RD Sharma Class 12 Ex 29.15 Solutions Chapter 29 The Plane

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter29
Exercise29.15
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 29.15 Solutions Chapter 29 The Plane

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

Question 8. Find the image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0.

Solution:

According to the question we have to find the image of point P(1, 3, 4)

in the plane 2x – y + z +3 = 0

Now let us assume that Q be the image of the point.

Here, the direction ratios of normal to plane = 2, -1, 1

The direction ratios of PQ which is parallel to the normal to the plane 

is proportional to 2, -1, 1 and the line PQ is passing through point P(1, 3, 4).

Thus, equation of the line PQ is: 

\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ \frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{-1}=λ

Now, the general point on the line PQ = (2λ + 1, -λ + 3, λ + 4)

Let Q = (2λ + 1, -λ + 3, λ + 4)          -Equation(1)

Here, Q is the image of P, so R is the mid point of PQ

Coordinates of R = \left(\frac{2λ+1+1}{2},\frac{-λ+3+3}{2},\frac{λ+4+4}{2}\right)\\ =\left(\frac{2λ+2}{2},\frac{-λ+6}{2},\frac{λ+8}{2}\right)\\ =\left(λ+1,\frac{-λ+6}{2},\frac{λ+8}{2}\right)

Point R is lies on the plane 2x – y + z + 3 = 0

= 2(λ + 1) – \left(\frac{-λ+6}{2}+\frac{λ+8}{2}\right) = 0

4λ + 4 + λ – 6 + λ + 8 + 6 = 0

6λ = -12

λ = -2

Now, put the value of λ in equation(1), we get

= (-4 + 1, 2 + 3, -2 + 4)

= (-3, 5, 2)

Hence, the image of point P(1, 3, 4) is (-3, 5, 2)

Question 9. Find the distance of the point with position vector -\hat{i}-5\hat{j}-10\hat{k}    from the point of intersection of the line \vec{r}=(2\hat{i}-\hat{j}+2\hat{k})+λ(3\hat{i}+4\hat{j}+12\hat{k})    with the plane \vec{r}.(\hat{i}-\hat{j}+\hat{k})=5  .

Solution:

According to the question we have to find distance of a point A with position 

vector (-\hat{i}-5\hat{j}-10\hat{k})    from the point of intersection of 

line \vec{r}=(2\hat{i}-\hat{j}+2\hat{k})+λ(3\hat{i}+4\hat{j}+12\hat{k})

with plane \vec{r}.(\hat{i}-\hat{j}+\hat{k})=5

Let the point of intersection of line and plan be B(\vec{b})

The line and the plane will intersect when,

[(2\hat{i}-\hat{j}+2\hat{k})+λ(3\hat{i}+4\hat{j}+12\hat{k})](\hat{i}-\hat{j}+\hat{k})=5\\ [(2+3λ)\hat{i}+(-1+4λ)\hat{j}+(2+12λ)\hat{k}](\hat{i}-\hat{j}+\hat{k})=5\\

(2 + 3λ)(1) + (-1 + 4λ)(-1) + (2 + 12λ)(1) = 5

2 + 3λ + 1 – 4λ + 2 + 12λ = 5

11λ = 5 – 5

λ = 0

So, the point B is given by

\vec{b}=(2\hat{i}-\hat{j}+2\hat{k})+(0)(3\hat{i}+4\hat{j}+12\hat{k})\\ \vec{b}=(2\hat{i}-\hat{j}+2\hat{k})
\vec{AB}=\vec{b}-\vec{a}
=(2\hat{i}-\hat{j}+2\hat{k})-(-\hat{i}-5\hat{j}-10\hat{k})
=(2\hat{i}-\hat{j}+2\hat{k}+\hat{i}+5\hat{j}+10\hat{k})=(3\hat{i}+4\hat{j}+12\hat{k})\\ |\vec{AB}|=\sqrt{(3)^2+(4)^2+(12)^2}=\sqrt{9+16+144}=\sqrt{169}=13

The required distance is 13 units.

Question 10. Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane \vec{r}.(\hat{i}-2\hat{j}+4\hat{k})+5=0  .

Solution: 

Plane = x – 2y + 4z + 5 = 0          -Equation(1)

Point = (1, 1, 2)

D = \left|\frac{1-2+8+5}{\sqrt{1+4+16}}\right|

= 12/√21

The length of the perpendicular from the given point to the plane = 12/√21

Let us assume that the foot of perpendicular be (x, y, z). 

