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):

$\frac{\mathrm{x}-{\mathrm{x}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}=\frac{\mathrm{y}-{\mathrm{y}}_1}{{\mathrm{y}}_2-{\mathrm{y}}_1}=\frac{\mathrm{z}-{\mathrm{z}}_1}{{\mathrm{z}}_2-{\mathrm{z}}_1}$

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,

$\frac{\mathrm{x}-4}{3-4}=\frac{\mathrm{y}-7}{5-7}=\frac{\mathrm{z}-1}{3-1}$

$\Rightarrow \frac{\mathrm{x}-4}{-1}=\frac{\mathrm{y}-7}{-2}=\frac{\mathrm{z}-1}{2}=\mathrm{k}$ (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 = $\frac{7}{3}$

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

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

Additional Information

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

$\frac{\mathrm{x}-{\mathrm{x}}_1}{\mathrm{l}}=\frac{\mathrm{y}-{\mathrm{y}}_1}{\mathrm{m}}=\frac{\mathrm{z}-{\mathrm{z}}_1}{\mathrm{n}}$

If lines are parallel, $\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}=\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}$

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) =  $\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}},\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}},\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}$

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 $⟨1,-{\displaystyle \frac{1}{2}},1⟩$

∴ The direction ratios of normal to the plane 2x - y + 2z + 1 = 0 is $⟨1,-{\displaystyle \frac{1}{2}},1⟩$

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,

$\frac{|ap+bq+cr+d|}{\sqrt{a^2+b^2+c^2}}$

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 $\frac{|1(1)+1(1)+1(1)+k|}{\sqrt{1^2+1^2+1^2}}$ = $\frac{|3+k|}{\sqrt{3}}$

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

⇒ $\sqrt{3}=\frac{1}{2}\frac{|3+k|}{\sqrt{3}}$

⇒ |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:

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)

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  $\frac{\mathrm{x}+4}{2}=\frac{4-\mathrm{y}}{-2}=\frac{2\mathrm{z}-4}{2\mathrm{k}}$ and $\frac{\mathrm{x}+3}{-\mathrm{k}}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}+1}{5}$ 

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

$\Rightarrow \frac{\mathrm{x}+4}{2}=\frac{-(\mathrm{y}-4)}{-2}=\frac{2(\mathrm{z}-2)}{2\mathrm{k}}\iff \frac{\left(\mathrm{x}+4\right)}{2}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-2}{\mathrm{k}}$

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

$\frac{\mathrm{x}+3}{-\mathrm{k}}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}+1}{5}$

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