2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry

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Karnataka 2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry

2nd PUC Maths Three Dimensional Geometry One Marks Questions and Answers

Question 1.
If a line makes angles 90°, 135°, 45° with the positive X, Y and Z – axes respectively, find its direction cosines.
Answer:
Let direction cosines of the line be l, m and n with the X, Y, and Z-axes respectively, and given that .
α = 90°, β = 135° and γ = 45°
Then, l = cos α = cos 90° = 0, m = cos β = cos 135° = -1/√2
and n = cos γ = cos 45° = 1/√2
Therefore, the direction cosines of the line are 0, \(-\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{2}}\).

Question 2.
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Answer:
Let the line makes an angle α with each of the three coordinate axes, then its direction cosines are
l = cos α, m = cos α, n = cos α
We know that l² + m² + n² = 1
⇒ cos² α + cos² α + cos² α = 1 [∵ l = m = n = cos α]
⇒ 3cos² α = 1 ⇒ cos² α = \(\frac { 1 }{ 3 }\) ⇒ cos θ = ± \(\frac{1}{\sqrt{3}}\)
∴ Direction cosines of the line are either \(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\) to \(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\)

Question 3.
If a line has the direction ratios -18, 12, – 4, then what are its direction cosines?
Answer:
Given, direction’ratios are -18, 12, -4.
Here a = – 18, b = 12 and c = – 4, then direction cosines of a line are

Question 4.
Find the intercepts cut-off by the plane 2x + y – z = 5.
Answer:
Given plane 2x + y – z = 5 can be written as

It is known that the equation of a plane in intercept form is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1, where a, b, c are the intercepts cut-off by the plane at X, Y and Z-axes respectively.
Therefore, for the given equation, a = \(\frac { 5 }{ 2 }\) ,b = 5 and c = -5,
Hence, the intercepts of the plane are \(\frac { 5 }{ 2 }\), 5 and -5.

Question 5.
Find the equation of the plane with intercept 3 on the Y-axis and parallel to ZOX- plane.
Answer:
Equation of the ZOX-plane is y = k
Any plane parallel to it is of the form y = 0
Since, the y-intercept of the plane is 3.
∴ K = 3
Thus, the equation of the required plane is y = 3.

Question 6.
In the following problems find the distance of each of the given points from the corresponding given plane:

Answer:

(a) The given point is (0, 0, 0) and the plane is 3x – 4y + 12z – 3 = 0
∴ From Eq. (i),

(b) The given point is (3, -2, 1) and the plane is 2x – y + 2z + 3 = 0
∴ From Eq. (i),

(c) The given point is (2, 3, -5) and the plane is x + 2y – 2z = 9 = 0
∴ From Eq. (i),

(d) The given point is (-6, 0, 0) and the plane is 2x – 3y + 6z – 2 = 0
∴ From Eq. (i),

2nd PUC Maths Three Dimensional Geometry Two Marks Questions and Answers

Question 1.
Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.
Answer:
Let the given points be A, B and C respectively, then A(2, 3,4), B(-1,-2, 1) and C(5, 8, 7)
It is known that the direction ratios of line joining the points, (x₁, y₁, z₁) and (x₂, y₂, z₂) are given by (x₂ – x₁, y₂ – y₁, z₂ – z₁).
Then, direction ratios of AB are (-1 -2, -2 -3, 1 – 4) ⇒ (-3, -5, -3)
direction ratios of AC are (5 – 2, 8 – 3, 7 – 4) ⇒ (3, 5, 3)
As \(\frac{-3}{5}=\frac{-5}{5}=\frac{-3}{3}\), therefore the direction ratios of AB and AC are proportional.
Hence, AB and AC are parallel, but these have a point A in common, therefore AB and AC are along the same line i.e., A, B and C are collinear.

Question 2.
Find the direction cosines of the sides of the triangle whose vertices are (3,5,-4), (-1,1,2) and (-5,-5,-2).
Answer:
Let the vertices of the triangle be A(3,5,-4), B(-1,1,2) and C(-5, -5,-2) respectively.
The direction ratios of side AB are (-1,-3, 1 – 5, 2 – (-4)) (-4, -4, 2 + 4), i.e., (-4, -4, 6).
[∵ If A(x₁, y₁, z₁) and B(x₂, y₂, z₂) then dr’s of AB = (x₂ – x₁, y₂ – y₁, z₂ – z₁)]
Then, magnitude of AB,

∵ If a, b, c are direction ratios, then direction cosines are

Similary, the direction ratios of side BC are (-5 – (-1)), (-5 -1) and (-2 -2) i.e., (-4, -6, -4).

