Three Dimensional Geometry MCQ Quiz - Objective Question with Answer for Three Dimensional Geometry - Download Free PDF

Calculation:

Given,

The plane passes through the point \(P(1, 1, 1)\) and is perpendicular to the line with direction ratios \( (3, 2, 1) \).

The general equation of the plane is:

\(Ax + By + Cz = D\), where \( (A, B, C) \) are the direction ratios of the normal to the plane.

Since the plane is perpendicular to the line, the normal vector to the plane is \( (3, 2, 1) \).

Thus, the equation of the plane becomes:

\( 3x + 2y + z = D \).

To find \( D \), substitute the point \( (1, 1, 1) \) into the equation of the plane:

\( 3(1) + 2(1) + 1 = D \), which simplifies to:

\( 3 + 2 + 1 = D \),

\( D = 6 \).

∴ The equation of the plane is \( 3x + 2y + z = 6 \).

Hence, the correct answer is Option 2.

Calculation:

Given,

The line equation is: \( \frac{x+1}{p} = \frac{y-1}{q} = \frac{z-2}{r} \),  where p = 2q = 3r .

The line makes an angle θ  with the positive direction of the y-axis.

Using the relation between the direction cosines

\( l = \frac{p}{\sqrt{p^2 + q^2 + r^2}}, m = \frac{q}{\sqrt{p^2 + q^2 + r^2}}, n = \frac{r}{\sqrt{p^2 + q^2 + r^2}} \),

we substitute p = 3r  and  q = \(\frac{3r}{2} \)

Direction cosine with y -axis is given by \( m = \frac{q}{\sqrt{p^2 + q^2 + r^2}} \).

\( m = \frac{\frac{3r}{2}}{\sqrt{(3r)^2 + \left(\frac{3r}{2}\right)^2 + r^2}} = \frac{\frac{3r}{2}}{\sqrt{9r^2 + \frac{9r^2}{4} + r^2}} = \frac{\frac{3r}{2}}{\sqrt{\frac{49r^2}{4}}} = \frac{3}{7} \).

Now, using the double angle identity for cosine \( \cos 2θ = 2 \cos^2 θ - 1 \):

\( \cos 2θ = 2 \left( \frac{3}{7} \right)^2 - 1 = 2 \times \frac{9}{49} - 1 = \frac{18}{49} - 1 = \frac{-31}{49} \).

∴ The value of \( \cos 2θ \) is \( -\frac{31}{49} \).

Hence, the correct answer is Option 1.

Calculation:

We know that \( \cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1 \) ... (i)

And \( \cos(\alpha + \beta) \cdot \cos(\alpha - \beta) = \cos^{2} \alpha - \sin^{2} \beta \)

Substitute and simplify:

\( = \cos^{2} \alpha + \cos^{2} \beta - 1 = 1 - \cos^{2} \gamma \quad \Rightarrow \quad - \cos^{2} \gamma \)

Hence, the correct answer is Option 2.

Calculation: 

Equation of plane is 

(x - 2y - z - 5) + b(x + y + 3z - 5) = 0 

\(\left|\begin{array}{lll} 1+b & -2+b & -1+3 b \\ 1 & 1 & 2 \\ 2 & 3 & 1 \end{array}\right|=0\)

⇒ b = 12

∴ plane is 13x + 10y + 35z = 65

Distance from given point to plane = 9 

Hence, the correct answer is 9. 

Calculation:

The vector \(\vec{PQ} \)is:

\( \vec{PQ} = (4 - (-3), 6 - 2, -2 - 3) = (7, 4, -5) \)

The direction vector of the line\(\vec{d} \) is (3, 3, -1).

The cross product \(\vec{PQ} \times \vec{d} \) is:

\( \vec{PQ} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 4 & -5 \\ 3 & 3 & -1 \end{vmatrix} = (11, -8, 9) \)

The magnitude of the cross product is:

\( |\vec{PQ} \times \vec{d}| = \sqrt{11^2 + (-8)^2 + 9^2} = \sqrt{266} \)

The magnitude of the direction vector \(\vec{d}\) = (3, 3, -1) is:

\( |\vec{d}| = \sqrt{3^2 + 3^2 + (-1)^2} = \sqrt{19} \)

Now, the distance d is:

\( d = \frac{|\vec{PQ} \times \vec{d}|}{|\vec{d}|} = \frac{\sqrt{266}}{\sqrt{19}} = \sqrt{\frac{266}{19}} = \sqrt{14} \)

Hence, the correct answer is Option 4.

Concept:

The direction cosines of the vector are the cosines of angles that the vector forms with the coordinate axes.

Calculation:

Z-axis makes an angle 90° with X-axis, 90° with Y-axis, and 0° with Z-axis

∴ Direction cosines of Z-axis: cos 90, cos 90, cos 0

i.e., 0, 0, 1

Now sum of the direction cosine of z-axis = 0 +  0 + 1 = 1

Hence, option (3) is correct. 

