Question 1. Find the image of the point (0, 0, 0) in the plane 3x + 4y - 6z + 1 = 0.
Solution:
According to the question we have
Plane = 3x + 4y - 6z + 1 = 0
Line passing through origin and perpendicular to plane is given by
\frac{x}{3}=\frac{y}{4}=\frac{z}{-6}=r So, let the image of (0, 0, 0) = (3r, 4r, -6r)
The midpoint of (0, 0, 0) and (3r, 4r, -6r) lies on the given plane
3(3r/2) + 2(4r) - 3(-6y) + 1 = 0
30.5y = -1
r = -2/61
So, the image is (-6/61, -8/61, 12/61)
Question 2. Find the reflection of the point (1, 2, -1) in the plane 3x - 5y + 4z = 5
Solution:
According to the question we have to find the reflection of
the point P(1, 2, -1) in the plane 3x - 5y + 4z = 5
So, let Q = reflection of the point P
R = midpoint of PQ.
Then, R lies on the plane 3x - 5y + 4z = 5.
Now, the direction ratios of PQ are proportional to 3, -5, 4 and
PQ is passing through (1, 2, -1).
So, equation of PQ is,
\frac{x-1}{3}=\frac{y-2}{-5}=\frac{z+1}{4}=λ Let Q be (3λ + 1, -5λ + 2, 4λ - 1)
The coordinates of R are =
\left(\frac{3λ+1+1}{2},\frac{-5λ+2+2}{2},\frac{4λ-1-1}{2}\right)=\left(\frac{3λ+2}{2},\frac{-5λ+4}{2},\frac{4λ-2}{2}\right) Since, R lies on the given plane i.e., 3x - 5y + 4z = 5
Therefore,
3\left(\frac{3λ+2}{2}\right)-5\left(-\frac{5λ+4}{2}\right)+4\left(\frac{4λ-2}{2}\right)=5 9λ + 6 + 25λ - 20 + 16λ - 8 = 10
50λ - 22 = 10
50λ = 32
λ = 16/25
Q = (3λ + 1, -5λ + 2, 4λ -1) -Equation(1)
Now, put the value of λ in equation (1), we get,
= (3(16/25)+1, -5(16/25)+2, 4(16/25)-1)
= ((48/25)+1, (-16/5)+2, (64/25)-1)
= (73/25, -6/5, 39/25)
Hence, the reflection of point (1, 2, -1) = (73/25, -6/5, 39/25)
Question 3. Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line \frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1} . Hence or otherwise deduce the length of the perpendicular.
Solution:
According to the question we have to find foot of the perpendicular, say Q,
drawn from point P(5, 4, 2) to the line
\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}=λ So, Let us assume Q = (2λ - 1, 3λ + 3, -λ + 1) -Equation(1)
Direction ratio of line PQ are = (2λ - 6, 3λ - 1, -λ - 1)
Here, the line PQ is perpendicular to line the given line AB
So,
a1a2 + b1b2 + c1c2 = 0
(2λ - 6)(2) + (3λ - 1)(3) + (-λ - 1)(-1) = 0
4λ - 12 + 9λ - 3 + λ + 1 = 0
14λ - 14 = 0
λ = 14/14
λ = 1
So, put the value of λ in equation(1), we get
= (2(1) - 1, 3(1) + 3, -(1) + 1)
= (2 - 1, 3 + 3, -1 +1)
= (1, 6, 0)
Now, we find the length of perpendicular PQ using distance formula
= \sqrt{(5-1)^2+(4-6)^2+(2-0)^2}\\ = \sqrt{16+4+4}\\ = √24
= 2√6
So, the foot of the perpendicular is (1, 6, 0)
Length of the perpendicular is 2√6 units.
Question 4. Find the image of the point with position vector 3\hat{i}+\hat{j}+2\hat{k} in the plane \vec{r}.(2\hat{i}-\hat{j}+\hat{k})=4 . Also find the position vector of the foot of the perpendicular and the equation of the perpendicular line through 3\hat{i}+\hat{j}+2\hat{k} .
Solution:
According to the question we have to find image of the point P(3, 1, 2)
in the plane
\vec{r}.(2\hat{i}-\hat{j}+\hat{k})=4 or 2x - y + z = 4.Let Q be the image of the point P.
So,
The direction ratios of normal to the point = 2, -1, 1
The direction ratios of line PQ perpendicular to 2, -1, 1 and
PQ is passing through (3, 1, 2)
So equation of PQ is
\frac{x-3}{2}=\frac{y-1}{-1}=\frac{z-2}{1}=λ General point on the line PQ is = (2λ + 3, -λ + 1, λ + 2)
Let us assume Q = (2λ + 3, -λ + 1, λ + 2) -Equation(1)
Let R be the mid point of PQ. Then,
Coordinates of R =
\left(\frac{2λ+3+3}{2},\frac{-λ+1+1}{2},\frac{λ+2+2}{2}\right)=\left(\frac{2λ+6}{2},\frac{-λ+2}{2},\frac{λ+4}{2}\right) Since, R lies on the plane 2x - y + z = 4, we get
2\left(\frac{2λ+6}{2}\right)-\left(\frac{-λ+2}{2}\right)+\left(\frac{λ+4}{2}\right)=4 4λ + 12 + λ - 2 + λ + 4 = 8
6λ = 8 - 14
λ = -6/6
λ = -1
So, put the value of λ in equation(1), we get
Image of P = Q(2 (-1) + 3, - (-1) + 1, -1 + 2)
Image of P = (1, 2, 1)
Equation of the perpendicular line through
3\hat{i}+\hat{j}+2\hat{k} is
\vec{r}=3\hat{i}+\hat{j}+2\hat{k}+λ(2\hat{i}-\hat{j}+\hat{k}) Position vector of the image point is
3\hat{i}+\hat{j}+2\hat{k}+λ(2\hat{i}-\hat{j}+\hat{k})=(3+2λ)\hat{i}+(1-λ)\hat{j}+(2+k)\hat{k} Position vector of the foot of the perpendicular is
\frac{[(3+2λ)\hat{i}+(1-λ)\hat{j}+(2+k)\hat{k}]+[3\hat{i}+\hat{j}+2\hat{k}]}{2}
=(3+λ)\hat{i}+(1-\frac{λ}{2})\hat{j}+(2+\frac{λ}{2})\hat{k} By putting the value of λ in the position vector of the foot of the perpendicular is
2\hat{i}+\frac{3}{2}\hat{j}+\frac{3}{2}\hat{k}
Question 5. Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x - 2y + 4z + 5 = 0. Also, find the length of the perpendicular.
