# RBSE Maths Class 12 Chapter 14: Three Dimensional Geometry Important Questions and Solutions

RBSE Class 12 Maths Chapter 14 – Three Dimensional Geometry Important questions and solutions are provided here. All the important questions and solutions given here have stepwise solutions and contain required formulas to understand the steps clearly. The RBSE Class 12 important questions and solutions available at BYJU’S will help the students in clearing their exams with flying colours.

Chapter 14 of RBSE Class 12 contains seven exercises all which have questions on 2D geometry with the combination of vectors. Enough practice questions are available in this chapter on finding the direction ratios, cosines, angle between two vectors, equations and so on. Besides, the concepts of perpendicular lines, intersecting lines and parallel planes also exist in this chapter.

### RBSE Maths Chapter 14: Exercise 14.1 Textbook Important Questions and Solutions

Question 1: Find the direction cosines of the line passing through two points (4, 2, 3) and (4, 5, 7).

Solution:

If P(x1, y1, z1) and Q(x2, y2, z2) are the line joining two points, then the direction cosines are:

(x2 – x1)/PQ = (y2 – y1)/PQ = (z2 – z1)/PQ

Here, PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2

From the given,

x1 = 4, y1 = 2, z1 = 3

x2 = 4, y2 = 5, z2 = 7

PQ = √[(4 – 4)2 + (5 – 2)2 + (7 – 3)2

= √(0 + 9 + 16

= √(25)

= 5

Therefore, the required direction cosines are:

(4 – 4)/5, (5 – 2)/5, (7 – 3)/5

= 0, 3/5, 4/5

Question 2: If the direction ratios of the line are 2, -1, -2, then find the direction cosines.

Solution:

Let the given direction ratios be: a = 2, b = -1, c = -2

Direction cosines of line are l, m and n.

l = ±a/√(a2 + b2 + c2), m = ±b/√(a2 + b2 + c2), n = ±c/√(a2 + b2 + c2)

l = ±2/√[(2)2 + (-1)2 + (-2)2]

= ±2/√(4 + 1 + 4)

= ±2/√9

= (±2)/(±3)

= 2/3

Similarly,

m = -(±1)/(±3) = -1/3

n = 1(±2)/(±3) = -2/3

### RBSE Maths Chapter 14: Exercise 14.2 Textbook Important Questions and Solutions

Question 3: Find the equation of the line which passes through the point (5, -2, 4) and is parallel to the vector $2\hat{i}-\hat{j}+3\hat{k}$.

Solution:

Given,

The points through which the line passes are (5, -2, 4).

The position vector of the given point is $\vec{a}=5\hat{i}-2\hat{j}+4\hat{k}$

Given vector is $\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$

Vector equation of a line that passes through a given point whose position vector is $\vec{a}$ and parallel to a given vector $\vec{b}$ is

$\vec{r}=\vec{a}+\lambda \vec{b}\\\vec{r}=(5\hat{i}-2\hat{j}+4\hat{k})+\lambda(2\hat{i}-\hat{j}+3\hat{k})$

Question 4: The cartesian equation of a line is 3x + 1 = 6y – 2 = 1 – z. Find the point through which it passes and find the direction ratios and vector equation.

Solution:

Given line is:

3x + 1 = 6y – 2 = 1 – z

3[x + (1/3)] = 6[y – (1/3)] = 1 – z

[x + (1/3)]/ (1/3) = [y – (1/3)]/ (1/6) = (z – 1)/ (-1)

Therefore, the line passes through the point (-1/3, 1/3, 1).

Direction cosines of the line are 1/3, 1/6, -1 or 2, 1, -6.

Position vector of (-1/3, 1/3, 1) is $\vec{a}=-\frac{1}{3}\hat{i}+\frac{1}{3}\hat{i}+\hat{k}$

And

$\vec{m}=2\hat{i}+\hat{i}-6\hat{k}$

Hence, the required vector equation is:

$\vec{r}=\vec{a}+\lambda\vec{m}\\\vec{r}=(-\frac{1}{3}\hat{i}+\frac{1}{3}\hat{j}+\hat{k})+\lambda (2\hat{i}+\hat{i}-6\hat{k})$

### RBSE Maths Chapter 14: Exercise 14.3 Textbook Important Questions and Solutions

Question 5: Show that the line passing through the points (1, -1, 2) and (3, 4, -2) is perpendicular to the line passing through the points (0, 3, 2) and (3, 5, 6).

