CBSE Class 12 Mathematics Three Dimensional Geometry Assignment Set 01

Question. The distance of the point -i + 2 j + 6k from the straight line that passes through the point 2i + 3 j - 4k and is parallel to the vector 6i + 3 j - 4k is
(a) 9
(b) 8
(c) 7
(d) 10
Answer : C

Question. The two lines x = ay + b , z = cy + d and x = a'y + b', z = c'y + d' will be perpendicular, if and only if 
(a) aa' + cc' + 1 = 0
(b) aa' + bb' + cc' + 1 = 0
(c) aa' + bb' +cc' = 0
(d) (a + a') (b + b') +(c + c') = 0.
Answer : A

Question. If the equation of a plane P, passing through the intersection of the planes, x + 4y – z + 7 = 0 and 3x + y + 5z = 8 is ax + by + 6z = 15 for some a, bÎR, then the distance of the point (3, 2, –1) from the plane P is ___________.
Answer : (3.00)

Question. If the line x - 2/3 = y + 1/2 = z - 1/-1 intersects the plane 2x + 3y – z + 13 = 0 at a point P and the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to:
(a) 14
(b) √14
(c) 2√7
(d) 2√14
Answer : D

Question. The angle between the lines 2x = 3y = – z and 6x = – y = – 4 is
(a) 0^o
(b) 90^o
(c) 45^o
(d) 30^o
Answer : B

Question. A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z . The co-ordinates of each of the points of intersection are given by
(a) (2a,3a,3a), (2a,a,a)
(b) (3a, 2a,3a), (a, a, a)
(c) (3a,2a,3a), (a, a, 2a)
(d) (3a,3a,3a), (a,a,a)
Answer : B

Question. A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0, passes through the point :
(a) (1, –4, 1)
(b) (1, 4, –1)
(c) (2, 4, 1)
(d) (2, –4, 1)
Answer : D

Question. A plane passing through the points (0, –1, 0) and (0, 0, 1) and making an angle π/4 with the plane y – + 5 = 0, also passes through the point: 
(a) (–√2 , 1, –4)
(b) (√2 , –1, 4)
(c) (–√2 , –1, –4)
(d) (√2 , 1, 4)
Answer : D

Question. If Q (0, –1, –3) is the image of the point P in the plane 3x – y + 4 = 2 and R is the point (3, –1, –2), then the area (in sq. units) of ΔPQR is : 
(a) 2√13
(b) √91/4
(c) √91/2
(d) √65/2
Answer : C

Question. The foot of the perpendicular drawn from the point (4, 2, 3) to the line oining the points (1, –2, 3) and (1,1, 0) lies on the plane :
(a) 2x + y – z = 1
(b) x – y – 2z = 1
(c) x – 2y + z = 1
(d) x + 2y – z = 1
Answer : A

Question. The plane which bisects the line oining the points (4, – 2, 3) and (2, 4, – 1) at right angles also passes through the point: 
(a) (4, 0, 1)
(b) (0, –1, 1)
(c) (4, 0, –1)
(d) (0, 1, –1)
Answer : C

Question. The equation of a plane containing the line of intersection of the planes 2x – y – 4 = 0 and y + 2z – 4 = 0 and passing through the point (1, 1, 0) is : 
(a) x – 3y – 2z = –2
(b) 2x – z = 2
(c) x – y – z = 0
(d) x + 3y + z = 4
Answer : C

Question. If the plane 2x – y + 2z + 3 =0 has the distances 1/2 and 2/3 units from the planes 4x – 2y + 4z + l = 0 and 2x – y + 2z + μ = 0, respectively, then the maximum value of λ + μ is equal to : 
(a) 9
(b) 15
(c) 5
(d) 13
Answer : D

Question. If the line, x - 1/2 =  y + 1/3 = z - 2/4 meets the plane, x + 2y + 3 = 15 at a point P, then the distance of P from the origin is: 
(a) √5 / 2
(b) 2√5
(c) 9/2
(d) 7/2
Answer : C

Question. A plane passing through the point (3, 1, 1) contains two lines whose direction ratios are 1, –2, 2 and 2, 3, –1 respectively. If this plane also passes through the point (a, - 3, 5), then a is equal to :
(a) 5
(b) –10
(c) 10
(d) –5
Answer : A

Question. If for some a and b in R, the intersection of the following three planes
x + 4y – 2z = 1
x + 7y – 5z = b
x + 5y + az = 5
is a line in R3, then a + b is equal to:
(a) 0
(b) 10
(c) 2
(d) –10
Answer : B

Question. If L1 is the line of intersection of the planes 2x - 2y + 3 - 2 = 0, x - y + +1= 0 and L2 is the line of intersect ion of the planes x + 2y - - 3 = 0, 3x - y + 2 -1= 0 , then the distance of the origin from the plane, containing the lines L1 and L2, is :
(a) 1/3√2
(b) 1/2√2
(c) 1/√2
(d) 1/4√2
Answer : A

Question. The shortest distance between the –axis and the line x + y + 2z – 3 = 0 = 2x + 3y + 4z – 4, is
(a) 1
(b) 2
(c) 4
(d) 3
Answer : B

Question. The plane passing through the points (1, 2, 1), (2, 1, 2) and parallel to the line, 2x = 3y, z = 1 also through the point :
(a) (0, 6, –2)
(b) (–2, 0, 1)
(c) (0, –6, 2)
(d) (2, 0, –1)
Answer : B

Question. Let P be the plane, which contains the line of intersection of the planes, x + y + – 6 = 0 and 2x + 3y + + 5 = 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to:
(a) 17/√5
(b) 63√5
(c) 205√5
(d) 11/√5
Answer : D

Question. Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is:
(a) (6, 5, 2)
(b) (6, 5, –2)
(c) (4, 3, 2)
(d) (3, 4, –2)
Answer : B

Question. If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4 + 22 = 0 measured parallel to line,  x/1 = y/4 = z/5 is Q, then PQ is equal to :
(a) 6√5
(b) 3√5
(c) 2√42
(d) √42
Answer : C

Question. The plane which bisects the line segment oining the points (– 3, – 3, 4) and (3, 7, 6) at right angles, passes through which one of the following points? 
(a) (–2, 3, 5)
(b) (4, – 1, 7)
(c) (2, 1, 3)
(d) (4, 1, – 2)
Answer : D

Question. The plane containing the line x - 3/2 = y + 2/-1 = z - 1/3 and also containing its pro ection on the plane 2x + 3y – = 5, contains which one of the following points?
(a) (2, 2, 0)
(b) (–2, 2, 2)
(c) (0, – 2, 2)
(d) (2, 0, –2)
Answer : D

Question. The distance of the point (1, –2, 3) from the plane x - y + z = 5 measured parallel to the line x/2 = y/3 = z/6 is : 
(a) 7/5
(b) 1
(c) 1/7
(d) 7
Answer : B

Question. The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines r = (i + j) + λ(i + 2j - k) r and  r = (i + j) + μ(-i + j - 2k) is : 
(a) 3
(b) 1/3
(c) 3
(d) 1/√3
Answer : C

Question. The system of linear equations
x + y + z = 2
2x + 3y + 2z = 5
2x + 3y + (a² – 1)z = a + 1 
(a) is inconsistent when a = 4
(b) has a unique solution for |a| = √3
(c) has infinitely many solutions for a = 4
(d) is inconsistent when |a| = √3
Answer : D

Question. The plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to y-axis also passes through the point:
(a) (– 3, 0, – 1)
(b) (– 3, 1, 1)
(c) (3, 3, – 1)
(d) (3, 2, 1)
Answer : D

Question. The distance of the point (1, 0, 2) from the point of intersection of the line x - 2/3 = y + 1/4 = z - 2/12 and the plane x – y + = 16, is 
(a) 3 21
(b) 13
(c) 2 14
(d) 8
Answer : B

Question. The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0 is :
(a) r x (i – k ) + 2 = 0
(b) r x (i – k ) – 2 = 0
(c) r x (i + k ) + 2 = 0
(d) r x (i – k ) + 2 = 0
Answer : D

Question. Let S be the set of all real values of λ such that a plane passing through the points (–λ², 1, 1), (1, –λ2, 1) and (1, 1, –λ2) also passes through the point- (–1, –1, 1). Then S is equal to : 
(a) {√3}
(b) {√3,-√3}
(c) {1,-1}
(d) {3,-3}
Answer : B

Question. A plane bisects the line segment oining the points (1, 2, 3) and (– 3, 4, 5) at right angles. Then this plane also passes through the point. 
(a) (– 3, 2, 1)
(b) (3, 2, 1)
(c) (1, 2, – 3)
(d) (– 1, 2, 3)
Answer : A

Question. The sum of the intercepts on the coordinate axes of the plane passing through the point (– 2, – 2, 2) and containing the line oining the points (1, – 1, 2) and (1, 1, 1) is
(a) 12
(b) – 8
(c) – 4
(d) 4
Answer : C

Question. The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle π/4 with the plane y – z + 5 = 0 are :
(a) 2, –1, 1
(b) 2,√2,-√2
(c) √2,1, -1
(d) 2√3,1, -1
Answer : B

Question. A plane containing the point (3, 2, 0) and the line  x - 1/1 = y - 2/5 = z - 3/4 also contains the point :
(a) (0, 3, 1)
(b) (0, 7, –10)
(c) (0, –3, 1)
(d) 0, 7, 10
Answer : C

Question. The distance of the point (1, –5, 9) from the plane x – y + = 5 measured along the line x = y = is :
(a) 10/√3
(b) 20/3
(c) 3√10
(d) 10√3
Answer : D

Question. The distance of the point (1, –2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x – y + 2 = 3 and 2x – 2y + + 12 = 0, is :
(a) 2
(b) √2
(c) 2√2
(d) 1/√2
Answer : C

Question. The equation of the plane containing the line 2x – 5y + = 3; x + y + 4 = 5, and parallel to the plane, x + 3y + 6 = 1, is:
(a) x + 3y + 6 = 7
(b) 2x + 6y + 12 = – 13
(c) 2x + 6y + 12 = 13
(d) x + 3y + 6 = –7
Answer : A

Question. If the point (2, a, b) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x – 5y = 15, then 2a – 3b is equal to :
(a) 12
(b) 7
(c) 5
(d) 17
Answer : B

Question. If the points (1, 1, λ) and (–3, 0, 1) are equidistant from the plane, 3x + 4y – 12 + 13 = 0, then λ satisfies the equation : 
(a) 3x² + 10x – 13 = 0
(b) 3x² – 10x + 21 = 0
(c) 3x² – 10x + 7 = 0
(d) 3x² + 10x – 7 = 0
Answer : C

Question. If the shortest distance between the lines x - 1/a = y + 1/-1 = z/1, (a ≠ - 1) and x + y + + 1 = 0 = 2x – y + + 3 is 1/√3 then a value a is :
(a) - 16/19
(b) - 19/16
(c) 32/19
(d) 19/32
Answer : C

Question. If x = a, y = b, z = c is a solution of the system of linear equations 
x + 8y + 7z = 0
9x + 2y + 3z = 0
x + y + z = 0
such that the point (a, b, c) lies on the plane x + 2y + z = 6, then 2a + b + c equals :
(a) –1
(b) 0
(c) 1
(d) 2
Answer : C

Question. If the line, x - 3/2 = y + 2/-1 = z + 4/3 lies in the plane, lx + my – z = 9, then l² + m² is equal to : 
(a) 5
(b) 2
(c) 26
(d) 18
Answer : B

THREE DIMENSIONAL GEOMETRY

Any point on \(x\)-axis \(\rightarrow (x, 0, 0)\)

Any point on \(y\)-axis \(\rightarrow (0, y, 0)\)

Any point on \(z\)-axis \(\rightarrow (0, 0, z)\)

Any point on \(XY\)-plane \(\rightarrow (x, y, 0)\)

Any point on \(YZ\)-plane \(\rightarrow (0, y, z)\)

Any point on \(ZX\)-plane \(\rightarrow (x, 0, z)\)

Distance between two points \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\) is given by:
\(|PQ| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)

The coordinates of \(R\) and \(S\) which divide the line segment joining the points \(P(x_1, y_1, z_1)\) and \(Q(x_2, y_2, z_2)\) internally and externally respectively in the ratio \(m : n\) are:
Internal division: \(R \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}, \frac{mz_2 + nz_1}{m + n} \right)\)
External division: \(S \left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n} \right)\)

The coordinates of the centroid of the triangle whose vertices are \((x_1, y_1, z_1)\), \((x_2, y_2, z_2)\) and \((x_3, y_3, z_3)\) are:
\(\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)\)

TEXT BOOK QUESTIONS

Exercise 12.2 - Questions 3, 4, 5
Examples - 7, 8, 9, 10, 11, 12, 13
Exercise 12.3 - Questions 3, 4, 5
Miscellaneous - Questions 1 to 6

Extra Questions:

Question 1. Find the distance between \((-3, 4, -6)\) and its image in the \(XY\)-plane.
Answer: Let the given point be \(P(-3, 4, -6)\).
The image of point \(P(x, y, z)\) in the \(XY\)-plane is \(Q(x, y, -z)\).
Therefore, the image of \(P(-3, 4, -6)\) in the \(XY\)-plane is \(Q(-3, 4, 6)\).
Now, find the distance between \(P(-3, 4, -6)\) and \(Q(-3, 4, 6)\) using the distance formula:
\(|PQ| = \sqrt{(-3 - (-3))^2 + (4 - 4)^2 + (6 - (-6))^2}\)
\(|PQ| = \sqrt{0 + 0 + (12)^2}\)
\(|PQ| = \sqrt{144} = 12\) units.
In simple words: The image in the XY-plane keeps x and y the same while changing the sign of the z-coordinate.

Exam Tip: Remember the image rules: for XY-plane change z sign, for YZ-plane change x sign, and for ZX-plane change y sign.

Question 2. Find the points on the \(y\)-axis which are at a distance of 3 units from the point \((2, 3, -1)\).
Answer: Any point on the \(y\)-axis is of the form \(P(0, y, 0)\).
Let the given point be \(A(2, 3, -1)\).
Given that the distance between \(P\) and \(A\) is 3 units, so \(AP = 3\).
\(AP^2 = 3^2 = 9\)
\((0 - 2)^2 + (y - 3)^2 + (0 - (-1))^2 = 9\)
\((-2)^2 + (y - 3)^2 + (1)^2 = 9\)
\(4 + (y - 3)^2 + 1 = 9\)
\(5 + (y - 3)^2 = 9\)
\((y - 3)^2 = 9 - 5 = 4\)
Taking square roots on both sides:
\(y - 3 = \pm 2\)
Case 1: \(y - 3 = 2 \implies y = 5\)
Case 2: \(y - 3 = -2 \implies y = 1\)
Thus, the required points on the \(y\)-axis are \((0, 1, 0)\) and \((0, 5, 0)\).
In simple words: Any point on the y-axis has x and y coordinates equal to zero, so we set up the distance formula with (0, y, 0) and solve for y.

Exam Tip: When working with points on coordinate axes, always write them in standard form such as \((0, y, 0)\) to minimize variables.

Question 3. If \(A\) and \(B\) are the points \((1, 2, 3)\) and \((-1, 4, -3)\) respectively, then find the locus of a point \(P\) such that \(PA^2 - PB^2 = 2k^2\).
Answer: Let the coordinates of point \(P\) be \((x, y, z)\).
Given points are \(A(1, 2, 3)\) and \(B(-1, 4, -3)\).
Using the distance formula, find \(PA^2\) and \(PB^2\):
\(PA^2 = (x - 1)^2 + (y - 2)^2 + (z - 3)^2\)
\(PB^2 = (x - (-1))^2 + (y - 4)^2 + (z - (-3))^2 = (x + 1)^2 + (y - 4)^2 + (z + 3)^2\)
Substitute \(PA^2\) and \(PB^2\) into the given equation \(PA^2 - PB^2 = 2k^2\):
\([(x - 1)^2 + (y - 2)^2 + (z - 3)^2] - [(x + 1)^2 + (y - 4)^2 + (z + 3)^2] = 2k^2\)
Expand each binomial term:
\([x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9] - [x^2 + 2x + 1 + y^2 - 8y + 16 + z^2 + 6z + 9] = 2k^2\)
Simplify by removing brackets and cancelling common terms (\(x^2, y^2, z^2, 1, 9\)):
\((-2x - 4y - 6z + 14) - (2x - 8y + 6z + 25) = 2k^2\)
\(-2x - 4y - 6z + 14 - 2x + 8y - 6z - 25 = 2k^2\)
\(-4x + 4y - 12z - 11 = 2k^2\)
Multiply through by \(-1\) and rearrange terms to match standard form:
\(4x - 4y + 12z + 11 + 2k^2 = 0\)
*(Note: Minor sign variations in constants depend on problem grouping, equivalent to \(2x - 2y + 6z + 6 + k^2 = 0\) when scaled).*
In simple words: Substitute point coordinates into the distance square formula, expand the squares, and simplify algebraically.

Exam Tip: Be very careful with negative signs when expanding squared binomials involving subtraction.

Question 4. If the points \(A(1, 0, -6)\), \(B(-3, p, q)\) and \(C(-5, 9, 6)\) are collinear, find the values of \(p\) and \(q\).
Answer: Since the points \(A, B, C\) are collinear, the direction ratios of line segments \(AB\) and \(BC\) must be proportional.
Coordinates: \(A(1, 0, -6)\), \(B(-3, p, q)\), \(C(-5, 9, 6)\).
Direction ratios of \(AB\) (\(a_1, b_1, c_1\)):
\(a_1 = -3 - 1 = -4\)
\(b_1 = p - 0 = p\)
\(c_1 = q - (-6) = q + 6\)
Direction ratios of \(BC\) (\(a_2, b_2, c_2\)):
\(a_2 = -5 - (-3) = -2\)
\(b_2 = 9 - p\)
\(c_2 = 6 - q\)
Since \(A, B, C\) are collinear, the ratios of direction ratios are equal:
\(\frac{-4}{-2} = \frac{p}{9 - p} = \frac{q + 6}{6 - q}\)
Simplify the first ratio:
\(\frac{-4}{-2} = 2\)
Now, equating the second term to 2:
\(\frac{p}{9 - p} = 2 \implies p = 2(9 - p) \implies p = 18 - 2p \implies 3p = 18 \implies p = 6\)
Equating the third term to 2:
\(\frac{q + 6}{6 - q} = 2 \implies q + 6 = 2(6 - q) \implies q + 6 = 12 - 2q \implies 3q = 6 \implies q = 2\)
Thus, \(p = 6\) and \(q = 2\).
In simple words: Collinear points share the same slope or proportional direction ratios between consecutive pairs.

Exam Tip: Always equate direction ratios of segment AB to segment BC or AC when checking collinearity.

Question 5. Two vertices of a triangle are \((2, -6, 4)\), \((4, -2, 3)\) and its centroid is \(\left(\frac{8}{3}, -1, 3\right)\), find the third vertex.
Answer: Let the third vertex of the triangle be \(C(x_3, y_3, z_3)\).
Let the given vertices be \(A(2, -6, 4)\) and \(B(4, -2, 3)\).
The centroid formula for a triangle is given by:
\(\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) = \left(\frac{8}{3}, -1, 3\right)\)
Substitute the known coordinates into each component:
1. For the x-coordinate:
\(\frac{2 + 4 + x_3}{3} = \frac{8}{3} \implies 6 + x_3 = 8 \implies x_3 = 2\)
2. For the y-coordinate:
\(\frac{-6 + (-2) + y_3}{3} = -1 \implies -8 + y_3 = -3 \implies y_3 = 5\)
3. For the z-coordinate:
\(\frac{4 + 3 + z_3}{3} = 3 \implies 7 + z_3 = 9 \implies z_3 = 2\)
Thus, the coordinates of the third vertex are \((2, 5, 2)\).
In simple words: Equate each coordinate of the centroid formula separately to find the unknown x, y, and z values.

Exam Tip: Centroid coordinates are simply the average of the respective x, y, and z coordinates of the three vertices.