Three Dimensional Geometry MCQ Quiz in मल्याळम - Objective Question with Answer for Three Dimensional Geometry - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

A = (1, 0, 3) B = (4, 7, 1) & C = (3, 5, 3)

Concept:

Line in 3D: 

Line passing through two points P(x1, y1, z1) and Q(x2, y2, z2):

\(\frac{{{\rm{x}} - {{\rm{x}}_1}}}{{{{\rm{x}}_2} - {{\rm{x}}_1}}} = \frac{{{\rm{y}} - {{\rm{y}}_1}}}{{{{\rm{y}}_2} - {{\rm{y}}_1}}} = \frac{{{\rm{z}} - {{\rm{z}}_1}}}{{{{\rm{z}}_2} - {{\rm{z}}_1}}}\)

Note:

Let direction ratios of a line be (l1, m1, n1) and for another line be (l2,

m2, n2) then:

If lines are perpendicular, l1l2 + m1m2 + n1n2 = 

Calculation:

Let D be foot of perpendicular.

The direction ratio of line BC is

3 - 4, 5 - 7, 3 - 1 = -1, -2, 2

Equation of BC,

\(\frac{{{\rm{x}} - 4}}{{3 - 4}} = \frac{{{\rm{y}} - { 7} }}{{5-7}} = \frac{{{\rm{z}} - 1}}{{3-1}}\)

\(⇒ \frac{{{\rm{x}} - 4}}{{-1}} = \frac{{{\rm{y}} - { 7} }}{{-2}} = \frac{{{\rm{z}} - 1}}{{2}} =\rm k\) (Let)

⇒ x = -k + 4, y = -2k + 7 and z = 2k + 1

General point on BC be D = (-k + 4, -2k + 7, 2k + 1) 

Direction ratio of line AD will be

-k + 4 - 1, -2k + 7 - 0, 2k + 1 - 3

⇒ -k + 3, -2k + 7, 2k - 2 

Now, since the line BC and AD are perpendicular,

(-k + 3) × -1 + (-2k + 7) × -2 + (2k - 2) × 2 = 0

⇒ k - 3 + 4k - 14 + 4k - 4 = 0

⇒ 9k = 21

⇒ k = \(\frac 73\)

⇒ D = (-7/3 + 4, -2 × (7/3) + 7, 2 × (7/3) + 1)

∴  D = \(\left( \frac 5 3 , \frac 7 3 , \frac {17}{3}\right)\)

Additional Information

Equation of line passing through a point P(x1, y1, z1) with direction cosine l, m, n:

\(\frac{{{\rm{x}} - {{\rm{x}}_1}}}{{\rm{l}}} = \frac{{{\rm{y}} - {{\rm{y}}_1}}}{{\rm{m}}} = \frac{{{\rm{z}} - {{\rm{z}}_1}}}{{\rm{n}}}\)

If lines are parallel, \(\frac{{{{\rm{l}}_1}}}{{{{\rm{l}}_2}}} = \frac{{{{\rm{m}}_1}}}{{{{\rm{m}}_2}}} = \frac{{{{\rm{n}}_1}}}{{{{\rm{n}}_2}}}\)

Concept:

The distance between two parallel planes ax + by + cz +d₁ = 0 and ax + by + cz +d2 = 0 is given by: \(D= \left | \frac{d_{1}-d_{2}}{\sqrt{a^2+b^2+c^2}} \right |\)

Calculation:

Given: 2x + y - 2z + 6 = 0 and 4x + 2y - 4z - 6 = 0 are two planes.

Here, we can rewrite the equation of plane 2x + y - 2z + 6 = 0 as 4x + 2y - 4z + 12 = 0 by multiplying both the sides of 2x + y - 2z + 6 = 0 with 2.

As we can see that, the plane 4x + 2y - 4z + 12 = 0 and 4x + 2y - 4z - 6 = 0 are parallel planes.

As we know that, the distance between two parallel planes ax + by + cz +d1 = 0 and ax + by + cz +d2 = 0 is given by: \(D= \left | \frac{d_{1}-d_{2}}{\sqrt{a^2+b^2+c^2}} \right |\)

Here, a = 4, b = 2, c = - 4, d₁ = 12 and d₂ = - 6.

​⇒ \(D= \left | \frac{d_{1}-d_{2}}{\sqrt{a^2+b^2+c^2}} \right |\)​

⇒ \(D= \left | \frac{12-(-6)}{\sqrt{4^2+2^2+(-4)^2}} \right |\)

⇒ ​\(D= \left | \frac{18}{\sqrt{16+4+16}} \right |\)

⇒ \(D= \left | \frac{18}{6} \right |=3\)

Hence, option 4 is correct.

Concept:

If a, b and c are  direction ratios of a line, then direction cosines are given by:

⇒(l, m, n) =  \(\frac{{\rm{a}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }},\frac{{\rm{b}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }},\frac{{\rm{c}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }}\)

Calculation:

Given, direction ratios are (1, 2, 3)

Here, a = 1, b = 2 and c = 3, then direction cosines of a line

⇒(l, m, n) \( = \left( {\frac{{1}}{{\sqrt {{{\left( {1} \right)}^2} + {{\left( {2} \right)}^2} + {{\left( { 3} \right)}^2}} }},\frac{{2}}{{\sqrt {{{\left( {1} \right)}^2} + {{\left( {2} \right)}^2} + {{\left( { 3} \right)}^2}} }},\frac{{ 3}}{{\sqrt {{{\left( { 1} \right)}^2} + {{\left( {2} \right)}^2} + {{\left( { 3} \right)}^2}} }}} \right)\)