Therefore, the direction cosines of BC are

The direction ratios of side CA are (-5 -3), (-5 -5) and (-2 -(-4) i.e., (-8, -10, 2) or (8, 10, -2).

Therefore, the direction cosines of AC are

Question 3.
Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular to the line through the points (0,3,2) and (3,5,6).
Answer:
The given points are A(1, -1, 2), B(3, 4, -2), C(0, 3, 2) and D(3, 5, 6) direction ratios of AB are
(3 – 1, 4- (-1), -2 – 2) = (2, 4 + 1, – 4) = (2, 5, -4)
and direction ratios of CD are (3 – 0, 5 – 3 ,6 – 2) or (3,2,4).
We know that two lines AB and Cd with direction ratios, a₁, b₁ c₁, and a₂, b₂, c₂ are perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0
∴ 2 × 3 + 5 × 2 + (- 4) × 4 = 6 + 10 – 16 = 0
Therefore, the lines AB and CD are at right angles.

Question 4.
Show that the line through the points (4,7,8), (2,3,4) is parallel to the line through the points (-1,-2,1), (1,2,5).
Answer:
Let the given points be A(4,7,8), B(2,3,4), C(-1,-2, 1) and D(1 , 2, 5)
Direction ratios of AB (a₁, b₁, c₁,) are (2 – 4), (3 – 7), (4 – 8) ⇒ (- 2, – 4, – 4)
∴ a₁ = – 2, b₁ = – 4, C₁ = -4
Direction ratios of CD (a₂, b₂, c₁) are (1 + 1), (2 + 2), (5 – 1) ⇒ (2,4,4)
∴ a₂ = 2, b₂ = 4, c₂ = 4.
AB is parallel to CD if

Question 5.
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3î + 2ĵ – 2k̂.
Answer:
It is given that the line passes through the point A( 1,2,3).
Therefore, the position vect of A is 3î + 2ĵ – 2k̂ and parallel to point B (1, 2, 3), where position vector b = 3î + 2ĵ – 2k̂.
It is known that the line which passes through point a and parallel to b is given by r = a + λb,
where A, is a constant.
∴ r = î + 2ĵ + 3k̂ + λ (3î + 2ĵ – 2k̂)
This is the required equation of the line.

Question 6.
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector and 2î – ĵ + 4k̂ is in the direction î + 2ĵ – k̂.
Answer:
It is given that the line passes through the point with position vector \(\vec{a}\) = 2î – ĵ + 4k̂ and \(\vec{b}\) = î + 2ĵ – k̂.
It is known that a line through a point with position vector a and parallel to b i.e., indirection of b is given by the equation, r = a + λb.
∴ \(\vec{r}\) = 2î – ĵ + 4k̂ + λ(î + 2ĵ – k̂)
This is the required equation of the line in vector form.

Eliminating λ, we obtain the Cartesian form equation as

This is the required equation of the line in Cartesian form.

Question 7.
Find the Cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\).
Answer:
It is given that the line passes through the point (-2, 4, -5) and is parallel to

It is known that the equation of the line passes through the point (x₁, y₁, z₁) and with direction ratios a, b, c, is given by

Also, the line passes through (-2, 4, -5) and having direction ratios (3, 5, 6) therefore, the equation of the line (in Cartesian form) is

which is the equation of required line.

Question 8.
The Cartesian equation of a line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\) write its vector form.
Answer:
The given Cartesian equation is

The above line passes through the point (5, -4, 6). The position vector of this point is \(\vec{a}\) = 5î – 4ĵ + 6k̂.
Also, the direction ratios of the given line are 3, 7 and 2. This means that the line is in the direction of vector, b = 3î + 7ĵ + 2k̂
It is known that the line through position vector a and in the direction of the vector b is given by the equation,

This is the required equation of the given line in vector form.

Question 9.
Find the vector and the Cartesian equations of the line the passes through the origin and (5, -2, 3).
Answer:
Let a and b be the position vectors of points (0, 0, 0) and (5, -2, 3) respectively
\(\vec{a}\) = 0î + 0ĵ + 0k̂ and \(\vec{b}\) = 5î – 2ĵ + 3k̂.
We know that the vector equation of a line passing through the points having position vectors a and b is r

Which is Cartesian form of the required line.