Concept:

1. Direction angles: If α, β, and γ are the angles made by the line segment with the coordinate axis then these angles are termed to be the direction angles.
2. Direction cosines: The cosines of direction angles are the direction cosines of the line. Hence, cos α, cos β and cos γ are called as the direction cosines

It is denoted by l, m and n. ⇔ l = cos α, m = cos β and n = cos γ

3. The sum of squares of the direction cosines of a line is equal to unity.
4. l² + m² + n² = 1 or cos² α + cos² β + cos² γ = 1
4. Direction ratios: Any numbers which are proportional to the direction cosines of a line are called as the direction ratios. It is denoted by ‘a’, ‘b’ and ‘c’.
5. a ∝ l, b ∝ m and c ∝ n ⇔ a = kl, b = km and c = kn Where k is a constant.

Calculation:

We have to find the value of sin² α + sin² β + sin² γ

We know that sum of squares of the direction cosines of a line is equal to unity.

⇒ cos² α + cos² β + cos² γ = 1

⇒ 1 - sin² α + 1 - sin² β + 1 - sin² γ = 1 (∵ sin² θ + cos² θ = 1)

⇒ 3 – (sin² α + sin² β + sin² γ) = 1

⇒ 3 – 1 = sin² α + sin² β + sin² γ

CONCEPT:

• Let A₁x + B₁y + C₁z + D₁ = 0 and A₂x + B₂y + C₂ z + D₂ = 0 are the equations of two planes aligned at an angle θ where A₁, B₁, C₁ and A₂, B₂, C₂ are the direction ratios of the normal to the planes, then the cosine of the angle between the two planes is given by:

\(cos\theta = \left| {\frac{{{A_1}{A_2} + {B_1}{B_2} + {C_1}{C_2}}}{{\sqrt {A_1^2 + B_1^2 + C_1^2} \sqrt {A_2^2 + B_2^2 + C_2^2} }}} \right|\)

CALCULATION:

Given planes are x + 2y + z = 7 and 2x – y + z = 13.

\(cos\theta = \left| {\frac{{1.2 - 2.1 + 1.1}}{{\sqrt {{1^2} + {2^2} + {1^2}} \sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {1^2}} }}} \right| = \frac{1}{6}\)

Concept:

Let θ be the angle between two lines of slope m₁ and m₂, then the acute angle between the lines is given by:

\(\tan θ = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1} \cdot \ {m_2}}}} \right|\)

Calculations:

Consider, the slope of the line (x - 2 = 0) is m₁

So, m₁ = ∞ .

And, the slope of the line (√3x - y - 2 = 0) is m₂

So, m₂ = √3.

Now, the angle between the given lines is θ.

⇒ \(\tan θ = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1} \cdot \ {m_2}}}} \right|\)

⇒ \(\tan θ = \left| {\frac{{\frac{{{m_2}}}{{{m_1}}} - 1}}{{\frac{1}{{{m_1}}} + {m_2}}}} \right|\)

⇒ \(tan \ θ = \frac{1}{\sqrt 3}\)

⇒ θ = 30°

Hence, the correct option is 2.

Alternate Method

For line 1:x – 2 = 0Hence, x = 2

Since it is a vertical line, the slope is undefined.

Hence, θ₁ = 90°

For line 2:

√3x - y - 2 = 0

Compare this equation with, y = mx + c

It becomes, y = √3x + 2

Hence, m = √3 = tan θ₂

Hence, θ₂ = \(tan^{-1}{\sqrt 3}\) = 60°

Now, the graph of intersection between two lines can be drawn as

Hence, acute angle between the lines = θ₁ – θ₂ = 90° – 60° = 30°

Concept:

Let two lines having direction ratios a1, b1, c1, and a2, b2, c2 respectively.

Condition for perpendicular lines: a1a2 + b1b2 + c1c2 = 0

Condition for parallel lines: \(\rm \frac {a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2}\)

Calculation:

Given lines are \(\frac{{{\rm{x}} - 2}}{{2{\rm{k}}}} = \frac{ {{\rm{y}-3}}}{3} = \frac{{{\rm{z}} + 2}}{{ - 1}}\) and \(\frac{{{\rm{x}} - 2}}{8} = \frac{ {{\rm{y}-3}}}{6} = \frac{{{\rm{z}} + 2}}{{ - 2}}\)

The direction ratio of the first line is (2k, 3, -1) and the direction ratio of second line is (8, 6, -2)

Lines are parallel;

So, \(\rm \frac {2k}{8} = \frac {3}{6} = \frac {-1}{-2}\)

⇒ \(\rm \frac {k}{4} = \frac {1}{2} = \frac {1}{2}\)

∴ k = 2

CONCEPT:

The equation of a line with direction ratio  that passes through the point (x1, y1, z1) is given by the formula:

\(\rm \frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\)

CALCULATION:

Given: Equation of line is 2x = 3y = 5 - 4z

We will first be converting the above expression in standard form for comparison, i.e. we need to get rid of coefficients of x, y, and z, i.e. 2, 3, and 4 respectively.

