Three-dimensional Geometry Test - 1

Concept:

Standard equations for some special plane:

• A plane parallel to the x-y-plane must have a standard equation z = d, where d = distance of a plane from xy plane
• A plane parallel to the y-z-plane has equation x = d, where d = distance of a plane from yz plane
• A plane parallel to the x-z-plane has equation y = d, where d = distance of a plane from xz plane

Calculation:

Equation of xz plane is y = 0.

We know, a plane parallel to the x-z-plane has equation y = d

Here, d = 1

∴The equation of plane parallel to xz plane at unit distance will be, y = 1

Hence, option (2) is correct.

Concept:

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

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

Calculation:

Given: d.r's are 1. − 2. and 3

So, a = 1, b = -2 and c = 3

\(\rm \sqrt{a^2+b^2+c^2}= \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{14}\)

Now, d.c's of line = (l, m, n) =  \(\frac{{\rm\pm{a}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }},\frac{{\rm\pm{b}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }},\frac{{\rm\pm{c}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} }}\)

= \(\pm {1 \over \sqrt14}. \mp{2 \over \sqrt14}. \pm {3 \over \sqrt14}\)

Concept:

The equation of a line with direction ratio (a, b, c) that passes through the point (x₁, y₁, z₁) is given by the formula: \(\rm \frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\)

The direction ratios of z-axis are <0, 0, 1>

Calculation:

Given:

Line passes through (a, b, c) and parallel to the z-axis.

As we know, equation of line is \(\rm \frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\)

Therefore, the equation to the straight line passing through (a, b, c) and parallel to z-axis is  \(\rm \dfrac{x-a}{0}=\dfrac{y-b}{0}=\dfrac{z-c}{1}\)

Concept:

The equation of a line with a direction ratio (1, 1, 2) that passes through the point (4, 2, k) is given by the formula

\(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\)

Equation of plane in 3-D: 2x - 4y + z  = 7,

Since the given line is lying on the given plane so the point through which the line is passing ie (4, 2, k) will also lie on the given plane.

Calculation:

Given:

Here, equation of plane 2x - 4y + z  = 7,

Equation of the straight line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\)

Put the point  (4, 2, k) on the given plane as it will also lie on this plane.

⇒ 2×4 - 4×2 + k  = 7

⇒ k = 7

So, the correct answer will be option 2.

Concept:

Section Formula: Section formula is used to determine the coordinate of a point that divides a line into two parts such that ratio of their length is m : n

⇒ Let A and B be the given two points (x₁, y₁) and (x₂, y₂) respectively and C(x, y) be the point dividing the line- segment AB internally in the ratio m: n

I. Internal Section Formula: When the line segment is divided internally in the ratio m : n, we use this formula. ⇔ \(\left( {{\rm{x}},{\rm{\;y}}} \right) = \left( {\frac{{{\rm{m}}{{\rm{x}}_2}{\rm{\;}} + {\rm{\;n}}{{\rm{x}}_1}}}{{{\rm{m}} + {\rm{\;n}}}},\frac{{{\rm{m}}{{\rm{y}}_2}{\rm{\;}} + {\rm{\;n}}{{\rm{y}}_1}}}{{{\rm{m}} + {\rm{n}}}}} \right)\)

II. External Section Formula: When point C lies on the external part of the line segment. ⇔ \(\left( {{\rm{x}},{\rm{\;y}}} \right) = \left( {\frac{{{\rm{m}}{{\rm{x}}_2} - {\rm{n}}{{\rm{x}}_1}}}{{{\rm{m}} - {\rm{\;n}}}},\frac{{{\rm{m}}{{\rm{y}}_2} - {\rm{n}}{{\rm{y}}_1}}}{{{\rm{m}} - {\rm{\;n}}}}} \right)\)
Calculation:

Let XY plane divides the line joining the points A(2, 3, -5) and B(-1, -2, -3) in the ratio k : 1

Using the section formula, the coordinate of the point of intersection is given by

\(\rm \left(\frac{-k+2}{k+1},\frac{-2k+3}{k+1},\frac{-3k-5}{k+1} \right )\)

As we know, on the XY plane z-coordinate is zero.

Therefore, \(\rm \frac{-3k-5}{k+1} = 0\)

⇒ -3k - 5 = 0

⇒ -3k = 5

\(\rm \Rightarrow \frac{k}{1} = \frac{-5}{3}\)

Therefore, the ratio is 5 : 3 externally

CONCEPT:

The distance between two parallel lines such as  \(\vec r = \;\overrightarrow {{a_1}} + \lambda \times \vec b\;and\;\vec r = \;\overrightarrow {{a_2}} + μ \times \;\vec b\) is given by \(d = \left| {\frac{{\vec b \times \left( {\overrightarrow {{a_2}} - \;\overrightarrow {{a_1}} } \right)}}{{\left| {\vec b} \right|}}} \right|\)

CALCULATION:

Given: Equation of lines L1 and L2 are \(\vec r = \;\hat i + \hat j + \hat k + \;\lambda \times \left( {3\hat i - \hat j} \right)\;and\;\vec r = 4\hat i-\hat k + \mu \times \left( {3\hat i - \hat j} \right)\)

As we can see that, the given equation of lines are parallel.