So DR’s are in proportional

\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-2}{4}=k

x = k + 1

y = -2k + 1

z = 4k + 2

Substitute (x, y, z) = (k + 1, -2k + 1, 4k + 2) in the plane equation(1)

k + 1 + 4k – 2 + 16k + 8 + 5 = 0

21k = -12

k = -12/21 = -4/7 

Hence, the coordinate of the foot of the perpendicular  = (3/7, 15/7, -2/7) 

Question 11. Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3, 2, 1) from the plane 2x – y + z + 1 = 0. Find also the image of the point in the plane.

Solution: 

Given:

Plane = 2x – y + z + 1 = 0          -Equation(1)

Point P = (3, 2, 1)

D =\left|\frac{6-2+1+1}{\sqrt{4+1+1}}\right|=\frac{6}{\sqrt{6}}=\sqrt{6}

The perpendicular distance of the point P from the plane(D) = √6

Let us assume that the foot of perpendicular be (x, y, z). 

So DR’s are in proportional

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

x = 2k + 3

y = -k + 2

z = k + 1

Substitute (x, y, z) = (2k + 3, -k + 2, k + 1) in the plane equation(1)

4k + 6 + k – 2 + k + 1 + 1 = 0

6k = -6

k = -6/6 = -1

The coordinate of the foot of the perpendicular = (1, 3, 0) 

Question 12. Find the direction cosines of the unit vector perpendicular to the plane \vec{r}.(6\hat{i}-3\hat{j}-2\hat{k})+1=0     passing through the origin.

Solution: 

Given:

Equation of the plane \vec{r}.(6\hat{i}-3\hat{j}-2\hat{k})+1=0

Thus, the direction ratios normal to the plane are 6, -3 and -2

Hence, the direction cosines to the normal to the plane are

\frac{6}{\sqrt{6^2+(-3)^2+(-2)^2}},\frac{-3}{\sqrt{6^2+(-3)^2+(-2)^2}},\frac{-2}{\sqrt{6^2+(-3)^2+(-2)^2}}\\

= 6/7, -3/7, -2/7

= -6/7, 3/7, 2/7

The direction cosines of the unit vector perpendicular to the plane 

are same as the direction cosines of the unit vector perpendicular

to the plane are: -6/7, 3/7, 2/7 

Question 13. Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 4z – 6 = 0.

Solution: 

According to the question,

Plane = 2x – 3y + 4z – 6 = 0

The direction ratios of the normal to the plane are 2, -3 and 4.

Thus, the direction ratios of the line perpendicular to the plane are 2, -3 and 4.

The equation of the line passing (x1, y1, z1) having direction ratios a, b and c is

\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}

Thus, the equation of the line passing through the origin 

with direction ratios 2, -3 and 4 is

\frac{x-0}{2}=\frac{y-0}{-3}=\frac{z-0}{4}\\ \frac{x}{2}=\frac{y}{-3}=\frac{z}{4}=r

Here, r is same constant.

Any point on the line is of the form 2r, -3r, and 4r, 

if the point P(2r, -3r, 4r) lies on the plane 2x – 3y + 4z – 6 = 0.

Thus, we have,

2(2r) – 3(-3r) + 4(4r) – 6 = 0

4r + 9r + 16r – 6 = 0

29r = 6

r = 6/29

Thus, the coordinates of the point of intersection of the perpendicular 

from the origin and the plane are:

P(2×6/29, -3×629, 4×6/29) = P(12/29, -18/29, 24/29) 

Question 14. Find the length and the foot of the perpendicular from the point (1, 3/2, 2) to the plane 2x – 2y + 4z +5 = 0.

Solution:

Given: 

Point = (1, 3/2, 2) 

Plane = 2x – 2y + 4z + 5 = 0

D = \left|\frac{2-3+8+5}{\sqrt{4+4+16}}\right|=\frac{12}{2\sqrt{6}}

= √6

So, the length of the perpendicular from the point to the plane(D) = √6

Let the foot of perpendicular be (x, y, z). So, DR’s are in proportional 

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

x = 2k + 1

y = -2k + 3/2

z = 4k + 2

So, using the values of x, y, z in equation of the plane we have,

2(2k + 1) – 2(-2k + 2/3) +4(4k + 2) + 5 = 0

4k + 2 + 4k – 3 + 16k + 8 + 5 = 0

24k = -12

k = -1/2

So, the coordinate of the foot of the perpendicular = (0, 5/2, 0)

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