Class 12 RD Sharma Solutions- Chapter 29 The Plane – Exercise 29.15 | Set 2

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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)

My Personal Notes

Last Updated : 18 Dec, 2021

Related Articles

Class 12 RD Sharma Solutions- Chapter 29 The Plane – Exercise 29.15 | Set 2

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

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

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

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.

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.

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

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.

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Related Articles

Class 12 RD Sharma Solutions- Chapter 29 The Plane – Exercise 29.15 | Set 2

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

Question 9. Find the distance of the point with position vector from the point of intersection of the line with the plane .

Question 10. Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane .

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.

Question 12. Find the direction cosines of the unit vector perpendicular to the plane passing through the origin.

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

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.

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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

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

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.