Solution:

The equation of line passing through points (1,-1,2) and (3, 4,-2) is:

(x – 1)/ (3 – 1) = [y – (-1)]/ [4 – (-1)] = (z – 3)/ (-2 – 2)

(x – 1)/2 = (y + 1)/5 = (z – 2)/-4

Thus, l1 = 2, m1 = 5, n1 = -4

And the equation of the line passing through the points (0, 3, 2) and (3, 5, 6) is:

(x – 0)/ (3 – 0) = (y – 3)/ (5 – 3) = (z – 2)/ (6 – 2)

x/3 = (y – 3)/2 = (z – 2)/4

Thus, l2 = 3, m2 = 2, n2 = 4

Now,

l1l2 + m,m2 + n1n2 = 2 × 3 + 5 × 2 + (-4) × 4

= 6 + 10 – 16

= 0

Therefore, the lines passing through the given points are perpendicular.

Question 6: If the lines (x – 1)/-3 = (y – 2)/2k = (z – 3)/2 and (x – 1)/3k = (y – 1)/1 = (z – 6)/-5 are mutually perpendicular, then find the value of k.

Solution:

The direction cosine of (x – 1)/-3 = (y – 2)/2k = (z – 3)/2 are:

l1 = -3, m1 = 2k, n1 = 2

And the direction cosines of (x – 1)/3k = (y – 1)/1 = (z – 6)/-5 are:

l2 = 3k, m2 = 1, n2 = -5

Given that, the lines are perpendicular.

⇒ l1l2 + m1m2 + n1n2 = 0

⇒ (-3) × (3k) + (2k) × 1 + 2 × (-5) = 0

⇒ -9k + 2k – 10 = 0

⇒ -7k = 10

⇒ k = -10/7

### RBSE Maths Chapter 14: Exercise 14.4 Textbook Important Questions and Solutions

Question 7: Show that the lines (x – 1)/2 = (y – 2)/3 = (z -3)/4 and (x – 4)/5 = (y – 1)/2 = z are mutually intersecting. Find the point of intersection.

Solution:

Let (x – 1)/2 = (y – 2)/3 = (z -3)/4 = r1

Thus, the coordinates of any point on this line are (2r1 + 1, 3r1 + 2, 4r1 + 3).

And let (x – 4)/5 = (y – 1)/2 = z or (x – 4)/5 = (y – 1)/2 = (z – 0)/1 = r2

Thus, the coordinates of any point on this line are (5r2 + 4, 2r2 + 1, r2).

If the lines intersect each other, then any one point of both the lines will coincide.

Therefore,

2r1 + 1 = 5r2 + 4 ….(i)

3r1 + 2 = 2r2 + 1 ….(ii)

4r1 + 3 = r2 ….(iii)

From (i) and (ii),

2r1 – 5r2 = 3 ….(iv)

3r1 – 2r2 = -1 ….(v)

Solving (iv) and (v),

r1 = -1 and r2 = -1

Therefore, the point of intersection is (-1, -1, -1).

Question 8: Find the foot of perpendicular from the point (2, 3, 4) to the line (4 – x)/2 = y/6 = (1 – z)/3. Also, find the perpendicular distance of the line from the point.

Solution:

Given,

The equation of line AB is: (4 – x)/2 = y/6 = (1 – z)/3

⇒(x – 4)/-2 = (y – 0)/6 = (z – 1)/-3 = λ (say) ….(i)

Coordinates of point M on the line AB are (-2λ + 4, 6λ + 0, -3λ + 1)

Direction cosines of the perpendicular LM

l1, m1, n1 = x2 – x1, y2 – y1, z2 – z1

= -2λ + 4 – 2, 6λ + 0 – 3, -3λ + 1 – 4

= -2λ + 2, 6λ – 3, -3λ – 3

And the direction cosines of the line AB are:

l2, m2, n2 = -2, 6, -3

AB and LM are perpendicular to each other.

⇒ l1l2 + m1m2 + n1n2 = 0

(-2λ + 2)(-2) + (6λ – 3)(6) + (-3λ – 3) = 0

4λ – 4 + 36λ – 18 + 9λ + 9 = 0

49λ – 13 = 0

49λ = 13

λ = 13/49

Substituting the value λ = 13/49 in point M.

M = (170/49, 78/49, 10/49)

Hence, the length of the perpendicular is:

PQ = √[(170/49 – 2)2 + (78/49 – 3)2 + (10/49 – 4)2]

= (3/7) √101 units

### RBSE Maths Chapter 14: Exercise 14.5 Textbook Important Questions and Solutions

Question 9: Find the shortest distance between the lines

(x – 1)/2 = (y + 1)/3 = z and (x + 1)/3 = (y – 2)/1 = 2

Also, find the equations of the line of shortest distance.