LCM of 2, 3, and 4 is 12.

∴ Dividing the above equation by 12, we get:

\(⇒ \frac{{2x}}{{12}} = \frac{{3y}}{{12}} = \frac{{5\ -\ 4z}}{{12}}\;\)

\(⇒ \frac{x}{6} = \frac{y}{4} = \frac{{z - \frac{5}{4}}}{{ - 3}}\;\)

By comparing the above equation with \(\rm \frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\) we get

⇒ a = 6, b = 4 and c = -3

So, the direction ratios of the given line is: <6, 4, - 3>

Hence, the correct option is 2.

Concept:

Let two lines having direction ratios a_1, b₁, c_1, and a_2, b₂, c₂ respectively.

Condition for perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0

Calculation:

Given lines are  \(\frac{{2{\rm{x}} - 2}}{{2{\rm{k}}}} = \frac{{4 - {\rm{y}}}}{3} = \frac{{{\rm{z}} + 2}}{{ - 1}}\) and \(\frac{{{\rm{x}} - 5}}{1} = \frac{{\rm{y}}}{{\rm{k}}} = \frac{{{\rm{z}} + 6}}{4}\) 

Write the above equation of a line in the standard form of lines

\( \Rightarrow \frac{{2\left( {{\rm{x}} - 1} \right)}}{{2{\rm{k}}}} = \frac{{ - \left( {{\rm{y}} - 4} \right)}}{3} = \frac{{{\rm{z}} + 2}}{{ - 1}} \Leftrightarrow \frac{{\left( {{\rm{x}} - 1} \right)}}{{\rm{k}}} = \frac{{{\rm{y}} - 4}}{{ - 3}} = \frac{{{\rm{z}} + 2}}{{ - 1}}\)

So, the direction ratio of the first line is (k, -3, -1)

\(\frac{{{\rm{x}} - 5}}{1} = \frac{{\rm{y}}}{{\rm{k}}} = \frac{{{\rm{z}} + 6}}{4}\)

So, direction ratio of second line is (1, k, 4)

Lines are perpendicular,

∴ (k × 1) + (-3 × k) + (-1 × 4) = 0

⇒ k – 3k – 4 = 0

⇒ -2k – 4 = 0

∴ k = -2

Concept:

The angle between the lines:

The angle between the lines \(\rm \frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\;and\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}\) is given by:

\(\rm \cos θ = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right) ⋅ \left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\), where a₁, b₁, c₁, a₂, b₂ and c₂ are the direction ratios

Calculation:

Given:  \(\frac {x + 1}{2} = \frac {y - 2}{5} = \frac {z + 3}{4}\) and \(\frac {x - 1}{1} = \frac {y + 2}{2} = \frac {z - 3}{-3}\)

Direction ratios of lines are a₁ = 2, b₁ = 5, c₁ = 4 and a₂ = 1, b₂ = 2 , c₂ = -3

As we know, The angle between the lines is given by \(\rm \cos θ = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right) ⋅ \left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\)

\(\Rightarrow \rm \cos θ = \frac{{{2} \times {1} + {5}\times{2} + {4}\times{-3}}}{{\left( {\sqrt {2^2 + 5^2 + 4^2} } \right) ⋅ \left( {\sqrt {1^2 + 2^2 + (-3)^2} } \right)}} = 0\)

∴ θ = 90° 

Concept:

Perpendicular Distance of a Point from a Plane 

Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d

And a point whose coordinate is, (x1, y1, z1)

Now, distance = \(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

Calculation:

We have to find the distance of the point (1, 2, 3) form the plane x – 3y + 2z + 13 = 0

Distance \(= \left| {\frac{{1\; \times \;1\; + \left( { - 3} \right)\; \times \;2\; + \;2\; \times \;3\; + \;13}}{{\sqrt {{1^2}\; + \;\;{{\left( { - 3} \right)}^2}\; + \;{{\left( 2 \right)}^2}} }}} \right|\)

\(= \left| {\frac{{1\; - 6\; + \;6\; + \;13}}{{\sqrt {1\; + \;\;9\; + \;4} }}} \right|\)

\(= \;\left| {\frac{14}{{\sqrt {14} }}} \right|\) = √14

Concept:

• The intercept form of the plane is given by ⇔ \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\)

Where ‘a’ is the x- intercept, ‘b’ is the y- intercept and ‘c’ is the z- intercept.

Calculation:

Given:

Equation of plane 5x + 2y + z – 13 = 0

⇒ 5x + 2y + z = 13

\(\Rightarrow \frac{{5{\rm{x\;}} + {\rm{\;}}2{\rm{y\;}} + {\rm{\;z\;}}}}{{13}} = 1\)

\(\Rightarrow \frac{x}{{\frac{{13}}{5}}} + \frac{y}{{\frac{{13}}{2}}} + \frac{z}{{\frac{{13}}{1}}} = 1\)