The **important questions for class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry **are covered here. The important questions are taken from the previous year question papers and sample papers, which will help you to achieve more marks in the annual examinations. The class 11 Maths syllabus is framed as per the **CBSE Board. **Practice the problems provided here to obtain excellent marks in class 11 Maths final examination. Also, get all the chapters important questions for Maths here.

Class 11 Maths Chapter 12 – Introduction to 3D Geometry incorporates the following important concepts such as:

- Coordinate Points in the three-dimensional space
- Distance between two points
- Section Formula

**Also, Check: **

- Important 1 Mark Questions for CBSE Class 11 Maths
- Important 4 Marks Questions for CBSE Class 11 Maths
- Important 6 Marks Questions for CBSE Class 11 Maths

## Class 11 Chapter 12 – Introduction to Three Dimensional Geometry Important Questions with Solutions

Go through and practice the following important questions in class 11 Maths, that should help you to solve the problems faster in the final examination

**Question 1: **

Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, â€“1).

**Solution:**

Assume that P (x, y, z) be the point that is equidistant from two points A(1, 2, 3) and B(3, 2, â€“1).

Thus, we can say that, PA = PB

Take square on both the sides, we get

PA^{2 }= PB^{2}

It means that,

(x+1)^{2} + (y-2)^{2}+(z-3)^{2} = (x-3)^{2}+(y-2)^{2}+(z+1)^{2}

â‡’ x^{2} â€“ 2x + 1 + y^{2} â€“ 4y + 4 + z^{2} â€“ 6z + 9 = x^{2} â€“ 6x + 9 + y^{2} â€“ 4y + 4 + z^{2} + 2z + 1

Now, simplify the above equation, we get:

â‡’ â€“2x â€“4y â€“ 6z + 14 = â€“6x â€“ 4y + 2z + 14

â‡’ â€“ 2x â€“ 6z + 6x â€“ 2z = 0

â‡’ 4x â€“ 8z = 0

â‡’ x â€“ 2z = 0

Hence, the required equation for the set of points is x â€“ 2z = 0.

**Question 2: **

Given that P (3, 2, â€“4), Q (5, 4, â€“6) and R (9, 8, â€“10) are collinear. Find the ratio in which Q

divides PR.

**Solution: **

Assume that the point Q (5, 4, â€“6) divide the line segment joining points P (3, 2, â€“4) and R (9, 8, â€“10) in the ratio k:1.

Therefore, by using the section formula, we can write it as:

(5, 4, -6) = [ (k(9)+3)/(k+1), (k(8)+2)/(k+1), (k(-10)-4)/(k+1)]

â‡’(9k+3)/(k+1) = 5

Now, bring the L.H.S denominator to the R.H.S and multiply it

â‡’9k+3 = 5k+5

Now, simplify the equation to find the value of k.

â‡’4k= 2

â‡’k = 2/4

â‡’k=Â½

Therefore, the value of k is Â½.

Hence, the point Q divides PR in the ratio of 1:2

**Question 3:**

Prove that the points: (0, 7, 10), (â€“1, 6, 6) and (â€“4, 9, 6) are the vertices of a right-angled triangle

**Solution:**

Let the given points be A = (0, 7, 10), B = (â€“1, 6, 6), and C = (â€“4, 9, 6)

Now, find the distance between the points

**Finding for AB:**

AB = âˆš [(-1-0)^{2} + (6-7)^{2 }+(6-10)^{2}]

AB = âˆš [(-1)^{2} + (-1)^{2 }+(-4)^{2}]

AB = âˆš(1+1+16)

AB = âˆš18

AB = 3âˆš2 â€¦. (1)

**Finding for BC:**

BC= âˆš [(-4+1)^{2} + (9-6)^{2 }+(6-6)^{2}]

BC = âˆš [(-3)^{2} + (3)^{2 }+(-0)^{2}]

BC = âˆš(9+9)

BC = âˆš18

BC = 3âˆš2 â€¦..(2)

**Finding for CA:**

CA= âˆš [(0+4)^{2} + (7-9)^{2 }+(10-6)^{2}]

CA = âˆš [(4)^{2} + (-2)^{2 }+(4)^{2}]

CA = âˆš(16+4+16)

CA = âˆš36

CA = 6 â€¦..(3)

Now, by Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2 } â€¦..(4)

Now, substitute (1),(2), and (3) in (4), we get:

6^{2 } = ( 3âˆš2)^{2} + ( 3âˆš2)^{2}

36 = 18+18

36 = 36

The given points obey the condition of Pythagoras Theorem.

Hence, the given points are the vertices of a right-angled triangle.

**Question 4: **

Calculate the perpendicular distance of the point P(6, 7, 8) from the XY – Plane.

(a)8Â (b)7Â Â (c)6Â (d) None of the above

**Solution:**

A correct answer is an **option (A)**

**Explanation:**

Assume that A be the foot of perpendicular drawn from the point P (6, 7, 8) to the XY plane and the distance of this foot A from P is z-coordinate of P, i.e., 8 units

Hence, the correct answer is an option (a)

**Question 5:**

If a parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes, then find the length of edges of a parallelopiped and length of the diagonal

**Solution:**

Let A = (2, 3, 5), B = (5, 9, 7)

To find the length of the edges of a parallelopiped = 5 â€“ 2, 9 â€“ 3, 7 â€“ 5

It means that 3, 6, 2.

Now, to find the length of a diagonal = âˆš(3^{2} + 6^{2} + 2^{2})

= âˆš(9+36+4)

= âˆš49

= 7

Therefore, the length of a diagonal of a parallelopiped is 7 units.

### Practice Problems for Class 11 Maths Chapter 12

Solve chapter 12 important problems given below:

- Find the coordinates of a point equidistant from the four points O (0,0,0), A(a, 0, 0), B(0, b, 0) and C(0,0, c). (Solution: (a/2, b/2, c/2)).
- Find the distance between the points P(-2,4,1) and Q(1, 2, – 5). (Solution: 7 units).
- Find the locus of the point which is equidistant from the points A(0,2,3) and (2, -2, 1). (Solution: x – 2y – z +1 = 0).
- Prove that the points (a,b,c), (b,c,a) and (c,a,b) are the vertices of an equilateral triangle.
- Find the ratio in which the line joining (2,4,5) and (3,5,4) is divided by the yz-plane. (Solution: 2:3 externally)
- Write the distance of point P(2,3,5) from the xy-plane. (Solution: 5)
- If the origin is the centroid of a triangle ABC having vertices A(a,1,3), B(-2,b,-5) and C(4,7,c), find the values of a,b,c. (Solution: -6,5,-8).
- Show that the lines joining the vertices of a tetrahedron to the centroids of the opposite faces are concurrent.
- The midterms of the sides of a triangle are (1,5,-1), (0,4,-2) and (2,3,4). Find its vertices. (Solution: A (1,2,3), b(3,4,5), c(-1, 6, -7))
- Name the octants in which the following points lie: (-5,-4,7), (-7,2,-5) (Solutions – X’OY’Z , X’OYZ’)

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