Question 10.
Find the vector and the Cartesian equation of the line that passes through the points (3,-2, – 5), (3, – 2, 6).
Answer:
Let a and b be the position vectors of points (3, -2, – 5), and (3, – 2, 6) respectively.
∴ \(\vec{a}\) = 3î – 2ĵ – 5k̂ and b = 3î – 2ĵ + 6k̂
We know that the vector equation of a line passing through the points having position vectors a and b is

Which is Cartesian form of the required line.

Question 11.
Find the value of p so that the lines \(\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}\) and \(\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles.
Answer:
Equation of the line given lines can be written as standard form

Direction ratios of these lines are respectively

Two lines with direction ratios a₁, b₁, c₁, and a₂, b₂, c₂ are perpendicular to each other, if a₁a₂ + b₁b₂ + c₁c₂ = o.

Thus, the value of p = \(\frac{70}{11}\)

Question 12.
Show that the lines \(\frac{x-5}{7}=\frac{y+2}{-5}\) = \(\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{x}{3}\) are perpendicular to each other.
Answer:
The equations of the given lines are

direction ratios of the given lines are respectively 7, – 5,1 and 1,2, 3.
Two lines with direction ratios, a₁, b₁, c₁, and a₂, b₂, c₂ are perpendicular to each other, if
a₁a₂ + b₁b₂ + c₁c₂ = 0.
∴ (7) (1) + (-5) (2) + (1) (2) + (1) (3) = 7 – 10 + 3 = 0
Therefore, the given lines are perpendicular.

Question 13.
Find the Cartesian equation of the following planes:
(a) r.(î + ĵ – k̂)= 0
(b) r.(2î + 3ĵ – 4k̂) = 1
(c) r.[(s – 2t)î + (3 – t)ĵ + (25 + t)k̂] = 15.
Answer:
(a) Given vector reqution is r.(î + ĵ – k̂)= 0 …… (i)
For any arbitrary point P(x, y, z) on the plane, position vector r is given by xî + yĵ + zk̂.
Substituting the value of pin Eq. (i), we obtain (xî + yĵ + zk̂).(î + ĵ – zk̂) = 2 ⇒ x + y – z = 2
Which is the required Cartesian equation,

(b) Given vector reqution is r.(2î + 3ĵ – 4k̂) = 1 ….. (ii)
For any arbitrary point P(x, y, z) on the plane, position vector r is given by xî + yĵ + zk̂.
Substituting the value of r in Eq. (i), we obtain

Which is the required Cratesian from.

(c) Given vector reqution is r.[(s – 2t)î + (3 – t)ĵ + (25 + t)k̂] = 15. ….. (iii)
For any arbitraiy point P(x, y, z) on the plane, position vector r is given by xî + yĵ + zk̂.
Substituting the value of r in Eq. (i), we obtain

This is the Cartesian equation of the plane.

2nd PUC Maths Three Dimensional Geometry Three Marks Questions and Answers

Question 1.
Find the shortest distance between the lines.

Answer:

(b)

Answer:

Question 2.
Find the angle between the following pairs of lines:

Answer:

(b)

Answer:

Answer:

(d)

Answer:

(e)

Answer:

Question 3.
Find the distance between the parallel lines \(\vec{r}\) = î + 2ĵ – 4k̂ + m(2î + 3ĵ + 6k̂) and
\(\vec{r}\) = 3î + 3ĵ – 5k̂ + n(2î + 3ĵ + 6k̂)
Answer:
The two lines are parallel we have

Question 4.
Find the angle between the planes whose vector equations are

Answer:

(b) 3x – 6y + 2z = 7 and 2x + 2y – 2z = 5
Answer:

Question 5.
Find the angle between the line \(\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}\) and the plane 10x + 2y – 11z = 0.
Note : The angle between a line and a plane is the complement of the angle between \(\vec{b}\) and \(\vec{n}\)
Answer:

Question 6.
In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin:
(a) 2x + 3y + 4z – 12 = 0
(b) 3y + 4z – 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
Answer:
(a) Given plane is 2x + 3y + 4z – 12 = 0 …… (i)
The direction ratios of normal are 2, 3 and 4.