So, by comparing the given equation of lines with  \(\vec r = \;\overrightarrow {{a_1}} + \lambda \times \vec b\;and\;\vec r = \;\overrightarrow {{a_2}} + μ \times \;\vec b\)

Here, we have \(\vec b = \;3\hat i - \hat j,\;\overrightarrow {{a_1}} = \;\hat i + \hat j\ + \hat k \ and\;\overrightarrow {{a_2}} = \;4\hat i - \;\hat k\)

As we know that, distance between two parallel lines such as  \(\vec r = \;\overrightarrow {{a_1}} + \lambda \times \vec b\;and\;\vec r = \;\overrightarrow {{a_2}} + μ \times \;\vec b\) is given by \(d = \left| {\frac{{\vec b \times \left( {\overrightarrow {{a_2}} - \;\overrightarrow {{a_1}} } \right)}}{{\left| {\vec b} \right|}}} \right|\) 

⇒ \(\overrightarrow {{a_2}} - \overrightarrow {{a_1}} = 3\hat i - \hat j - 2\hat k\) and 

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

\(|\vec b| = \sqrt{10}\)

\(\vec b \times \left( {\overrightarrow {{a_2}} - \;\overrightarrow {{a_1}} } \right) = \left[ {\begin{array}{*{20}{c}} i&j&k\\ 3&-1&0\\ 3&-1&-2 \end{array}} \right]\)

⇒ \(\vec b \times \left( {\overrightarrow {{a_2}} - \;\overrightarrow {{a_1}} } \right) = 2\hat i + 6\hat j\)

\(\vec b \times \left( {\overrightarrow {{a_2}} - \;\overrightarrow {{a_1}} } \right) =\frac{2i+6j}{\sqrt{10}}\)

\(\vec b \times \left( {\overrightarrow {{a_2}} - \;\overrightarrow {{a_1}} } \right) =\frac{\sqrt{2^2+6^2}}{\sqrt{10}}\)

\(\vec b \times \left( {\overrightarrow {{a_2}} - \;\overrightarrow {{a_1}} } \right) =\frac{{2\sqrt{10}}}{\sqrt{10}}\)

⇒ \(d = 2\ units \)

Hence, option D is the correct answer.

Concept:

If θ is the angle between the line \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) and the plane a2 x + b2 y + c2 z + d = 0 then \(\sin \theta = \frac{{{|a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}|}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right)\left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\)

Calculation:

Given: Equation of line is \(\frac{{x + 1}}{2} = \frac{{y}}{{\;3}} = \frac{{z - 3}}{6}\) and equation of plane is 10x + 2y - 11z - 3 = 0

As we know that the angle between the line \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) and the plane a2 x + b2 y + c2 z + d = 0 is given by: \(\sin \theta = \frac{{{|a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}|}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right)\left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\)

Here, a1 = 2, b1 = 3, c1 = 6, a2 = 10, b2 = 2 and c2 = - 11.

⇒ a1 ⋅ a2 + b1 ⋅ b2 + c1 ⋅ c2 = 20 + 6 - 66 = - 40

\(⇒ \sqrt {a_1^2 + b_1^2 + c_1^2} = 7 \;and\;\sqrt {a_2^2 + b_2^2 + c_2^2} = 15\)

\(\Rightarrow \sin \theta = \frac{40}{7 \times 15} = \frac{8}{21}\)

\(\Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{8}{21}} \right)\)

Hence, option A is the correct answer.

Concept:

• Equation of XY plane  is z = o
• Equation of YZ plane  is x = o
• Equation of XZ plane  is y = o

Explanation:

Given that,

z = 7

We know that

z = 0 represents the XY-plane.

∴ z  = 7 represents a plane parallel to XY-plane.

Concept:

 The equation of a line with direction ratio (a, b, c) that passes through the point (x1, y1, z1) is given by the formula: \(\rm \frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\)

If two planes intersect each other, the intersection will always be a line.

Calculation:

Given:

x = 3z + 4 and y = 2z - 3

⇒ x - 4 = 3z and y + 3 = 2z

\(\rm \Rightarrow \frac{x - 4}{3} = z \;\text{and} \ \frac{y + 3}{2} = z\)

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

If two planes intersect each other, the intersection will always be a line.

Now,  direction ratios of the line are 〈3, 2, 1〉

CONCEPT:

If two lines \(\vec r = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \;and\;\vec r = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \) are coplanar then \(\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \;\overrightarrow {{b_2}} } \right) = 0\)

EXPLANATION:

As we know that, if two lines \(\vec r = \overrightarrow {{a_1}} + \lambda \overrightarrow {{b_1}} \;and\;\vec r = \overrightarrow {{a_2}} + \mu \overrightarrow {{b_2}} \) are coplanar then \(\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \;\overrightarrow {{b_2}} } \right) = 0\)

Hence, option A is the correct answer