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

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  • Last Updated : 28 Mar, 2021
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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.


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