Equation of the line through origin and at right angles to Eq. (i) are

Any point on this line is (2t, 3t, 4t). This point lies in the plane (i)
If 2(2t) + 3(3t) + 4(4t) – 12 = 0 i.e, t = \(\frac { 12 }{ 29 }\)
∴ The required foot of perpendicular is

(b) Given plane is 3y + 4z – 6 = 0 ……. (ii)
∴ Dr’s of any line perpendicular to plane (ii) are 0,3,4.
Hence, equation of the line through origin and at right angles to

Any point on this line is (0,3t, 4t). This point lies in the plane (ii),
If 0 + 3 × (3t) + 4 × (4t) – 6 = 0 i.e., t = \(\frac { 6 }{ 25 }\)
Hence, the required foot of perpendicular is

(c) Given plane is x + y + z = 1 ……. (iii)
∴ Dr’s of any line perpendicular to plane (ii) are (1, 1,1)
Hence, equation of the line through origin and at right angles to

Any point on this line is (0, 3t, 4t). This point lies in the plane (iii), t + t + t = 1 i.e., t =\(\frac { 1 }{ 3 }\)

(d) Given plane is 5y + 8 = 0 …… (iv)
Dr’s of any line perpemlicular to plane (vi) are 0,5, 0.
Hence, equation of the line through origin and at right angles to

Any point on this line is (0, 5t, 0). This lies in the plane (vi) if 5 × 5t + 8 = 0 i.e., t = \(\frac { -8 }{ 25 }\)

Question 7.
Find the vector and Cartesian equations of the planes.
(a) That passes through the point (1,0, – 2) and the normal to the plane is î – 2ĵ + k̂.
(b) That passes through the point (1, 4, 6) and the normal to the plane is î – 2ĵ + k̂.
Answer:
(a) The position vector of point (1, 0, -2) is a = î – 2k̂
The normal vector N perpendicular to the plane is N = î + ĵ – k̂
The vector equation of the plane is given by (r – a). N = 0
⇒ [r – (î + 0ĵ – 2k̂)].(î + ĵ – k̂) = 0 ….. (i)
where, r is the position vector of any point P(x, y, z) in the plane.
i.e., r = xî + yĵ – zk̂
Therefore, Eq. (i) becomes

This is the Cartesian equation of the required plane.

(b) The position vector of point (1,4,6) is a = î + 4ĵ + 6k̂
The normal vector N perpendicular to the plane is N = î – 2ĵ + 6k̂
The vector equation of the plane is given by, (r – a). N = 0

⇒ (x – 1) + y – (z + 2) = 0
⇒ (x – 1) – 2 (y – 4) +(z – 6) = 0 ⇒ x – 2y + z + 1 = 0
This is the Cartesian equation of the required plane.

Question 8.
Find the equation of the planes that passes through the sets of three points,
(a) (1,1, – 1), (6, 4, – 5) and (-4, – 2, 3)
(b) (1, 1,0), (1,2, 1), (- 2, 2,-1).
Answer:
(a) The given points are A(1, 1, – 1), B(6, 4, – 5) and C(- 4, – 2,3).
Before determine the equation of the plane, firstly check the condition of collinear.

– 1(18 – 20) – 1(-12 + 16) = 2 + 2 – 4 = 0
Since, A, B, C re collinear points there will be infinite number of planes passing through the given points.

(b) The given points are A( 1, 1, 0), B( 1, 2, 1) and C(- 2, 2, – 1).
Firstly check the condition of collinear point.

-1 (- 2 – 2) – 1(-1 + 2) + 0(2 + 4) = – 5 ≠ 0
Therefore, a plane will pass through the points A, B and C.
It is known that the equation of the plane through the points (x₁, y₁, z₁,), (x₂, y₂, z₂) and (x₃, y₃, z₃) is

⇒ -2(x – 1) – 3 (y – 1) + 3z – 0 ⇒ -2x + 2 – 3y + 3 + 3z = 0
⇒ 2x + 3y – 3z – 5 = 0 ⇒ 2x + 3y- 3z = 5.

2nd PUC Maths Three Dimensional Geometry Five Marks Questions and Answers

Question 1.
Obtain the equation of a line passing through a point A with Position vector a and parallel to \(\vec{a}\) vector \(\vec{b}\) both in vector and cartesian form. (July 2014) (March 2015)
Answer:
Vector Form
Let \(\vec{a}\) be the position vector of the given point A with respect to the origin O of the rectangular coordinate system. Let l be the line which passes through the point A and is parallel to a given vector \(\vec{b}\). Let \(\vec{r}\) be the position vector of an arbitrary point P on the line (Fig 11.4).