Solution:

Let the given lines be:

(x – 1)/2 = (y + 1)/3 = (z – 0)/1 = r1 ….(i)

(x + 1)/3 = (y – 2)/1 = (z – 2)/0 = r2 ….(ii)

Thus, any point on the line (i) is P(2r1 + 1, 3r1 – 1, r1)

And any point on the line (ii) is Q(3r2 – 1, r2 + 2, r2 + 2)

Direction cosines of the line PQ = 3r2 – 2r1 – 2, r2 – 3r1 + 3, r2 + 2 – r1

Line PQ is perpendicular to the line (i).

⇒ l1l2 + m1m2 + n1n2 = 0

⇒ 2(3r2 – 2r1 – 2) + 3(r2 – 3r1 + 3) + 1(2 – r1) = 0

⇒ 9r2 – 14r1 = 7 ….(iii)

Now, the line PQ is perpendicular to line (ii).

⇒ l1l2 + m1m2 + n1n2 = 0

⇒ 3(3r2 – 2r1 – 2) + 1(r2 – 3r1 + 3) + 0(2 – r1) = 0

⇒ 10r2 – 9r1 – 3 = 0 ….(iv)

By solving (iii) and (iv),

r1 = 97/59 and r2 = 105/59

By substituting the values of r1 and r2 in P and Q.

P = (253/59, 232/59, 97/59)

Q = (256/59, 223/59, 2)

Distance between P and Q is:

PQ = √[(256/59 – 253/59)2 + (223/59 – 232/59)2 + (2 – 97/59)2]

= √[(3/59)2 + (-9/59)2 + (21/59)2]

= (3/59) √(1 + 9 + 49)

= (3/59) √59

= 3/√59 units

### RBSE Maths Chapter 14: Exercise 14.6 Textbook Important Questions and Solutions

Question 10: Find the equation of plane passing through the point (2, -1, 3) and perpendicular to the X-axis.

Solution:

Equation of plane passing through point (2,-1,3)

a(x – 2) + b(y + 1) + c(z – 3) = 0

Given that the plane is perpendicular to the X-axis.

⇒ b = 0, c = 0

Therefore, the equation of plane

a(x – 2) + 0(y + 1) + 0(z – 3) = 0

⇒ a(x – 2) = 0

⇒ x – 2 = 0 (∵a ≠ 0)

Question 11: Find the normal form of the equation of the plane 2x – 3y + 6z + 14 = 0.

Solution:

Given,

Equation of the plane is 2x – 3y + 6z + 14 = 0

Direction cosines of the normal plane are:

2/√(4 + 9 + 36), -3/√(4 + 9 + 36), 6/√(4 + 9 + 36)

= 2/√49, -3/√49, 6/√49

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

By dividing the given plane by 7,

(2/7)x – (3/7)y + (6/7)z + (14/7) = 0

⇒ (-2/7)x + (3/7)y – (6/7)z = 2

### RBSE Maths Chapter 14: Exercise 14.7 Textbook Important Questions and Solutions

Question 12: Find the angle between the planes: x + y + 2z = 9 and 2x – y + z = 15

Solution:

Given planes are: x + y + 2z = 9 and 2x – y + z = 15

Comparing with a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0,

a1 = 1, b1 = 1, c1 = 2

a2 = 2, b2 = -1, c2 = 1

cos θ = (a1a2 + b1b2 + c1c2)/ [√(a12 + b12 + c12) √(a22 + b22 + c22)]

cos θ = [1 × 2 + 1 × (-1) + 2 × 1]/ [√(1 + 1 + 4) √(4 + 1 + 1)]

= (2 – 1 + 2)/ (√6 √6)

= 3/6

= 1/2

cos θ = cos 60°

cos θ = cos π/3

Therefore, θ = π/3

Question 13: If the line $\vec{r}=\hat{i}+\lambda (2\hat{i}-m\hat{j}-3\hat{k})$ is parallel to the plane $\vec{r}.(m\hat{i}+3\hat{j}+\hat{k})=4$, then find the value of m.

Solution:

Given that, the line $\vec{r}=\hat{i}+\lambda (2\hat{i}-m\hat{j}-3\hat{k})$ is parallel to the plane $\vec{r}.(m\hat{i}+3\hat{j}+\hat{k})=4$.

$(2\hat{i}-m\hat{j}-3\hat{k}).(m\hat{i}+3\hat{j}+\hat{k})=0$

2 × m + (-m) × 3 + (-3) × 1 = 0

2m – 3m – 3 = 0

-m – 3 = 0

m = -3

### RBSE Maths Chapter 14: Additional Important Questions and Solutions

Question 1: If the equation lx + my + nz = p is the normal form of the plane, then which of the following is true or false.

(A) l, m, n are the d.c’s of the normal to the plane

(B) p is the perpendicular distance from the origin to the plane

(C) for every value of p, the plane passes through the origin

(D) l2 + m2 + n2 = 1

Solution: