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# JEE Advanced Maths Three Dimensional Geometry Previous Year Questions with Solutions

Three-dimensional geometry JEE Advanced previous year questions with solutions are given on this page. Direction ratios and direction cosines of a line, angle between two lines in terms of direction cosines and direction ratios, projection of a point on a line, angle between two lines, distance between two parallel lines, equation of a plane in different forms, etc are the important topics in three-dimensional geometry. These solutions are prepared by our team of experts. Learning these solutions will definitely help you to easily crack the problems from three-dimensional geometry for the JEE Advanced exam.

Question 1: Perpendiculars are drawn from points on the line (x+2)/2 = (y+1)/-1 = z/3 to the plane x+y+z = 3. The feet of perpendiculars lie on the line is

(a) x/5 = (y-1)/8 = (z-2)/-13

(b) x/2 = (y-1)/3 = (z-2)/-5

(c) x/4 = (y-1)/3 = (z-2)/-7

(d) x/2 = (y-1)/-7 = (z-2)/5

Solution:

Given that the equation of the line is (x+2)/2 = (y+1)/-1 = z/3 = λ

So any point P on the line is x = 2λ-2, y = -λ-1, z = 3λ ..(i)

It lies on the plane x+y+z = 3

=> (2λ-2) + (-λ-1) + 3λ = 3

=> 4λ – 6 = 0

=> λ = 3/2

Substitute λ in (i) and get P

So P = (1, -5/2, 9/2) ..(ii)

We can observe that (-2, -1, 0) is a point on the line.

Let (x, y, z) be the foot of the perpendicular from point (-2, -1, 0) on the plane x+y+z = 3.

=> (x+2)/1 = (y+1)/1 = (z-0)/1

= -(1(-2) + 1(-1) + 0(1) – 3)/(12+12+12)

=> Q(x, y, z) = (0, 1, 2) ..(iii)

Direction ratios of PQ = (1, -7/2, 5/2) (from (ii) and (iii))

= (2, -7, 5) ..(iv)

From (iii) and (iv), equation of required line is x/2 = (y-1)/-7 = (z-2)/5

Hence option d is the answer.

Question 2: Two lines L1: x = 5, y/(3-α) = z/-2 and L2: x = α, y/-1 = z/(2-α) are coplanar. Then α can take values

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

Given lines L1: x = 5, y/(3-α) = z/-2 and

L2: x = α, y/-1 = z/(2-α)

L1 and L2 are coplanar.

So

$$\begin{array}{l}\begin{vmatrix} 5-\alpha & 0 & 0\\ 0&3 -\alpha &-2 \\ 0& -1& 2-\alpha \end{vmatrix}=0\end{array}$$

=> (5-α)(3-α)(2-α) – 2 = 0

=> (5-α)(6 – 3α – 2α + α2 – 2) = 0

=> (5-α)(α-1)(α-4) = 0

=> α = 1, 4, 5

Hence option a and d is the answer.

Question 3: Let P be the image of the point (3, 1, 7) with respect to the plane x-y+z = 3. Then the equation of the plane passing through P and containing the straight line x/1 = y/2 = z/1 is

(a) x+y-3z = 0

(b) 3x+z = 0

(c) x-4y+7z = 0

(d) 2x-y = 0

Solution:

Equation of line passing through P is (x-3)1 = (y-1)/-1 = (z-7)/1

Distance of point P from the given plane = -2(6)/3 = -4

(x-3)1 = (y-1)/-1 = (z-7)/1 = -4

=> x = -1, y = 5, z = 3

=> P= (-1, 5, 3)

Equation of plane passing through P is a(x+1) + b(y-5) + c(z-3) = 0

Normal of the plane is perpendicular to the line from which this plane passes through.

So a+2b+c = 0 ..(i)

The plane will also pass through the origin since the line passes through the origin.

So a-5b-3c = 0 ..(ii)

Solving (i) and (ii)

a/1 = b/-4 = c/7

Required equation of plane is (x+1) -4(y-5) +7(z-3) = 0

=> x-4y+7z = 0

Hence option c is the answer.

Question 4: The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y – 2z = 5 and 3x – 6y – 2z = 7 is

(a) 14x+2y+15z = 31

(b) 14x+2y-15z = 1

(c) -14x+2y+15z = 3

(d) 14x-2y+15z = 27

Solution:

Let plane P1 => 2x + y – 2z = 5

P2 => 3x – 6y – 2z = 7

Let P be the plane perpendicular to P1 and P2

Also P passes through (1, 1, 1).

Hence required equation =

$$\begin{array}{l}\begin{vmatrix} x-1 & y-1 & z-1\\ 2 & 1 &-2 \\ 3 & -6 & -2 \end{vmatrix}=0\end{array}$$

=> (x-1)(-2-12) – (y-1)(-4+6) + (z-1)(-12-3) = 0

=> 14x-14+2y-2+15z-15 = 0

=> 14x + 2y +15z = 31

Hence option a is the answer.

Question 5: If for some α∈ R , the lines L1 : (x+1)/2 = (y-2)/-1 = (z-1)/1 and L2 : (x+2)/α = (y+1)/(5-α) = (z+1)/1 are coplanar, then the line L2 passes through the point:

(a) (2, -10, -2)

(b) (10, -2, -2)

(c) (10, 2, 2)

(d) (-2, 10, 2)

Solution:

A(-1,2,1), B(-2,-1,-1)

$$\begin{array}{l}\left [ \overrightarrow{AB} \overrightarrow{b_{1}} \overrightarrow{b_{2}}\right ]=0\end{array}$$
$$\begin{array}{l}\begin{vmatrix} -1 & -3&-2 \\ 2&-1 & 1\\ \alpha & 5-\alpha & 1 \end{vmatrix}=0\end{array}$$

-1(-1+α-5)+3(2-α)-2(10-2α+α) = 0

6-α+6-3α+2α-20 = 0

-8-2α = 0

α = -4

L2: (x+2)/-4 = (y+1)/9 = (z+1)/1

Check options. (2, -10, -2) satisfies above equation.