Conversely, for each value of the parameter λ, this equation gives the position vector of a point P on the line. Hence, the vector equation of the line is given by

Cartesian Form
Let the coordinates of the given point A be (x, y, z) and the direction ratios of the line be a, b, c. Consider the coordinates of any point P be (x. y, z). Then

Substituting these values in (1) and equating the coefficients of î, ĵ, and k̂, we get
x = x₁ + λa ; y = y₁ + λ b; z = z₁ + λc …… (2)
These are parametric equations of the line. Eliminating the parameter λ from (2), we get

This is the Cartesian equation of the line.

Question 2.
Derive the equation of a line in 3D passing through two points A and B with position vectors \(\vec{a}\) and \(\vec{b}\) respectively both in vector and Cartesian form.
Answer:
Vector Form

Question 3.
Derive the formula for the shortest distance between skew lines \(\vec{r}=\vec{a}+\lambda \vec{b}\) and \(\vec{r}=\vec{a}_{2}+\lambda \overrightarrow{b_{2}}\) in vector form.
Answer:

Question 4.
Derive the formula for the distance between two parallel lines \(\vec{r}=\overrightarrow{a_{1}}+\lambda \vec{b}\) and \(\overrightarrow{\mathbf{r}}=\overrightarrow{\mathbf{a}_{2}}+\mu \overrightarrow{\mathbf{b}}\) in vector form.
Answer:

FIG
If two lines l₁, and l₂ are parallel, then they are coplanar. Let the lines be given by

As l₁, l₂ are coplanar, if the foot of the perpendicular from T on the line l₁, is P, then the distance between the lines l₁, and l₂ = |TP|

where n̂ is the unit vector perpendicular to the plane of the lines l₁, and l₂

Question 5.
Derive the equation of a plane in normal form (both in the vector and Cartesian form). (March 2014)
Answer:

Cartesian form
Equation (2) gives the vector equation of a plane, where n̂ is the unit vector normal to the plane. Let P (x, y, z) be any point on the plane. Then

This is the Cartesian equation of the plane in the normal form.

Question 6.
Derive the equation of a plane perpendicular to a given vector and passing through a given point in both vector form and Cartesian form. (July 2015)
Answer:

Vector Form
Let a plane pass through a point A with position vector \(\vec{a}\) and perpendicular to the vector \(\vec{N}\)
Let \(\vec{r}\) be the position vector of any point P(x,y,z) in the plane.

Cartesian form
Let the given pointAbe (x₁, y₁, z₁) , P be(x, y, z)

Question 7.
Obtain the equation of the plane passing through three non-collinear points whose position vectors are \(\vec{a}\),\(\vec{b}\),\(\vec{c}\) in vector and Cartesian form.
Answer:

Vector Form
Let R, S and T be three non collinear points with position vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) respectively.
The vectors \(\overrightarrow{\mathrm{RS}}\) and \(\overrightarrow{\mathrm{RT}}\) are in the given plane. Therefore, the vector \(\overrightarrow{\mathrm{RS}} \times \overrightarrow{\mathrm{RT}}\) is any perpendicular to the plane containing points R, S and T. Let \(\vec{r}\) be the position vector of any point P in the plane. Therefore, the equation of the plane passing through R and perpendicular to the vector

Cartesian Form
Let (x₁, y₁, z₁), (x₂,y₂,z₂) and (x₃, y₃,z₃) be the coordinates of the points R, S and T respectively. Let (x, y, z) be the coordinates of any point P on the plane with position vector \(\vec{r}\) Then

Substituting these values in equation (1) of the vector form and expressing it in the form of a determinant, we have

Question 8.
Derive the condition for the coplanarity of two lines in space both in vector form and in Cartesian form.
Answer:
Vector Form
Let the given lines be

Cartesian Form
Let (x₁, y₁, z₁,) and (x₂, y₂, z₂) be the corrdinates of the points A and B respectively.
Let a₁,b₁, c₁ and a₂, b₂, c₂ be the direction ratios of \(\vec{b}_{1}\) and \(\vec{b}_{2}\), respectively. Then

The given lines are coplanar if and only if \(\overrightarrow{\mathrm{AB}} \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)=0\)= 0. In the cartesian form, it can be expressed as