat{j}+2\hat{k})+\lambda(3\hat{i}+4\hat{j}+2\hat{k})

⇒ \vec{r}=(2+3\lambda)\hat{i}+(-1+4\lambda)\hat{j}+(2+2\lambda)\hat{k}

Coordinates of any point of line should be in the form of (2+3\lambda,-1+4\lambda,2+2\lambda)

We know, the intersection point of line and plane lies on the plane, using this,

⇒ [(2+3\lambda)\hat{i}+(-1+4\lambda)\hat{j}+(2+2\lambda)\hat{k}].(\hat{i}-\hat{j}+\hat{k})=5.

⇒ 2+3\lambda+1+4\lambda+2+2\lambda-5=0

⇒ \lambda=0



Therefore, coordinates of point is given by, putting \lambda=0   ,

⇒ (2, -1, 2)

Therefore, now distance between (-1, -5, -10) and (2, -1, 2) is,

⇒ \sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2}

⇒ \sqrt{9+16+144}  ⇒ 13 units

Question 4: Find the distance of point (2, 12, 5) from the point of intersection of line

 \vec{r}=(2\hat{i}-4\hat{j}+2\hat{k})+\lambda(3\hat{i}+4\hat{j}+2\hat{k}) and \vec{r}.(\hat{i}-2\hat{j}+\hat{k})=0.

Solution:

Given equation of line is \vec{r}=(2\hat{i}-4\hat{j}+2\hat{k})+\lambda(3\hat{i}+4\hat{j}+2\hat{k})

⇒ \vec{r}=(2+3\lambda)\hat{i}+(-4+4\lambda)\hat{j}+(2+2\lambda)\hat{k}

Coordinates of any point of line should be in the form of (2+3\lambda,-4+4\lambda,2+2\lambda)

We know, the intersection point of line and plane lies on the plane, using this,



⇒ [(2+3\lambda)\hat{i}+(-4+4\lambda)\hat{j}+(2+2\lambda)\hat{k}].(\hat{i}-2\hat{j}+\hat{k})=0 .

⇒ 2+3\lambda+8-8\lambda+2+2\lambda=0

⇒ \lambda=-4

Therefore, coordinates of point is given by, putting \lambda=-4 ,

⇒ (14, 12, 10).

Therefore, now distance between the points (2, 12, 5) and (14, 12, 10) is,

⇒ \sqrt{(14-2)^2+(12-12)^2+(10-5)^2}  

⇒ \sqrt{144+0+25}  ⇒ 13 units

Question 5: Find the distance of point (-1, -5, -10) from the point of intersection of the joining A(2, -1, 2) and B(5, 3, 4) with the plane x-y+z=5.

Solution:

Equation of line joining the points A(2, -1, 2) and B(5, 3, 4) is \frac{x-2}{5-2}=\frac{y-(-1)}{3-(-1)}=\frac{z-2}{4-2}



⇒ \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}

Let, \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda

⇒ x=3\lambda+2,y=4\lambda-1,z=2\lambda+2

Coordinates of any point on the line is given by (3\lambda+2,4\lambda-1,2\lambda+2)

We know, The intersection of line and plane lies on the plane, so,

⇒ 3\lambda+2-(4\lambda-1)+2\lambda+2=5

⇒ \lambda=0

Therefore, the coordinates of points is, putting \lambda=0

(2, -1, 2)

Now, the distance between the points (-1, -5, -10) and (2, -1, 2) is,



⇒ \sqrt{(2+1)^2+(-1+5)^2+(2+10)^2}

⇒ \sqrt{9+16+144}  ⇒ 13 units

Question 6: Find the distance of point (3, 4, 4) from the point, where the line joining the points A(3, -4, -5) and B(2, -3, 1) intersects the 2x+y+z=7.

Solution:

Equation of line passing through A(3, -4, -5) and B(2, -3, 1) is given by \frac{x-3}{2-3}=\frac{y-(-4)}{-3-(-4)}=\frac{z-(-5)}{1-(-5)}

⇒ \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}

Let \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda

⇒ x=-\lambda+3,y=\lambda-4,z=6\lambda-5

Coordinates of any point on the line is given by (-\lambda+3,\lambda-4,6\lambda-5)

We know, The intersection of line and plane lies on the plane, so,

⇒ 2(-\lambda+3)+(\lambda-4)+(6\lambda-5)=7



⇒ 5\lambda-3=7

⇒ \lambda=2

Therefore, the coordinates of points is, putting \lambda=2

(1, -2, 7)

Now, the distance between (3, 4, 4) and (1, -2, 7) is,

⇒ \sqrt{(3-1)^2+(4+2)^2+(4-7)^2}

⇒ \sqrt{4+36+9} = 7 units

Question 7: Find the distance of point (1, -5, 9) from the plane x- y+ z=5 measured along the line x=y=z.

Solution:

Given, The equation of line is x=y=z, it can also be written as,

\frac{x-0}{1}=\frac{y-0}{1}=\frac{z-0}{1}, where (1, 1, 1) are direction ratios of the line.

Here we have to measure the distance along the line, the equation of line parallel to x=y=z have same direction ratios (1, 1, 1),

So, the equation of line passing through (1, -5, 9) and having direction ratios (1, 1, 1) is,

⇒ \frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}

Let \frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda

Coordinates of any point on the line is given by (\lambda+1,\lambda-5,\lambda+9)

We know, The intersection of line and plane lies on the plane, so,

⇒ (\lambda+1)-(\lambda-5)+(\lambda+9)=5

⇒ \lambda=-10

Therefore, the coordinates of point is given by, putting \lambda=-10  = (-9, -15, -1)

Now, distance between the points (1, -5, 9) and (-9, -15, -1) is,

⇒ \sqrt{(1+9)^2+(-5+15)^2+(9+1)^2}

⇒ \sqrt{300} units.

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.




My Personal Notes arrow_drop_up
Recommended Articles
Page :