Hence option a is the answer.

Question 6: 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) 1/7

(b) 7

(c) 7/5

(d) 1

Solution:

Equation of line through (1,-2,3) whose d.r.s. are (2, 3, -6)

(x-1)/2 = (y+2)/3 = (z-3)/-6 = λ

Any point on the line (2λ+1, 3λ-2, -6λ+3)

Substitute in equation of plane

x-y+z = 5

2λ+1- 3λ+2 -6λ+3 = 5

-7λ = -1

λ = 1/7

Distance = √((2λ)2+(3λ)2+(6λ)2)

= √(4λ2+9λ2+36λ2)

= 7λ

= 1

Hence option d is the answer.

Question 7: Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in the plane x + y = 3) lies on the z-axis. Let the distance of P from the x axis be 5. If R is the image of P in the xy-plane, then the length of PR is

Solution:

Let the coordinates of P are (a, b, c).

Coordinates of Q are (0, 0, c) and coordinates of R are (a, b, -c).

Given PQ is perpendicular to the plane x+y = 3

So PQ parallel to the normal of the given plane.

(ai+bj) is parallel to (i+j). (i, j are unit vectors)

Comparing we get a = b.

Since the midpoint of PQ lies on plane x+y = 3

So a/2 + b/2 = 3

=> a+b = 6

=> a = 3, b = 3

So distance of P from x axis = √(b2+c2) = 5 (given)

=> (b2+c2) = 25

c2 = 25-9 = 16

=> c = ±4

So PR = |2c|

= 8

Question 8: Let α, β, γ, δ be real numbers such that α2+ β22 ≠0 and α + γ = 1. Suppose the point (3, 2, -1) is the mirror image of the point (1, 0, -1) with respect to the plane αx + βy + γz = δ. Then which of the following statements is/are TRUE?

(a) α + β = 2

(b) δ – γ = 3

(c) δ + β = 4

(d) α + β + γ = δ

Solution: Midpoint of PQ = A(2, 1, -1)

Direction ratios of PQ = 2, 2, 0

As PQ perpendicular to plane and midpoint lies on plane.

The equation of the plane is

2(x-2) + 2(y-1) + 0(z+1) = 0

=> x+y = 3

Comparing with αx + βy + γz = δ

We get α = 1, β = 1, γ = 0 and δ = 3

Hence options a, b, c are true.

Question 9: In R3, consider the planes P1: y = 0 and P2: x+z = 1. Let P3 be a plane, different from P1 and P2, which passes through the intersection of P1 and P2. If the distance of the point (0, 1, 0) from P3 is 1 and the distance of a point ( α, β, γ ) from P3 is 2, then which of the following relation is (are) true?

(a) 2α +β+2γ +2= 0

(b) 2α -β+2γ +4 = 0

(c) 2α +β-2γ -10= 0

(d) 2α -β+2γ -8 = 0

Solution:

Let P3 be P2 + λP1 = 0

=> (x+z-1) + λy = 0

=> x+λy+z-1 = 0

Distance of the point (0, 1, 0) from P3 is |(λ-1)/√(2+λ2)| = 1

=> (λ-1)2 = (2+λ2)

=> -2λ + 1 = 2

=> -2λ = 1

=> λ = -1/2

Distance of point (α, β, γ) from P3: |(α+λβ + γ-1) /√(2+λ2)| = 2

|(α-(β/2) + γ-1) /(3/2)| = ±2

α- (β/2) + γ-1 =± 3

=> 2α- β + 2γ-2 =± 6

=> 2α- β + 2γ-8 = 0 or 2α- β + 2γ + 4 = 0

Hence option b and d are the answers.

Question 10: If the straight lines (x-1)/2 = (y+1)/k = z/2 and (x+1)/5 = (y+1)/2 = z/k are coplanar, then the plane(s) containing these two lines is (are)

(a) y + 2z = -1

(b) y + z = -1

(c) y – z = -1

(d) y – 2z = -1

Solution:

Given that lines are coplanar.

$$\begin{array}{l}\begin{vmatrix} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1}\\ a_{1} &b_{1} & c_{1}\\ a_{2}&b_{2} & c_{2} \end{vmatrix}=0\end{array}$$

=>

$$\begin{array}{l}\begin{vmatrix} 2 & 0 &0 \\ 2&k &2 \\ 5 & 2 & k \end{vmatrix}=0\end{array}$$

=> k = ±2

For k = 2, equation of the plane is given by

$$\begin{array}{l}\begin{vmatrix} x-1 & y+1 &z \\ 2&2 &2 \\ 5 & 2 & 2 \end{vmatrix}=0\end{array}$$

=> y-z+1 = 0

=> y-z = -1

For k = -2, equation of the plane is given by

$$\begin{array}{l}\begin{vmatrix} x-1 & y+1 &z \\ 2&-2 &2 \\ 5 & 2 & -2 \end{vmatrix}=0\end{array}$$

=> y+z+1 = 0

=> y+z = -1

Hence option b and c are correct.

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