Solution:
According to the question we have,
Plane = 2x - 2y + 4z + 5 = 0 -Equation(1)
Point = (1, 1, 2)
and find the coordinates of the foot of the perpendicular
= \left|\frac{2-2+8+5}{\sqrt{1+1+4}}\right|=\frac{13}{\sqrt{6}} Let us assume that the foot of perpendicular = (x, y, z).
So, DR's are in proportional
\frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{4}=k x = 2k + 1
y = -2k + 1
z = 4k + 2
Substitute (x, y, z) = (2k + 1, -2k + 1, 4k + 2) in the equation(1), we get
2x - 2y + 4z + 5 = 0
4k + 2 + 4k - 2 + 16k + 8 + 5 = 0
24k = -13
k = -13/24
So, the coordinates of the foot of the perpendicular (x, y, z) = (-1/12, 5/3, -1/6)
Question 6. Find the distance of the point (1, -2, 3) from the plane x - y + z + 5 measured along a line parallel to \frac{x}{2}=\frac{y}{3}=\frac{z}{-6}
Solution:
According to the question, we have to find the distance of point P(1, -2, 3)
from the plane x - y + z = 5 measured
parallel to line AB,
\frac{x}{2}=\frac{y}{3}=\frac{z}{-6} Let us assume Q = Mid point of the line joining P to plane.
We have, PQ parallel to line AB
The direction ratios of line PQ are proportional to direction ratios of line AB
The direction ratios of line PQ = 2, 3, -6
PQ is passing through point P(1, -2, 3).
Thus, the equation of PQ is,
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=λ\\ The general point on the line PQ = (2λ + 1, 3λ - 2, -6λ + 3)
Suppose the coordinates of Q = (2λ + 1, 3λ - 2, -6λ + 3)
Thus, Q lies on the plane x - y + z = 5
(2λ + 1) - (3λ - 2) + (-6λ + 3) = 5
2λ + 1 - 3λ + 2 - 6λ + 3 = 5
-7λ = -1
λ = 1/7
Coordinate of Q = (2λ + 1, 3λ - 2, -6λ + 3) -Equation(1)
Now, put the value of λ in equation(1), we get
Q = (2(1/7)+1, 3(1/7)-2, -6(1/7)+3)
Q = (9/7, -11/7, 15/7)
Now, we find the distance between (1, -2, 3) and plane = PQ
=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}\\ =\sqrt{(1-\frac{9}{7})^2+(-2+\frac{11}{7})^2+(3-\frac{15}{7})^2}\\ =\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}\\ =\sqrt{\frac{49}{49}} = 1
Hence, the required distance is 1 unit.
Question 7. Find the coordinates of the foot the perpendicular from the point (2, 3, 7) to the plane 3x - y - z = 7. Also, find the length of the perpendicular.
Solution:
Let us assume that Q be the foot of the perpendicular.
Now, the direction ratios of normal plane is 3, -1, -1
Line PQ is parallel to normal to plane
Direction ratios of PQ are proportional to 3, -1, -1
PQ is passing through point P(2, 3, 7)
So,
\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \frac{x-2}{3}=\frac{y-3}{-1}=\frac{z-7}{-1}=λ The general point on the line PQ
= (3λ + 2, -λ + 3, -λ + 7)
Coordinates of Q = (3λ + 2, -λ + 3, -λ + 7) -Equation(1)
Point Q lies on the plane 3x - y - z = 7
Thus,
3(3λ + 2) - (-λ + 3) - (-λ + 7) = 7
9λ + 6 + λ - 3 + λ - 7 = 7
11λ = 7 + 4
11λ = 11
λ = 11/11
λ = 1
Now, put the value of λ in equation(1), we get
Q = (3(1) + 2, -(1) + 3, -(1) + 7)
Q = (5, 2, 6)
Find the length of the perpendicular PQ
=\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}\\ =\sqrt{(2-5)^2+(3-2)^2+(7-6)^2}\\ =\sqrt{9+1+1}\\ = √11
Summary
Exercise 29.15 Set 1 in RD Sharma Class 12 typically focuses on various aspects of planes in three-dimensional space. Key concepts include:
- General equation of a plane: ax + by + cz + d = 0
- Normal vector of a plane: (a, b, c)
- Point-normal form of a plane equation
- Planes parallel or perpendicular to coordinate axes
- Intercept form of a plane equation
- Distance of a point from a plane