⇒(l, m, n) \( = \left( {\frac{{ 1}}{{\sqrt {14} }},\frac{{2}}{{\sqrt {14} }},\frac{{ 3}}{{\sqrt {14} }}} \right)\)

Concept:

Calculation:

Given: 

Equation of plane is 2x - y + 2z + 1 = 0

Compare with standard equation of plane ax + by + cz + d = 0 

Therefore, a = 2, b = -1 and c = 2

〈 a, b, c 〉 = 〈 2, -1, 2 〉  = 2 \(\left\langle 1, - \dfrac{1}{2}, 1 \right\rangle\)

∴ The direction ratios of normal to the plane 2x - y + 2z + 1 = 0 is \(\left\langle 1, - \dfrac{1}{2}, 1 \right\rangle\)

CONCEPT:

Distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is given by: \(\left| {\frac{{{d_1} - {d_2}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)

CALCULATION:

Here, we have to find the between the planes 2x - 3y + 6z - 5 = 0 and 6x - 9y + 18z + 20 = 0

The plane 6x - 9y + 18z + 20 = 0 can be re-written as 2x - 3y + 6z + 20/3 = 0

So, the planes 2x - 3y + 6z - 5 = 0 and 2x - 3y + 6z + 20/3 = 0 are parallel

As we know that, distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is given by: \(\left| {\frac{{{d_1} - {d_2}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)

Here, d1 = - 5, d2 = 20/3, a = 2, b = - 3 and c = 6

So, the distance between the given parallel planes is \(\left| {\frac{{-5 - \frac{20}{3}}}{{\sqrt {{2^2} + {(-3)^2} + {6^2}} }}} \right| = \frac{5}{3} \ units\)

Hence, option B is the correct answer.

Confusion PointsBefore applying the formula of distance b/w two planes, we have to make sure that the coefficients of x, y, & z must be the same in both planes.  

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). Then the distance between the point and the plane is given by \(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

Calculation:

We have to find the distance of the point (2, 3, -5) from the plane x + 2y - 2z = 9 

As we know that,  the distance between the point and the plane is given by \(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)

Here, A = 1, B = 2, C = -2, d = 9, x1 = 2, y1 = 3 and z1 = - 5

So, the distance of the given point form the given plane \(= \left| {\frac{{2\; \times \;1\; + \ {2}\ \times \;3\; + \;2 \times 5\; - \;9}}{{\sqrt {{1^2}\; + \;\;{{\left( {-2} \right)}^2}\; + \;{{\left( 2\right)}^2}} }}} \right|\)

\(= \;\left| {\frac{9}{{\sqrt {9} }}} \right|=3\)

So, the distance between the plane and the point is 3 units

Hence, option D is the correct answer.

Concept:

Distance of point (p,q,r) from the plane ax + by + cz + d = 0 is given by,

\(|ap +bq+ cr + d| \over \sqrt{a^2 + b^2 + c^2}\)

Calculation:

Distance of (1,1,1) from origin (0,0,0) = \(\sqrt{(1-0)^2 + (1-0)^2+ (1-0)^2} = \sqrt{3}\)

And the distance of the point (1, 1, 1) from the plane x + y + z + k = 0

is \(|1(1)+ 1(1)+1(1)+k| \over \sqrt{1^2 + 1^2 + 1^2}\) = \(|3+k| \over \sqrt{3}\)

Given that the distance of the point (1, 1, 1) from the origin is half its distance from the plane

⇒ \(\sqrt{3} = {1\over 2}{|3+k| \over \sqrt{3}}\)

⇒ |3 + k| = 6

⇒ k = 6 - 3, - 6 - 3 = 3, -9

∴ The correct answer is option (4).

Concept:

Equation of line joining two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is given as

\(\frac{x-x_1}{x_2-x_1} =\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\)

A point lies on the line only when its coordinates satisfy the equation of the line.

Calculation:

Given:

P = (5,y,z)

The equation of line joining (3,2,-1) and (6,-4,-2) is 

\(\frac{x-3}{6-3} =\frac{y-2}{-4-2}=\frac{z+1}{-2+1}= \frac {x-3}{3}=\frac {y-2}{-6}=\frac {z+1}{-1}\)

so if point P lies on the line then it must satisfy the above equation

\( \frac {5-3}{3}=\frac {y-2}{-6}=\frac {z+1}{-1}\)

\( \frac {2}{3}=\frac {y-2}{-6}\)

⇒ - 4 = y - 2

⇒ y =  - 2

Hence y coordinate of P is -2

Concept:

Let the two lines have direction ratio’s a1, b1, c1 and a2, b2, c2 respectively.

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

Calculation:

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

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

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

So, the direction ratio of the first line is (2, 2, k)

\(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\)

So, direction ratio of second line is (-k, 2, 5)

Lines are perpendicular,

∴ (2 × -k) + (2 × 2) + (k × 5) = 0

⇒ -2k + 4 + 5k = 0

⇒ 3k + 4 = 0

∴ k = -4/3

Concept:

If the equation of the line through (1, 2, -1) and (-1, 0, 1) then these points will satisfy the given line,

Calculation:

The general equation of the line,

x = lt + 1, y = mt + 2, z = nt - 1

It is obvious that points (1, 2, -1) is satisfying the line,

For point (-1, 0, 1),

-1 = lt + 1, 0 = mt + 2, 1 = nt - 1

Equating the values of t,

lt = -2, mt = -2, nt = 2

\(\Rightarrow t=\frac{{ - 2}}{l} = \frac{{ - 2}}{m} = \frac{2}{n}\); 

∴ (l, m, n) are (1, 1, -1)