Solution: It is given that a cube with 2 units edge, one of whose vertices coincides with the origin and the 3 edges passing through the origin, coincides with the positive direction of the axes...
and 
is 
, then the value of 
is 
Solution: Option(B) $\pm 5$ Explanation: Suppose $P$ be the point whose coordinate is $(a, 0,1)$ and $Q$ represents the point $(0,$, $(1,2) .$ It is given that, $\mathrm{PQ}=\sqrt{27}$ From the...
from the origin 
is 
Solution: Option (A) $\sqrt{50}$ Explanation: Suppose $\mathrm{P}$ be the point whose coordinate is $(3,4,5)$ and $\mathrm{Q}$ represents the origin. From the distance formula it can be written as...
on 
-axis 
Solution: Option(B) $\sqrt{34}$ Explanation: As it is known that $y$-axis lies on $x$ y plane and $y z$. Therefore, its distance from $x y$ and $y z$ plane is 0 . $\therefore$ By the basic...
Solution: (A) 3 units Explanation: From basic ideas of three-dimensional geometry, it is known that $x$-coordinate of a point is its distance from $y z$ plane. $\therefore$ The distance of Point $P...
Solution: It is given that the three points $A(0,-1,-7), B(2,1,-9)$ and $C(6,5,-13)$ are collinear So it can be written as $\begin{array}{l} A...
Solution: It is given that the mid-point of the sides of a triangle are $(1,5,-1),(0,4,-2)$ and $(2,3,$, 4). Suppose the vertices be $A\left(x_{1}, y_{1}, z_{1}\right), B\left(x_{2}, y_{2},...
and 
are collinear and find the ratio in which 
divides 
.Solution: It is given that the three points are A $(2,3,4), \mathrm{B}(-1,2,-3)$ and $\mathrm{C}(-4,1,-10)$ We need to find collinear points, $\begin{array}{l}...
and C 
be the vertices of a triangle. The internal bisector of the angle 
meets 
at the point 
. Find the coordinates of .Solution: It is given $A(2,2,-3), B(5,6,9)$ and $C(2,7,9)$ are the vertices of a triangle. And it is also given that the internal bisector of the angle A meets BC at the point D....
Solution: It is given that the triangle ABC having vertices $A(a, 1,3), B(-2, b,-5)$ and $C(4,7, c)$ and origin is the centroid. The coordinates of the centroid for a triangle is given by the...
and 
Solution: It is given the line segment joining the points are A $(2,1,-3)$ and $B(5,-8,3)$ Now suppose $P\left(x_{1}, y_{1}, z_{1}\right)$ and $Q\left(x_{2}, y_{2}, z_{2}\right)$ be the points which...
Solution: It is given that the three consecutive vertices of a parallelogram ABCD are A $(1,2,3), B(-1,$, $-2,-1)$ and $C(2,3,2)$ Suppose the fourth vertex be $D(x, y, z)$. By using midpoint...
Solution: It is given that the mid-points of the sides of a triangle are $(5,7,11),(0,8,5)$ and $(2,3,-$ 1). Suppose the vertices be $A\left(x_{1}, y_{1}, z_{1}\right), B\left(x_{2}, y_{2},...
and 
4).Solution: It is given that: Mid-points of sides of triangle $\mathrm{DEF}$ are: $\mathrm{D}(1,2,-3), \mathrm{E}(3,0,1)$ and $\mathrm{F}(-1,1,-4)$ Using the geometry of centroid, It is known that the...
Solution: It is given the centroid is origin and two vertices are $(2,4,6)$ and $(0,-2,-5)$ Suppose the third vertex be $(x, y, z)$ The coordinates of the centroid for a triangle is given by the...
with vertices 
and 
is right angled.Solution: The given vertices are $A(0,4,1), B(2,3,-1)$ and $C(4,5,0)$ We need to prove right angled triangle, consider $\begin{array}{l}...
Solution: The three consecutive vertices of a parallelogram $A B C D$ are as given $A(6,-2,4), B$ (2, $4,-8), C(-2,2,4)$ Suppose the forth vertex be $D(x, y, z)$ Midpoint of diagonal $A...
and 
are collinear.Solution: The given points are $A(1,-1,3), B(2,-4,5)$ and $(5,-13,11)$. We need to prove collinear, $\begin{array}{l} \mathrm{AB}=\sqrt{(1-2)^{2}+(-1+4)^{2}+(3-5)^{2}}=\sqrt{1+9+4}=\sqrt{14} \\...
, then the point 
is at a distance 1 unit from the origin.Solution: It is given that $x^{2}+y^{2}=1 \Rightarrow 1-x^{2}-y^{2}=0$ Distance of the point $\left(x, y \sqrt{1-x^{2}-y^{2}}\right)$ from origin is $=$ $\begin{array}{l}...
Solution: The distance from the origin to (6, 6, 7) $=\sqrt{{6^2}+{6^2}+{7^2}}$ $=\sqrt{{36}+{36}+{49}}$ $=\sqrt{121}$ $=11$ units
Solution: The points $(2, 0, 0)$ and $(–3, 0, 0)$ are at a distance of $=$ $|2 − (−3)| = 5$ units.
Solution: (i) $(4, – 3, – 5):- A (4, −3, 0), B (0, −3, −5), C (4, 0, −5)$
Solution: (i) $(3, 4, 5):- A (3, 4, 0), B (0, 4, 5), C (3, 0, 5)$ (ii) $(–5, 3, 7):- A (−5, 3, 0), B (0, 3, 7), C (−5, 0, 7)$
Solution: (i) $(4, – 3, – 5):- A (4, 0, 0), B (0, −3, 0), C (0, 0, −5)$
Solution: (i) $(3, 4, 2):- A (3, 0, 0), B (0, 4, 0), C (0, 0, 2)$ (ii) $(–5, 3, 7):- A (−5, 0, 0), B (0, 3, 0), C (0, 0, 7)$
Solution: (i) $(2, – 4, – 7)$:- 8th Octant, (ii) $(– 4, 2, – 5)$:- 6th Octant.
Solution: (i) $(– 4, 2, 5)$:- 2nd Octant, (ii) $(–3, –1, 6)$:- 3rd Octant,
Solution: (i) $(4, –2, –5)$:- 8th Octant, (ii) $(4, 2, –5)$:- 5th Octant,
Solution: (i) $(1, 2, 3)$:- 1st Octant, (ii) $(4, – 2, 3)$:- 4th Octant,
Solution: (i) $(– 2, – 4, –7)$:- 7th octant, (ii) $(– 4, 2, – 5)$:- 6th octant.
Solution: (i) $(1, – 1, 3)$:- 4th octant, (ii) $(– 1, 2, 4)$:- 2nd octant,