Chapter 12: Introduction to Three Dimensional Geometry

Ans: Given: Edge length of the cube $= 2{\text{ units}}$

One of the vertices coincides with the origin.

Three edges passing through the origin coincides with the positive direction of the axes.

Draw a diagram of the cube according to the conditions given. Start from the origin and use the sign convention of the coordinates in a specific quadrant to find the vertices.

The diagram of the cube according to the given conditions can be shown below: -

Here one of the vertices is the origin $O$, so its coordinate is $\left( {0,{\text{ }}0,{\text{ }}0} \right)$.

Since point $A$ lies on the $x$ axis so its coordinate is $\left( {2,{\text{ }}0,{\text{ }}0} \right)$.

Point $C$ lies on the $y$ axis so its coordinate is $\left( {0,{\text{ }}2,{\text{ }}0} \right)$.

Point $G$ lies on the $z$ axis so its coordinate is $\left( {0,{\text{ }}0,{\text{ }}2} \right)$.

Since point $B$ lies in the $xy$ plane so its coordinate is $\left( {2,{\text{ }}2,{\text{ }}0} \right)$.

Point $D$ lies in the $yz$ plane so its coordinate is $\left( {0,{\text{ }}2,{\text{ }}2} \right)$.

Point $F$ lies in the $xz$ plane so its coordinate is $\left( {2,{\text{ }}0,{\text{ }}2} \right)$.

Now, the point $E$ lies in the $3 - D$ space so its coordinate is $\left( {2,{\text{ }}2,{\text{ }}2} \right)$.

Ans: Option (c)

Given:The point $P\left( {3,{\text{ }}4,{\text{ }}5} \right)$.

Use the fact that the distance of a point $P\left( {{x_1},{\text{ }}{y_1},{\text{ }}{z_1}} \right)$ from the $yz$ plane is its $x$ coordinate.

The distance of a point $P\left( {{x_1},{\text{ }}{y_1},{\text{ }}{z_1}} \right)$ from the $yz$ plane is its $x$ coordinate ${x_1}$.

Therefore, the distance of the point $P\left( {3,{\text{ }}4,{\text{ }}5} \right)$ from the $yz$ plane $= 3{\text{ units}}$.

Ans: Option (b)

Given: The point $P\left( {3,{\text{ }}4,{\text{ }}5} \right)$.

The length of foot of perpendicular from a point $P\left( {{x_1},{\text{ }}{y_1},{\text{ }}{z_1}} \right)$ on the $y$ axis is $\sqrt {{x_1}^2 + {z_1}^2}$.

Therefore, the length of foot of perpendicular from the point $P\left( {3,{\text{ }}4,{\text{ }}5} \right)$ on the $y$ axis

$= \sqrt {{3^2} + {5^2}}$

$= \sqrt {34}$ 

Ans: Option (a)

Given:The origin and the point $\left( {3,{\text{ }}4,{\text{ }}5} \right)$.

Use the formula of the distance between the two points $\left( {{x_1},{\text{ }}{y_1},{\text{ }}{z_1}} \right)$ and $\left( {{x_2},{\text{ }}{y_2},{\text{ }}{z_2}} \right)$ given as $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$.

The origin is given by the coordinates $\left( {0,{\text{ }}0,{\text{ }}0} \right)$. Let us assume the distance between the points $\left( {0,{\text{ }}0,{\text{ }}0} \right)$ and $\left( {3,{\text{ }}4,{\text{ }}5} \right)$ is $d$.

The distance between the two points having coordinates $\left( {{x_1},{\text{ }}{y_1},{\text{ }}{z_1}} \right)$ and $\left( {{x_2},{\text{ }}{y_2},{\text{ }}{z_2}} \right)$ is given as $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$.

$\Rightarrow d = \sqrt {{{\left( {3 - 0} \right)}^2} + {{\left( {4 - 0} \right)}^2} + {{\left( {5 - 0} \right)}^2}}$

$\Rightarrow d = \sqrt {9 + 16 + 25}$

$\Rightarrow d = \sqrt {50}$ 

Ans: Option (b)

Given: The points $\left( {a,{\text{ }}0,{\text{ }}1} \right)$ and $\left( {0,{\text{ }}1,{\text{ }}2} \right)$.

Distance between the two points $= \sqrt {27}$

The distance between the two points having coordinates $\left( {{x_1},{\text{ }}{y_1},{\text{ }}{z_1}} \right)$ and $\left( {{x_2},{\text{ }}{y_2},{\text{ }}{z_2}} \right)$ is given as $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$. Therefore, for the points $\left( {a,{\text{ }}0,{\text{ }}1} \right)$ and $\left( {0,{\text{ }}1,{\text{ }}2} \right)$,

$\Rightarrow d = \sqrt {27}$

$\Rightarrow \sqrt {{{\left( {0 - a} \right)}^2} + {{\left( {1 - 0} \right)}^2} + {{\left( {2 - 1} \right)}^2}}  = \sqrt {27}$

$\Rightarrow {a^2} + 2 = 27$

$\Rightarrow{a^2} = 25$ 

Taking square root both the sides,

$\Rightarrow a =  \pm 5$

Ans: Option (a)

Given: $x -$ axis is the intersection of two planes.

Three dimensional geometry involves the mathematics of shapes in $3D$ space and the three coordinates in $XYZ$ plane.

On the $xy$ and $xz$ planes , the line of intersection is the $x -$ axis.

Hence, the two planes are $xy$ and $xz$.

Ans: Option (c)

Given: $y -$ axis.

Three dimensional geometry involves the mathematics of shapes in $3D$ space and the three coordinates in $XYZ$ plane.

On the $y -$ axis: $z = 0$, $x = 0$

Hence $z = 0$, $x = 0$.

Ans: Option (b)

Given: $( - 2, - 3, - 4)$

Three dimensional geometry involves the mathematics of shapes in $3D$ space and the three coordinates in $XYZ$ plane.

$( - 2, - 3, - 4)$ lies in the seventh octant.

Hence seventh octant is the correct answer.

Ans: Option (a)

Given: A plane is parallel to $yz -$ plane.

Three dimensional geometry involves the mathematics of shapes in $3D$ space and the three coordinates in $XYZ$ plane.

As the plane is parallel to $yz -$ plane, it is perpendicular to $x -$ axis.

Hence, the plane is perpendicular to $x -$ axis.

Ans: Option (a)

Given: $y = 0$, $z = 0$.

Three dimensional geometry involves the mathematics of shapes in $3D$ space and the three coordinates in $XYZ$ plane.

$y = 0$, $z = 0$ is the equation of $x -$ axis.

Hence, the locus is the equation of $x -$ axis.